Patikrinimas ar Diferencialinių binomų integravimas yra teisingas.
Apie patį integravimą:
"Diferencialinių binomų integravimas
Integralas
∫
x
m
(
a
+
b
x
n
)
p
d
x
,
{\displaystyle \int x^{m}(a+bx^{n})^{p}dx,}
m, n, p - racionalieji skaičiai, vadinamas integralu su binominiu diferencialu. Šį integralą elementariosiomis funkcijomis įmanoma išreikšti tik trimis atvejais:
I. p - sveikasis skaičius. Jei
p
>
0
,
{\displaystyle p>0,}
Niutono binomo formulę . Jei
p
<
0
,
{\displaystyle p<0,}
x
=
t
k
,
{\displaystyle x=t^{k},}
k - bendras trupmenų m ir n vardiklis. Pavyzdžiui, trupmenų
1
4
{\displaystyle {\tfrac {1}{4}}}
2
3
{\displaystyle {\tfrac {2}{3}}}
⋅
{\displaystyle \cdot }
II.
m
+
1
n
{\displaystyle {\frac {m+1}{n}}}
a
+
b
x
n
=
t
α
,
{\displaystyle a+bx^{n}=t^{\alpha },}
α
{\displaystyle \alpha }
p vardiklis.
III.
m
+
1
n
+
p
{\displaystyle {\frac {m+1}{n}}+p}
a
+
b
x
n
=
t
α
x
n
,
{\displaystyle a+bx^{n}=t^{\alpha }x^{n},}
α
{\displaystyle \alpha }
p vardiklis." Pasirenkame pavyzdį:
∫
x
1
3
(
1
−
x
2
3
)
−
1
2
d
x
=
−
3
∫
x
1
3
⋅
(
t
2
)
−
1
2
⋅
x
1
3
⋅
t
d
t
=
−
3
∫
x
2
3
⋅
t
−
1
⋅
t
d
t
=
{\displaystyle \int x^{\frac {1}{3}}(1-x^{\frac {2}{3}})^{-{\frac {1}{2}}}dx=-3\int x^{\frac {1}{3}}\cdot (t^{2})^{-{\frac {1}{2}}}\cdot x^{\frac {1}{3}}\cdot t\;dt=-3\int x^{\frac {2}{3}}\cdot t^{-1}\cdot t\;dt=}
=
−
3
∫
(
1
−
t
2
)
t
t
d
t
=
−
3
∫
(
1
−
t
2
)
d
t
=
−
3
(
t
−
t
3
3
)
+
C
=
−
3
(
1
−
x
2
3
−
(
1
−
x
2
3
)
3
3
)
+
C
=
{\displaystyle =-3\int (1-t^{2}){\frac {t}{t}}\;dt=-3\int (1-t^{2})\;dt=-3(t-{\frac {t^{3}}{3}})+C=-3({\sqrt {1-x^{\frac {2}{3}}}}-{\frac {\sqrt {(1-x^{\frac {2}{3}})^{3}}}{3}})+C=}
=
−
3
(
1
−
x
2
3
)
1
2
+
(
1
−
x
2
3
)
3
2
+
C
=
−
3
1
−
x
2
3
+
1
−
3
x
2
3
+
3
x
4
3
−
x
2
+
C
,
{\displaystyle =-3(1-x^{\frac {2}{3}})^{\frac {1}{2}}+(1-x^{\frac {2}{3}})^{\frac {3}{2}}+C=-3{\sqrt {1-x^{\frac {2}{3}}}}+{\sqrt {1-3x^{\frac {2}{3}}+3x^{\frac {4}{3}}-x^{2}}}+C,}
m
=
1
3
;
{\displaystyle m={\frac {1}{3}};}
n
=
2
3
;
p
=
−
1
2
;
m
+
1
n
=
2
;
1
−
x
2
3
=
t
2
;
−
2
3
x
−
1
3
d
x
=
2
t
d
t
;
d
x
=
−
3
x
1
3
t
d
t
;
x
2
3
=
1
−
t
2
.
{\displaystyle n={\frac {2}{3}};\;p=-{\frac {1}{2}};\;{\frac {m+1}{n}}=2;\;1-x^{\frac {2}{3}}=t^{2};\;-{\frac {2}{3}}x^{-{\frac {1}{3}}}dx=2t\;dt;\;dx=-3x^{\frac {1}{3}}tdt;\;x^{\frac {2}{3}}=1-t^{2}.}
Toliau patikriname kokia yra išvestinė gauto atsakymo
f
′
(
x
)
=
(
−
3
(
1
−
x
2
3
)
1
2
+
(
1
−
x
2
3
)
3
2
)
′
=
−
3
⋅
1
2
(
1
−
x
2
3
)
−
1
2
⋅
(
−
2
3
⋅
x
−
1
3
)
+
3
2
(
1
−
x
2
3
)
1
2
⋅
(
0
−
2
3
⋅
x
−
1
3
)
=
{\displaystyle f'(x)=(-3(1-x^{\frac {2}{3}})^{\frac {1}{2}}+(1-x^{\frac {2}{3}})^{\frac {3}{2}})'=-3\cdot {1 \over 2}(1-x^{\frac {2}{3}})^{-{\frac {1}{2}}}\cdot (-{2 \over 3}\cdot x^{-{\frac {1}{3}}})+{\frac {3}{2}}(1-x^{\frac {2}{3}})^{\frac {1}{2}}\cdot (0-{2 \over 3}\cdot x^{-{\frac {1}{3}}})=}
=
x
−
1
3
1
−
x
2
3
−
x
−
1
3
1
−
x
2
3
.
{\displaystyle ={\frac {x^{-{\frac {1}{3}}}}{\sqrt {1-x^{\frac {2}{3}}}}}-x^{-{\frac {1}{3}}}{\sqrt {1-x^{\frac {2}{3}}}}.}
Toliau galime apskaičiuoti lanko L ilgį funkcijos
f
(
x
)
=
−
3
(
1
−
x
2
3
)
1
2
+
(
1
−
x
2
3
)
3
2
{\displaystyle f(x)=-3(1-x^{\frac {2}{3}})^{\frac {1}{2}}+(1-x^{\frac {2}{3}})^{\frac {3}{2}}}
L
=
∫
a
b
1
+
(
f
′
(
x
)
)
2
d
x
=
∫
0
0.9
1
+
(
x
−
1
3
1
−
x
2
3
−
x
−
1
3
1
−
x
2
3
)
2
d
x
=
{\displaystyle L=\int _{a}^{b}{\sqrt {1+(f'(x))^{2}}}dx=\int _{0}^{0.9}{\sqrt {1+({\frac {x^{-{\frac {1}{3}}}}{\sqrt {1-x^{\frac {2}{3}}}}}-x^{-{\frac {1}{3}}}{\sqrt {1-x^{\frac {2}{3}}}})^{2}}}dx=}
=
∫
0
0.9
1
+
(
x
−
1
3
1
−
x
2
3
)
2
−
2
⋅
x
−
1
3
1
−
x
2
3
⋅
x
−
1
3
1
−
x
2
3
+
(
x
−
1
3
1
−
x
2
3
)
2
d
x
=
{\displaystyle =\int _{0}^{0.9}{\sqrt {1+({\frac {x^{-{\frac {1}{3}}}}{\sqrt {1-x^{\frac {2}{3}}}}})^{2}-2\cdot {\frac {x^{-{\frac {1}{3}}}}{\sqrt {1-x^{\frac {2}{3}}}}}\cdot x^{-{\frac {1}{3}}}{\sqrt {1-x^{\frac {2}{3}}}}+(x^{-{\frac {1}{3}}}{\sqrt {1-x^{\frac {2}{3}}}})^{2}}}dx=}
=
∫
0
0.9
1
+
x
−
2
3
1
−
x
2
3
−
2
⋅
x
−
2
3
+
x
−
2
3
(
1
−
x
2
3
)
d
x
=
∫
0
0.9
1
+
x
−
2
3
1
−
x
2
3
−
2
⋅
x
−
2
3
+
x
−
2
3
−
1
d
x
=
{\displaystyle =\int _{0}^{0.9}{\sqrt {1+{\frac {x^{-{\frac {2}{3}}}}{1-x^{\frac {2}{3}}}}-2\cdot x^{-{\frac {2}{3}}}+x^{-{\frac {2}{3}}}(1-x^{\frac {2}{3}})}}dx=\int _{0}^{0.9}{\sqrt {1+{\frac {x^{-{\frac {2}{3}}}}{1-x^{\frac {2}{3}}}}-2\cdot x^{-{\frac {2}{3}}}+x^{-{\frac {2}{3}}}-1}}\;dx=}
=
∫
0
0.9
x
−
2
3
1
−
x
2
3
−
x
−
2
3
d
x
=
∫
0
0.9
x
−
2
3
−
x
−
2
3
(
1
−
x
2
3
)
1
−
x
2
3
d
x
=
∫
0
0.9
x
−
2
3
−
x
−
2
3
+
1
1
−
x
2
3
d
x
=
{\displaystyle =\int _{0}^{0.9}{\sqrt {{\frac {x^{-{\frac {2}{3}}}}{1-x^{\frac {2}{3}}}}-x^{-{\frac {2}{3}}}}}\;dx=\int _{0}^{0.9}{\sqrt {\frac {x^{-{\frac {2}{3}}}-x^{-{\frac {2}{3}}}(1-x^{\frac {2}{3}})}{1-x^{\frac {2}{3}}}}}\;dx=\int _{0}^{0.9}{\sqrt {\frac {x^{-{\frac {2}{3}}}-x^{-{\frac {2}{3}}}+1}{1-x^{\frac {2}{3}}}}}\;dx=}
=
∫
0
0.9
1
1
−
x
2
3
d
x
=
{\displaystyle =\int _{0}^{0.9}{\sqrt {\frac {1}{1-x^{\frac {2}{3}}}}}\;dx=}
Toliau taikome metodą iš diferencialinių binomų integravimo, turime, kad m=0, a=1, b=-1 ,
n
=
2
3
{\displaystyle n={2 \over 3}}
p
=
−
1
2
{\displaystyle p=-{1 \over 2}}
III atvejį, nes
m
+
1
n
+
p
=
0
+
1
2
3
−
1
2
=
1.5
−
0.5
=
1
{\displaystyle {\frac {m+1}{n}}+p={\frac {0+1}{2 \over 3}}-{1 \over 2}=1.5-0.5=1}
a
+
b
x
n
=
t
α
x
n
,
{\displaystyle a+bx^{n}=t^{\alpha }x^{n},}
α
{\displaystyle \alpha }
p vardiklis. Tada
1
−
x
2
3
=
t
2
x
2
3
.
{\displaystyle 1-x^{2 \over 3}=t^{2}x^{2 \over 3}.}
1
=
t
2
x
2
3
+
x
2
3
{\displaystyle 1=t^{2}x^{2 \over 3}+x^{2 \over 3}}
1
t
2
+
1
=
x
2
3
{\displaystyle {\frac {1}{t^{2}+1}}=x^{2 \over 3}}
x
=
1
(
t
2
+
1
)
3
2
{\displaystyle x={\frac {1}{(t^{2}+1)^{3 \over 2}}}}
d
x
=
−
3
2
(
t
2
+
1
)
−
5
2
⋅
2
t
d
t
{\displaystyle dx=-{\frac {3}{2}}(t^{2}+1)^{-{\frac {5}{2}}}\cdot 2t\;dt}
t
=
1
−
x
2
3
x
2
3
.
{\displaystyle t={\sqrt {\frac {1-x^{2 \over 3}}{x^{2 \over 3}}}}.}
∫
0
0.9
1
1
−
x
2
3
d
x
=
∫
0
0.9
−
3
2
(
t
2
+
1
)
−
5
2
⋅
2
t
1
−
(
(
t
2
+
1
)
−
3
2
)
2
3
d
t
=
∫
0
0.9
−
3
2
(
t
2
+
1
)
−
5
2
⋅
2
t
1
−
1
t
2
+
1
d
t
=
{\displaystyle \int _{0}^{0.9}{\sqrt {\frac {1}{1-x^{\frac {2}{3}}}}}\;dx=\int _{0}^{0.9}{\sqrt {\frac {-{\frac {3}{2}}(t^{2}+1)^{-{\frac {5}{2}}}\cdot 2t}{1-((t^{2}+1)^{-{3 \over 2}})^{\frac {2}{3}}}}}dt=\int _{0}^{0.9}{\sqrt {\frac {-{\frac {3}{2}}(t^{2}+1)^{-{\frac {5}{2}}}\cdot 2t}{1-{\frac {1}{t^{2}+1}}}}}\;dt=}
=
∫
0
0.9
−
3
2
(
t
2
+
1
)
−
5
2
⋅
2
t
t
2
+
1
−
1
t
2
+
1
d
t
=
∫
0
0.9
−
3
2
(
t
2
+
1
)
−
5
2
⋅
2
t
⋅
(
t
2
+
1
)
t
2
d
t
=
∫
0
0.9
−
3
2
(
t
2
+
1
)
−
1
2
⋅
2
t
t
d
t
=
{\displaystyle =\int _{0}^{0.9}{\sqrt {\frac {-{\frac {3}{2}}(t^{2}+1)^{-{\frac {5}{2}}}\cdot 2t}{\frac {t^{2}+1-1}{t^{2}+1}}}}\;dt=\int _{0}^{0.9}{\sqrt {\frac {-{\frac {3}{2}}(t^{2}+1)^{-{\frac {5}{2}}}\cdot 2t\cdot (t^{2}+1)}{t^{2}}}}\;dt=\int _{0}^{0.9}{\frac {\sqrt {-{\frac {3}{2}}(t^{2}+1)^{-{\frac {1}{2}}}\cdot 2t}}{t}}\;dt=}
=
∫
0
0.9
−
3
(
t
2
+
1
)
−
1
2
d
t
=
−
3
∫
0
0.9
(
t
2
+
1
)
−
1
4
d
t
{\displaystyle =\int _{0}^{0.9}{\sqrt {-3(t^{2}+1)^{-{\frac {1}{2}}}}}\;dt={\sqrt {-3}}\int _{0}^{0.9}(t^{2}+1)^{-{\frac {1}{4}}}\;dt}
Šis pavyzdis taip paprastai neišsisprendžia. Bet galima patikrinti ar
x
1
3
(
1
−
x
2
3
)
−
1
2
{\displaystyle x^{\frac {1}{3}}(1-x^{\frac {2}{3}})^{-{\frac {1}{2}}}}
x
−
1
3
1
−
x
2
3
−
x
−
1
3
1
−
x
2
3
.
{\displaystyle {\frac {x^{-{\frac {1}{3}}}}{\sqrt {1-x^{\frac {2}{3}}}}}-x^{-{\frac {1}{3}}}{\sqrt {1-x^{\frac {2}{3}}}}.}
x
1
3
(
1
−
x
2
3
)
−
1
2
=
0.5
1
3
(
1
−
0.5
2
3
)
−
1
2
=
0.793700526
⋅
(
1
−
0.629960525
)
−
1
2
=
{\displaystyle x^{\frac {1}{3}}(1-x^{\frac {2}{3}})^{-{\frac {1}{2}}}=0.5^{\frac {1}{3}}(1-0.5^{\frac {2}{3}})^{-{\frac {1}{2}}}=0.793700526\cdot (1-0.629960525)^{-{\frac {1}{2}}}=}
=
0.793700526
⋅
0.370039475
−
1
2
=
0.793700526
⋅
0.6083087
−
1
=
0.793700526
⋅
1.643902182
=
1.304766027.
{\displaystyle =0.793700526\cdot 0.370039475^{-{\frac {1}{2}}}=0.793700526\cdot 0.6083087^{-1}=0.793700526\cdot 1.643902182=1.304766027.}
x
−
1
3
1
−
x
2
3
−
x
−
1
3
1
−
x
2
3
=
0.5
−
1
3
1
−
0.5
2
3
−
0.5
−
1
3
1
−
0.5
2
3
=
{\displaystyle {\frac {x^{-{\frac {1}{3}}}}{\sqrt {1-x^{\frac {2}{3}}}}}-x^{-{\frac {1}{3}}}{\sqrt {1-x^{\frac {2}{3}}}}={\frac {0.5^{-{\frac {1}{3}}}}{\sqrt {1-0.5^{\frac {2}{3}}}}}-0.5^{-{\frac {1}{3}}}{\sqrt {1-0.5^{\frac {2}{3}}}}=}
=
0.793700526
−
1
1
−
0.629960525
−
0.793700526
−
1
1
−
0.629960525
=
1.25992105
0.370039475
−
1.25992105
0.370039475
=
{\displaystyle ={\frac {0.793700526^{-1}}{\sqrt {1-0.629960525}}}-0.793700526^{-1}{\sqrt {1-0.629960525}}={\frac {1.25992105}{\sqrt {0.370039475}}}-1.25992105{\sqrt {0.370039475}}=}
=
2.071186963
−
0.766420936
=
1.304766026.
{\displaystyle =2.071186963-0.766420936=1.304766026.}
Dar toks dalykas, integratorius http://integrals.wolfram.com/index.jsp?expr=x%5E%281%2F3%29%281-x%5E%282%2F3%29%29%5E%28-1%2F2%29&random=false pirmą funkciją integruoja taip:
∫
x
1
3
(
1
−
x
2
3
)
−
1
2
d
x
=
−
1
−
x
2
/
3
(
x
2
/
3
+
2
)
+
C
=
−
1
−
0.5
2
/
3
(
0.5
2
/
3
+
2
)
=
{\displaystyle \int x^{\frac {1}{3}}(1-x^{\frac {2}{3}})^{-{\frac {1}{2}}}dx=-{\sqrt {1-x^{2/3}}}(x^{2/3}+2)+C=-{\sqrt {1-0.5^{2/3}}}(0.5^{2/3}+2)=}
=
−
1
−
0.629960525
(
0.629960525
+
2
)
=
−
0.370039475
⋅
2.629960525
=
{\displaystyle =-{\sqrt {1-0.629960525}}(0.629960525+2)=-{\sqrt {0.370039475}}\cdot 2.629960525=}
=
−
0.6083087
⋅
2.629960525
=
−
1.599827869.
{\displaystyle =-0.6083087\cdot 2.629960525=-1.599827869.}
Bet atsakymai tokie patys
∫
x
1
3
(
1
−
x
2
3
)
−
1
2
d
x
=
−
3
(
1
−
x
2
3
)
1
2
+
(
1
−
x
2
3
)
3
2
+
C
=
−
3
(
1
−
0.5
2
3
)
1
2
+
(
1
−
0.5
2
3
)
3
2
=
{\displaystyle \int x^{\frac {1}{3}}(1-x^{\frac {2}{3}})^{-{\frac {1}{2}}}dx=-3(1-x^{\frac {2}{3}})^{\frac {1}{2}}+(1-x^{\frac {2}{3}})^{\frac {3}{2}}+C=-3(1-0.5^{\frac {2}{3}})^{\frac {1}{2}}+(1-0.5^{\frac {2}{3}})^{\frac {3}{2}}=}
=
−
3
(
1
−
0.629960525
)
1
2
+
(
1
−
0.629960525
)
3
2
=
−
3
⋅
0.370039475
1
2
+
0.370039475
3
2
=
{\displaystyle =-3(1-0.629960525)^{\frac {1}{2}}+(1-0.629960525)^{\frac {3}{2}}=-3\cdot 0.370039475^{\frac {1}{2}}+0.370039475^{\frac {3}{2}}=}
=
−
3
⋅
0.6083087
+
0.6083087
3
=
−
1.824926101
+
0.225098232
=
−
1.599827869.
{\displaystyle =-3\cdot 0.6083087+0.6083087^{3}=-1.824926101+0.225098232=-1.599827869.}
Tikslas yra patikrinti ar su internetiniu integratoriumi gauto rezultato išvestinė duos tokį patį atsakymą kaip neintegruotos pirmos funkcijos įstačius x=1/2, taigi:
(
−
1
−
x
2
/
3
(
x
2
/
3
+
2
)
)
′
=
−
1
2
(
1
−
x
2
/
3
)
−
1
2
⋅
(
−
2
3
x
−
1
3
)
(
x
2
/
3
+
2
)
−
1
−
x
2
/
3
⋅
2
3
x
−
1
3
=
{\displaystyle (-{\sqrt {1-x^{2/3}}}(x^{2/3}+2))'=-{1 \over 2}(1-x^{2/3})^{-{\frac {1}{2}}}\cdot (-{2 \over 3}x^{-{\frac {1}{3}}})(x^{2/3}+2)-{\sqrt {1-x^{2/3}}}\cdot {2 \over 3}x^{-{\frac {1}{3}}}=}
=
x
−
1
3
3
(
1
−
x
2
/
3
)
−
1
2
(
x
2
/
3
+
2
)
−
2
3
x
−
1
3
1
−
x
2
/
3
=
0.5
−
1
3
3
(
1
−
0.5
2
/
3
)
−
1
2
(
0.5
2
/
3
+
2
)
−
2
3
⋅
0.5
−
1
3
1
−
0.5
2
/
3
=
{\displaystyle ={x^{-{\frac {1}{3}}} \over 3}(1-x^{2/3})^{-{\frac {1}{2}}}(x^{2/3}+2)-{2 \over 3}x^{-{\frac {1}{3}}}{\sqrt {1-x^{2/3}}}={0.5^{-{\frac {1}{3}}} \over 3}(1-0.5^{2/3})^{-{\frac {1}{2}}}(0.5^{2/3}+2)-{2 \over 3}\cdot 0.5^{-{\frac {1}{3}}}{\sqrt {1-0.5^{2/3}}}=}
=
1.25992105
3
(
1
−
0.629960525
)
−
1
2
(
0.629960525
+
2
)
−
2
3
⋅
1.25992105
1
−
0.629960525
=
{\displaystyle ={1.25992105 \over 3}(1-0.629960525)^{-{\frac {1}{2}}}(0.629960525+2)-{2 \over 3}\cdot 1.25992105{\sqrt {1-0.629960525}}=}
=
0.419973680
⋅
0.370039475
−
1
2
⋅
2.629960525
−
0.839947366
⋅
0.370039475
=
{\displaystyle =0.419973680\cdot 0.370039475^{-{\frac {1}{2}}}\cdot 2.629960525-0.839947366\cdot {\sqrt {0.370039475}}=}
=
0.419973680
⋅
1.643902182
⋅
2.629960525
−
0.839947366
⋅
0.6083087
=
{\displaystyle =0.419973680\cdot 1.643902182\cdot 2.629960525-0.839947366\cdot 0.6083087=}
=
1.815713303
−
0.51094729
=
1.304766013.
{\displaystyle =1.815713303-0.51094729=1.304766013.}
Patikrinsime parbolės lanko ilgį padalindami parabolės šaką į 10 atkarpų-tiesių, kai 0<x<4. Kiekvienos atkarpos projekcijos į Ox ašį ilgis yra 0,4. Todėl reikiau gauti visas x reikšmes:
x
0
=
0
;
{\displaystyle x_{0}=0;}
x
1
=
0.4
;
{\displaystyle x_{1}=0.4;}
x
2
=
2
⋅
0.4
=
0.8
;
{\displaystyle x_{2}=2\cdot 0.4=0.8;}
x
3
=
3
⋅
0.4
=
1.2
;
{\displaystyle x_{3}=3\cdot 0.4=1.2;}
x
4
=
4
⋅
0.4
=
1.6
;
{\displaystyle x_{4}=4\cdot 0.4=1.6;}
x
5
=
5
⋅
0.4
=
2
;
{\displaystyle x_{5}=5\cdot 0.4=2;}
x
6
=
6
⋅
0.4
=
2.4
;
{\displaystyle x_{6}=6\cdot 0.4=2.4;}
x
7
=
7
⋅
0.4
=
2.8
;
{\displaystyle x_{7}=7\cdot 0.4=2.8;}
x
8
=
8
⋅
0.4
=
3.2
;
{\displaystyle x_{8}=8\cdot 0.4=3.2;}
x
9
=
9
⋅
0.4
=
3.6
;
{\displaystyle x_{9}=9\cdot 0.4=3.6;}
x
10
=
10
⋅
0.4
=
4.
{\displaystyle x_{10}=10\cdot 0.4=4.}
Dabar toliau reikia surasti visas y reikšmes, įstačius x reikšmes:
y
0
=
x
0
2
=
0
2
=
0
;
{\displaystyle y_{0}=x_{0}^{2}=0^{2}=0;}
y
1
=
x
1
2
=
0.4
2
=
0.16
;
{\displaystyle y_{1}=x_{1}^{2}=0.4^{2}=0.16;}
y
2
=
x
2
2
=
0.8
2
=
0.64
;
{\displaystyle y_{2}=x_{2}^{2}=0.8^{2}=0.64;}
y
3
=
x
3
2
=
1.2
2
=
1.44
;
{\displaystyle y_{3}=x_{3}^{2}=1.2^{2}=1.44;}
y
4
=
x
4
2
=
1.6
2
=
2.56
;
{\displaystyle y_{4}=x_{4}^{2}=1.6^{2}=2.56;}
y
5
=
x
5
2
=
2
2
=
4
;
{\displaystyle y_{5}=x_{5}^{2}=2^{2}=4;}
y
6
=
x
6
2
=
2.4
2
=
5.76
;
{\displaystyle y_{6}=x_{6}^{2}=2.4^{2}=5.76;}
y
7
=
x
7
2
=
2.8
2
=
7.84
;
{\displaystyle y_{7}=x_{7}^{2}=2.8^{2}=7.84;}
y
8
=
x
8
2
=
3.2
2
=
10.24
;
{\displaystyle y_{8}=x_{8}^{2}=3.2^{2}=10.24;}
y
9
=
x
9
2
=
3.6
2
=
12.96
;
{\displaystyle y_{9}=x_{9}^{2}=3.6^{2}=12.96;}
y
10
=
x
10
2
=
4
2
=
16.
{\displaystyle y_{10}=x_{10}^{2}=4^{2}=16.}
Dabar belieka surasti atkarpu ilgius kaip nuo taško (0; 0) iki taško (0,4; 0,16); nuo taško (0,4; 0,16) iki (0,8; 0,64) ir taip toliau:
a
1
=
(
x
1
−
x
0
)
2
+
(
y
1
−
y
0
)
2
=
(
0.4
−
0
)
2
+
(
0.16
−
0
)
2
=
0.16
+
0.0256
=
0.1856
=
0.430813184
;
{\displaystyle a_{1}={\sqrt {(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}}}={\sqrt {(0.4-0)^{2}+(0.16-0)^{2}}}={\sqrt {0.16+0.0256}}={\sqrt {0.1856}}=0.430813184;}
a
2
=
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
=
(
0.8
−
0.4
)
2
+
(
0.64
−
0.16
)
2
=
0.16
+
0.2304
=
0.3904
=
0.624819974
;
{\displaystyle a_{2}={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}={\sqrt {(0.8-0.4)^{2}+(0.64-0.16)^{2}}}={\sqrt {0.16+0.2304}}={\sqrt {0.3904}}=0.624819974;}
a
3
=
(
x
3
−
x
2
)
2
+
(
y
3
−
y
2
)
2
=
(
1.2
−
0.8
)
2
+
(
1.44
−
0.64
)
2
=
0.16
+
0.64
=
0.8
=
0.894427191
;
{\displaystyle a_{3}={\sqrt {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}}={\sqrt {(1.2-0.8)^{2}+(1.44-0.64)^{2}}}={\sqrt {0.16+0.64}}={\sqrt {0.8}}=0.894427191;}
a
4
=
(
x
4
−
x
3
)
2
+
(
y
4
−
y
3
)
2
=
(
1.6
−
1.2
)
2
+
(
2.56
−
1.44
)
2
=
0.16
+
1.2544
=
1.4144
=
1.1892855
;
{\displaystyle a_{4}={\sqrt {(x_{4}-x_{3})^{2}+(y_{4}-y_{3})^{2}}}={\sqrt {(1.6-1.2)^{2}+(2.56-1.44)^{2}}}={\sqrt {0.16+1.2544}}={\sqrt {1.4144}}=1.1892855;}
a
5
=
(
x
5
−
x
4
)
2
+
(
y
5
−
y
4
)
2
=
(
2
−
1.6
)
2
+
(
4
−
2.56
)
2
=
0.16
+
2.0736
=
2.2336
=
1.494523335
;
{\displaystyle a_{5}={\sqrt {(x_{5}-x_{4})^{2}+(y_{5}-y_{4})^{2}}}={\sqrt {(2-1.6)^{2}+(4-2.56)^{2}}}={\sqrt {0.16+2.0736}}={\sqrt {2.2336}}=1.494523335;}
a
6
=
(
x
6
−
x
5
)
2
+
(
y
6
−
y
5
)
2
=
(
2.4
−
2
)
2
+
(
5.76
−
4
)
2
=
0.16
+
3.0976
=
3.2576
=
1.804882268
;
{\displaystyle a_{6}={\sqrt {(x_{6}-x_{5})^{2}+(y_{6}-y_{5})^{2}}}={\sqrt {(2.4-2)^{2}+(5.76-4)^{2}}}={\sqrt {0.16+3.0976}}={\sqrt {3.2576}}=1.804882268;}
a
7
=
(
x
7
−
x
6
)
2
+
(
y
7
−
y
6
)
2
=
(
2.8
−
2.4
)
2
+
(
7.84
−
5.76
)
2
=
0.16
+
4.3264
=
4.4864
=
2.118112367
;
{\displaystyle a_{7}={\sqrt {(x_{7}-x_{6})^{2}+(y_{7}-y_{6})^{2}}}={\sqrt {(2.8-2.4)^{2}+(7.84-5.76)^{2}}}={\sqrt {0.16+4.3264}}={\sqrt {4.4864}}=2.118112367;}
a
8
=
(
x
8
−
x
7
)
2
+
(
y
8
−
y
7
)
2
=
(
3.2
−
2.8
)
2
+
(
10.24
−
7.84
)
2
=
0.16
+
5.76
=
5.92
=
2.433105012
;
{\displaystyle a_{8}={\sqrt {(x_{8}-x_{7})^{2}+(y_{8}-y_{7})^{2}}}={\sqrt {(3.2-2.8)^{2}+(10.24-7.84)^{2}}}={\sqrt {0.16+5.76}}={\sqrt {5.92}}=2.433105012;}
a
9
=
(
x
9
−
x
8
)
2
+
(
y
9
−
y
8
)
2
=
(
3.6
−
3.2
)
2
+
(
12.96
−
10.24
)
2
=
0.16
+
7.3984
=
7.5584
=
2.749254444
;
{\displaystyle a_{9}={\sqrt {(x_{9}-x_{8})^{2}+(y_{9}-y_{8})^{2}}}={\sqrt {(3.6-3.2)^{2}+(12.96-10.24)^{2}}}={\sqrt {0.16+7.3984}}={\sqrt {7.5584}}=2.749254444;}
a
10
=
(
x
10
−
x
9
)
2
+
(
y
10
−
y
9
)
2
=
(
4
−
3.6
)
2
+
(
16
−
12.96
)
2
=
0.16
+
9.2416
=
9.4016
=
3.066202863.
{\displaystyle a_{10}={\sqrt {(x_{10}-x_{9})^{2}+(y_{10}-y_{9})^{2}}}={\sqrt {(4-3.6)^{2}+(16-12.96)^{2}}}={\sqrt {0.16+9.2416}}={\sqrt {9.4016}}=3.066202863.}
Toliau reikia sudėti visų atkarpų ilgį, kad gauti parabolės šakos ilgį, kai x kinta nuo 0 iki 4. Gauname:
L
=
a
1
+
a
2
+
a
3
+
a
4
+
a
5
+
a
6
+
a
7
+
a
8
+
a
9
+
a
10
=
{\displaystyle L=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{10}=}
=
0.430813184
+
0.624819974
+
0.894427191
+
1.1892855
+
1.494523335
+
1.804882268
+
2.118112367
+
2.433105012
+
2.749254444
+
3.066202863
=
16.80542614.
{\displaystyle =0.430813184+0.624819974+0.894427191+1.1892855+1.494523335+1.804882268+2.118112367+2.433105012+2.749254444+3.066202863=16.80542614.}
Padalinus į daugiau dalių atsakymas taptų panašesnis į atsakymą gautą integravimo budu. Apskaičiuosime parabolės
y
=
x
2
{\displaystyle y=x^{2}}
0
≤
x
≤
4.
{\displaystyle 0\leq x\leq 4.}
Randame
y
′
=
(
x
2
)
′
=
2
x
{\displaystyle y'=(x^{2})'=2x}
integralų lentele
∫
x
2
+
a
d
x
=
x
2
a
+
x
2
+
a
2
ln
|
x
+
x
2
+
a
|
{\displaystyle \int {\sqrt {x^{2}+a}}\;{\mathsf {d}}x={\frac {x}{2}}{\sqrt {a+x^{2}}}+{\frac {a}{2}}\ln \left|x+{\sqrt {x^{2}+a}}\right|}
L
=
∫
0
4
1
+
y
′
2
d
x
=
∫
0
4
1
+
(
2
x
)
2
d
x
=
∫
0
4
1
+
4
x
2
d
x
=
4
∫
0
4
1
4
+
x
2
d
x
=
{\displaystyle L=\int _{0}^{4}{\sqrt {1+y'^{2}}}dx=\int _{0}^{4}{\sqrt {1+(2x)^{2}}}dx=\int _{0}^{4}{\sqrt {1+4x^{2}}}dx={\sqrt {4}}\int _{0}^{4}{\sqrt {{1 \over 4}+x^{2}}}dx=}
=
2
(
x
2
0
,
25
+
x
2
+
0
,
25
2
ln
|
x
+
x
2
+
0
,
25
|
)
|
0
4
=
{\displaystyle =2({x \over 2}{\sqrt {0,25+x^{2}}}+{\frac {0,25}{2}}\ln \left|x+{\sqrt {x^{2}+0,25}}\right|)|_{0}^{4}=}
=
2
(
4
2
0
,
25
+
4
2
+
0
,
25
2
ln
|
4
+
4
2
+
0
,
25
|
)
−
2
(
0
2
0
,
25
+
0
2
+
0
,
25
2
ln
|
0
+
0
2
+
0
,
25
|
)
=
{\displaystyle =2({4 \over 2}{\sqrt {0,25+4^{2}}}+{\frac {0,25}{2}}\ln \left|4+{\sqrt {4^{2}+0,25}}\right|)-2({0 \over 2}{\sqrt {0,25+0^{2}}}+{\frac {0,25}{2}}\ln \left|0+{\sqrt {0^{2}+0,25}}\right|)=}
=
2
(
2
16
,
25
+
0
,
125
ln
|
4
+
16
,
25
|
)
−
2
(
0
+
0
,
125
ln
|
0
+
0
,
25
|
)
=
{\displaystyle =2(2{\sqrt {16,25}}+0,125\ln \left|4+{\sqrt {16,25}}\right|)-2(0+0,125\ln \left|0+{\sqrt {0,25}}\right|)=}
≈
2
(
2
⋅
4
,
031128874
+
0
,
125
ln
|
8
,
031128874
|
)
−
2
(
0
,
125
⋅
ln
|
0
,
5
|
)
≈
{\displaystyle \approx 2(2\cdot 4,031128874+0,125\ln |8,031128874|)-2(0,125\cdot \ln |0,5|)\approx }
≈
2
(
8
,
062257748
+
0
,
125
⋅
2
,
0833251
)
−
2
(
0
,
125
⋅
(
−
0
,
69314718
)
)
=
{\displaystyle \approx 2(8,062257748+0,125\cdot 2,0833251)-2(0,125\cdot (-0,69314718))=}
=
2
(
8
,
062257748
+
0
,
260415637
)
−
2
(
−
0
,
086643397
)
=
16
,
64534677
+
0
,
173286795
=
16
,
81863357.
{\displaystyle =2(8,062257748+0,260415637)-2(-0,086643397)=16,64534677+0,173286795=16,81863357.}
Palyginimui, tiesios linijos ilgis nuo taško (0; 0) iki taško (4; 16) yra
c
=
a
2
+
b
2
=
4
2
+
16
2
=
272
=
16
,
4924225.
{\displaystyle c={\sqrt {a^{2}+b^{2}}}={\sqrt {4^{2}+16^{2}}}={\sqrt {272}}=16,4924225.}
Patikrinsime parbolės lanko ilgį padalindami parabolės šaką į 20 atkarpų-tiesių, kai 0<x<4. Kiekvienos atkarpos projekcijos į Ox ašį ilgis yra 0,2.
y reikšmes, įstačius x reikšmes:
y
0
=
x
0
2
=
0
2
=
0
;
{\displaystyle y_{0}=x_{0}^{2}=0^{2}=0;}
y
1
=
x
1
2
=
0.2
2
=
0.04
;
{\displaystyle y_{1}=x_{1}^{2}=0.2^{2}=0.04;}
y
2
=
x
2
2
=
0.4
2
=
0.16
;
{\displaystyle y_{2}=x_{2}^{2}=0.4^{2}=0.16;}
y
3
=
x
3
2
=
0.6
2
=
0.36
;
{\displaystyle y_{3}=x_{3}^{2}=0.6^{2}=0.36;}
y
4
=
x
4
2
=
0.8
2
=
0.64
;
{\displaystyle y_{4}=x_{4}^{2}=0.8^{2}=0.64;}
y
5
=
x
5
2
=
1
2
=
1
;
{\displaystyle y_{5}=x_{5}^{2}=1^{2}=1;}
y
6
=
x
6
2
=
1.2
2
=
1.44
;
{\displaystyle y_{6}=x_{6}^{2}=1.2^{2}=1.44;}
y
7
=
x
7
2
=
1.4
2
=
1.96
;
{\displaystyle y_{7}=x_{7}^{2}=1.4^{2}=1.96;}
y
8
=
x
8
2
=
1.6
2
=
2.56
;
{\displaystyle y_{8}=x_{8}^{2}=1.6^{2}=2.56;}
y
9
=
x
9
2
=
1.8
2
=
3.24
;
{\displaystyle y_{9}=x_{9}^{2}=1.8^{2}=3.24;}
y
10
=
x
10
2
=
2
2
=
4
;
{\displaystyle y_{10}=x_{10}^{2}=2^{2}=4;}
y
11
=
x
11
2
=
2.2
2
=
4.84
;
{\displaystyle y_{11}=x_{11}^{2}=2.2^{2}=4.84;}
y
12
=
x
12
2
=
2.4
2
=
5.76
;
{\displaystyle y_{12}=x_{12}^{2}=2.4^{2}=5.76;}
y
13
=
x
13
2
=
2.6
2
=
6.76
;
{\displaystyle y_{13}=x_{13}^{2}=2.6^{2}=6.76;}
y
14
=
x
14
2
=
2.8
2
=
7.84
;
{\displaystyle y_{14}=x_{14}^{2}=2.8^{2}=7.84;}
y
15
=
x
15
2
=
3
2
=
9
;
{\displaystyle y_{15}=x_{15}^{2}=3^{2}=9;}
y
16
=
x
16
2
=
3.2
2
=
10.24
;
{\displaystyle y_{16}=x_{16}^{2}=3.2^{2}=10.24;}
y
17
=
x
17
2
=
3.4
2
=
11.56
;
{\displaystyle y_{17}=x_{17}^{2}=3.4^{2}=11.56;}
y
18
=
x
18
2
=
3.6
2
=
12.96
;
{\displaystyle y_{18}=x_{18}^{2}=3.6^{2}=12.96;}
y
19
=
x
19
2
=
3.8
2
=
14.44
;
{\displaystyle y_{19}=x_{19}^{2}=3.8^{2}=14.44;}
y
20
=
x
20
2
=
4
2
=
16.
{\displaystyle y_{20}=x_{20}^{2}=4^{2}=16.}
Dabar belieka surasti atkarpu ilgius kaip nuo taško (0; 0) iki taško (0,2; 0,04); nuo taško (0,2; 0,04) iki (0,4; 0,16) ir taip toliau:
a
1
=
(
x
1
−
x
0
)
2
+
(
y
1
−
y
0
)
2
=
(
0.2
−
0
)
2
+
(
0.04
−
0
)
2
=
0.2
2
+
0.04
2
=
0.04
+
0.0016
=
0.0416
=
0.20396078
;
{\displaystyle a_{1}={\sqrt {(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}}}={\sqrt {(0.2-0)^{2}+(0.04-0)^{2}}}={\sqrt {0.2^{2}+0.04^{2}}}={\sqrt {0.04+0.0016}}={\sqrt {0.0416}}=0.20396078;}
a
2
=
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
=
(
0.4
−
0.2
)
2
+
(
0.16
−
0.04
)
2
=
0.2
2
+
0.12
2
=
0.04
+
0.0144
=
0.0544
=
0.233238075
;
{\displaystyle a_{2}={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}={\sqrt {(0.4-0.2)^{2}+(0.16-0.04)^{2}}}={\sqrt {0.2^{2}+0.12^{2}}}={\sqrt {0.04+0.0144}}={\sqrt {0.0544}}=0.233238075;}
a
3
=
(
x
3
−
x
2
)
2
+
(
y
3
−
y
2
)
2
=
(
0.6
−
0.4
)
2
+
(
0.36
−
0.16
)
2
=
0.2
2
+
0.2
2
=
0.04
+
0.04
=
0.08
=
0.282842712
;
{\displaystyle a_{3}={\sqrt {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}}={\sqrt {(0.6-0.4)^{2}+(0.36-0.16)^{2}}}={\sqrt {0.2^{2}+0.2^{2}}}={\sqrt {0.04+0.04}}={\sqrt {0.08}}=0.282842712;}
a
4
=
(
x
4
−
x
3
)
2
+
(
y
4
−
y
3
)
2
=
(
0.8
−
0.6
)
2
+
(
0.64
−
0.36
)
2
=
0.2
2
+
0.28
2
=
0.04
+
0.0784
=
0.1184
=
0.34409301
;
{\displaystyle a_{4}={\sqrt {(x_{4}-x_{3})^{2}+(y_{4}-y_{3})^{2}}}={\sqrt {(0.8-0.6)^{2}+(0.64-0.36)^{2}}}={\sqrt {0.2^{2}+0.28^{2}}}={\sqrt {0.04+0.0784}}={\sqrt {0.1184}}=0.34409301;}
a
5
=
(
x
5
−
x
4
)
2
+
(
y
5
−
y
4
)
2
=
(
1
−
0.8
)
2
+
(
1
−
0.64
)
2
=
0.2
2
+
0.36
2
=
0.04
+
0.1296
=
0.1696
=
0.411825205
;
{\displaystyle a_{5}={\sqrt {(x_{5}-x_{4})^{2}+(y_{5}-y_{4})^{2}}}={\sqrt {(1-0.8)^{2}+(1-0.64)^{2}}}={\sqrt {0.2^{2}+0.36^{2}}}={\sqrt {0.04+0.1296}}={\sqrt {0.1696}}=0.411825205;}
a
6
=
(
x
6
−
x
5
)
2
+
(
y
6
−
y
5
)
2
=
(
1.2
−
1
)
2
+
(
1.44
−
1
)
2
=
0.2
2
+
0.44
2
=
0.04
+
0.1936
=
0.2336
=
0.483321838
;
{\displaystyle a_{6}={\sqrt {(x_{6}-x_{5})^{2}+(y_{6}-y_{5})^{2}}}={\sqrt {(1.2-1)^{2}+(1.44-1)^{2}}}={\sqrt {0.2^{2}+0.44^{2}}}={\sqrt {0.04+0.1936}}={\sqrt {0.2336}}=0.483321838;}
a
7
=
(
x
7
−
x
6
)
2
+
(
y
7
−
y
6
)
2
=
(
1.4
−
1.2
)
2
+
(
1.96
−
1.44
)
2
=
0.2
2
+
0.52
2
=
0.04
+
0.2704
=
0.3104
=
0.557135531
;
{\displaystyle a_{7}={\sqrt {(x_{7}-x_{6})^{2}+(y_{7}-y_{6})^{2}}}={\sqrt {(1.4-1.2)^{2}+(1.96-1.44)^{2}}}={\sqrt {0.2^{2}+0.52^{2}}}={\sqrt {0.04+0.2704}}={\sqrt {0.3104}}=0.557135531;}
a
8
=
(
x
8
−
x
7
)
2
+
(
y
8
−
y
7
)
2
=
(
1.6
−
1.4
)
2
+
(
2.56
−
1.96
)
2
=
0.2
2
+
0.6
2
=
0.04
+
0.36
=
0.4
=
0.632455532
;
{\displaystyle a_{8}={\sqrt {(x_{8}-x_{7})^{2}+(y_{8}-y_{7})^{2}}}={\sqrt {(1.6-1.4)^{2}+(2.56-1.96)^{2}}}={\sqrt {0.2^{2}+0.6^{2}}}={\sqrt {0.04+0.36}}={\sqrt {0.4}}=0.632455532;}
a
9
=
(
x
9
−
x
8
)
2
+
(
y
9
−
y
8
)
2
=
(
1.8
−
1.6
)
2
+
(
3.24
−
2.56
)
2
=
0.2
2
+
0.68
2
=
0.04
+
0.4624
=
0.5024
=
0.708801805
;
{\displaystyle a_{9}={\sqrt {(x_{9}-x_{8})^{2}+(y_{9}-y_{8})^{2}}}={\sqrt {(1.8-1.6)^{2}+(3.24-2.56)^{2}}}={\sqrt {0.2^{2}+0.68^{2}}}={\sqrt {0.04+0.4624}}={\sqrt {0.5024}}=0.708801805;}
a
10
=
(
x
10
−
x
9
)
2
+
(
y
10
−
y
9
)
2
=
(
2
−
1.8
)
2
+
(
4
−
3.24
)
2
=
0.2
2
+
0.76
2
=
0.04
+
0.5776
=
0.6176
=
0.785875308
;
{\displaystyle a_{10}={\sqrt {(x_{10}-x_{9})^{2}+(y_{10}-y_{9})^{2}}}={\sqrt {(2-1.8)^{2}+(4-3.24)^{2}}}={\sqrt {0.2^{2}+0.76^{2}}}={\sqrt {0.04+0.5776}}={\sqrt {0.6176}}=0.785875308;}
a
11
=
(
x
11
−
x
10
)
2
+
(
y
11
−
y
10
)
2
=
(
2.2
−
2
)
2
+
(
4.84
−
4
)
2
=
0.2
2
+
0.84
2
=
0.04
+
0.7056
=
0.7456
=
0.863481325
;
{\displaystyle a_{11}={\sqrt {(x_{11}-x_{10})^{2}+(y_{11}-y_{10})^{2}}}={\sqrt {(2.2-2)^{2}+(4.84-4)^{2}}}={\sqrt {0.2^{2}+0.84^{2}}}={\sqrt {0.04+0.7056}}={\sqrt {0.7456}}=0.863481325;}
a
12
=
(
x
12
−
x
11
)
2
+
(
y
12
−
y
11
)
2
=
(
2.4
−
2.2
)
2
+
(
5.76
−
4.84
)
2
=
0.2
2
+
0.92
2
=
0.04
+
0.8464
=
0.8864
=
0.941488183
;
{\displaystyle a_{12}={\sqrt {(x_{12}-x_{11})^{2}+(y_{12}-y_{11})^{2}}}={\sqrt {(2.4-2.2)^{2}+(5.76-4.84)^{2}}}={\sqrt {0.2^{2}+0.92^{2}}}={\sqrt {0.04+0.8464}}={\sqrt {0.8864}}=0.941488183;}
a
13
=
(
x
13
−
x
12
)
2
+
(
y
13
−
y
12
)
2
=
(
2.6
−
2.4
)
2
+
(
6.76
−
5.76
)
2
=
0.2
2
+
1
2
=
0.04
+
1
=
1.04
=
1.019803903
;
{\displaystyle a_{13}={\sqrt {(x_{13}-x_{12})^{2}+(y_{13}-y_{12})^{2}}}={\sqrt {(2.6-2.4)^{2}+(6.76-5.76)^{2}}}={\sqrt {0.2^{2}+1^{2}}}={\sqrt {0.04+1}}={\sqrt {1.04}}=1.019803903;}
a
14
=
(
x
14
−
x
13
)
2
+
(
y
14
−
y
13
)
2
=
(
2.8
−
2.6
)
2
+
(
7.84
−
6.76
)
2
=
0.2
2
+
1.08
2
=
0.04
+
1.1664
=
1.2064
=
1.098362417
;
{\displaystyle a_{14}={\sqrt {(x_{14}-x_{13})^{2}+(y_{14}-y_{13})^{2}}}={\sqrt {(2.8-2.6)^{2}+(7.84-6.76)^{2}}}={\sqrt {0.2^{2}+1.08^{2}}}={\sqrt {0.04+1.1664}}={\sqrt {1.2064}}=1.098362417;}
a
15
=
(
x
15
−
x
14
)
2
+
(
y
15
−
y
14
)
2
=
(
3
−
2.8
)
2
+
(
9
−
7.84
)
2
=
0.2
2
+
1.16
2
=
0.04
+
1.3456
=
1.3856
=
1.177115118
;
{\displaystyle a_{15}={\sqrt {(x_{15}-x_{14})^{2}+(y_{15}-y_{14})^{2}}}={\sqrt {(3-2.8)^{2}+(9-7.84)^{2}}}={\sqrt {0.2^{2}+1.16^{2}}}={\sqrt {0.04+1.3456}}={\sqrt {1.3856}}=1.177115118;}
a
16
=
(
x
16
−
x
15
)
2
+
(
y
16
−
y
15
)
2
=
(
3.2
−
3
)
2
+
(
10.24
−
9
)
2
=
0.2
2
+
1.24
2
=
0.04
+
1.5376
=
1.5776
=
1.256025477
;
{\displaystyle a_{16}={\sqrt {(x_{16}-x_{15})^{2}+(y_{16}-y_{15})^{2}}}={\sqrt {(3.2-3)^{2}+(10.24-9)^{2}}}={\sqrt {0.2^{2}+1.24^{2}}}={\sqrt {0.04+1.5376}}={\sqrt {1.5776}}=1.256025477;}
a
17
=
(
x
17
−
x
16
)
2
+
(
y
17
−
y
16
)
2
=
(
3.4
−
3.2
)
2
+
(
11.56
−
10.24
)
2
=
0.2
2
+
1.32
2
=
0.04
+
1.7424
=
1.7824
=
1.335065541
;
{\displaystyle a_{17}={\sqrt {(x_{17}-x_{16})^{2}+(y_{17}-y_{16})^{2}}}={\sqrt {(3.4-3.2)^{2}+(11.56-10.24)^{2}}}={\sqrt {0.2^{2}+1.32^{2}}}={\sqrt {0.04+1.7424}}={\sqrt {1.7824}}=1.335065541;}
a
18
=
(
x
18
−
x
17
)
2
+
(
y
18
−
y
17
)
2
=
(
3.6
−
3.4
)
2
+
(
12.96
−
11.56
)
2
=
0.2
2
+
1.4
2
=
0.04
+
1.96
=
2
=
1.414213562
;
{\displaystyle a_{18}={\sqrt {(x_{18}-x_{17})^{2}+(y_{18}-y_{17})^{2}}}={\sqrt {(3.6-3.4)^{2}+(12.96-11.56)^{2}}}={\sqrt {0.2^{2}+1.4^{2}}}={\sqrt {0.04+1.96}}={\sqrt {2}}=1.414213562;}
a
19
=
(
x
19
−
x
18
)
2
+
(
y
19
−
y
18
)
2
=
(
3.8
−
3.6
)
2
+
(
14.44
−
12.96
)
2
=
0.2
2
+
1.48
2
=
0.04
+
2.1904
=
2.2304
=
1.493452376
;
{\displaystyle a_{19}={\sqrt {(x_{19}-x_{18})^{2}+(y_{19}-y_{18})^{2}}}={\sqrt {(3.8-3.6)^{2}+(14.44-12.96)^{2}}}={\sqrt {0.2^{2}+1.48^{2}}}={\sqrt {0.04+2.1904}}={\sqrt {2.2304}}=1.493452376;}
a
20
=
(
x
20
−
x
19
)
2
+
(
y
20
−
y
19
)
2
=
(
4
−
3.8
)
2
+
(
16
−
14.44
)
2
=
0.2
2
+
1.56
2
=
0.04
+
2.4336
=
2.4736
=
1.57276826.
{\displaystyle a_{20}={\sqrt {(x_{20}-x_{19})^{2}+(y_{20}-y_{19})^{2}}}={\sqrt {(4-3.8)^{2}+(16-14.44)^{2}}}={\sqrt {0.2^{2}+1.56^{2}}}={\sqrt {0.04+2.4336}}={\sqrt {2.4736}}=1.57276826.}
Toliau reikia sudėti visų atkarpų ilgį, kad gauti parabolės šakos ilgį, kai x kinta nuo 0 iki 4. Gauname:
L
=
a
1
+
a
2
+
a
3
+
a
4
+
a
5
+
a
6
+
a
7
+
a
8
+
a
9
+
a
10
+
{\displaystyle L=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{10}+}
+
a
11
+
a
12
+
a
13
+
a
14
+
a
15
+
a
16
+
a
17
+
a
18
+
a
19
+
a
20
=
{\displaystyle +a_{11}+a_{12}+a_{13}+a_{14}+a_{15}+a_{16}+a_{17}+a_{18}+a_{19}+a_{20}=}
=
0.0416
+
0.0544
+
0.08
+
0.1184
+
0.1696
+
0.2336
+
0.3104
+
0.4
+
0.5024
+
0.6176
+
{\displaystyle ={\sqrt {0.0416}}+{\sqrt {0.0544}}+{\sqrt {0.08}}+{\sqrt {0.1184}}+{\sqrt {0.1696}}+{\sqrt {0.2336}}+{\sqrt {0.3104}}+{\sqrt {0.4}}+{\sqrt {0.5024}}+{\sqrt {0.6176}}+}
+
0.7456
+
0.8864
+
1.04
+
1.2064
+
1.3856
+
1.5776
+
1.7824
+
2
+
2.2304
+
2.4736
=
{\displaystyle +{\sqrt {0.7456}}+{\sqrt {0.8864}}+{\sqrt {1.04}}+{\sqrt {1.2064}}+{\sqrt {1.3856}}+{\sqrt {1.5776}}+{\sqrt {1.7824}}+{\sqrt {2}}+{\sqrt {2.2304}}+{\sqrt {2.4736}}=}
=
4.643549801
+
12.17177616
=
16.81532597.
{\displaystyle =4.643549801+12.17177616=16.81532597.}
Padalinus į daugiau dalių atsakymas taptų panašesnis į atsakymą gautą integravimo budu. Apskaičiuosime parabolės
y
=
x
2
{\displaystyle y=x^{2}}
0
≤
x
≤
4.
{\displaystyle 0\leq x\leq 4.}
Randame
y
′
=
(
x
2
)
′
=
2
x
{\displaystyle y'=(x^{2})'=2x}
integralų lentele
∫
x
2
+
a
d
x
=
x
2
a
+
x
2
+
a
2
ln
|
x
+
x
2
+
a
|
{\displaystyle \int {\sqrt {x^{2}+a}}\;{\mathsf {d}}x={\frac {x}{2}}{\sqrt {a+x^{2}}}+{\frac {a}{2}}\ln \left|x+{\sqrt {x^{2}+a}}\right|}
L
=
∫
0
4
1
+
y
′
2
d
x
=
∫
0
4
1
+
(
2
x
)
2
d
x
=
∫
0
4
1
+
4
x
2
d
x
=
4
∫
0
4
1
4
+
x
2
d
x
=
{\displaystyle L=\int _{0}^{4}{\sqrt {1+y'^{2}}}dx=\int _{0}^{4}{\sqrt {1+(2x)^{2}}}dx=\int _{0}^{4}{\sqrt {1+4x^{2}}}dx={\sqrt {4}}\int _{0}^{4}{\sqrt {{1 \over 4}+x^{2}}}dx=}
=
2
(
x
2
0
,
25
+
x
2
+
0
,
25
2
ln
|
x
+
x
2
+
0
,
25
|
)
|
0
4
=
{\displaystyle =2({x \over 2}{\sqrt {0,25+x^{2}}}+{\frac {0,25}{2}}\ln \left|x+{\sqrt {x^{2}+0,25}}\right|)|_{0}^{4}=}
=
2
(
4
2
0
,
25
+
4
2
+
0
,
25
2
ln
|
4
+
4
2
+
0
,
25
|
)
−
2
(
0
2
0
,
25
+
0
2
+
0
,
25
2
ln
|
0
+
0
2
+
0
,
25
|
)
=
{\displaystyle =2({4 \over 2}{\sqrt {0,25+4^{2}}}+{\frac {0,25}{2}}\ln \left|4+{\sqrt {4^{2}+0,25}}\right|)-2({0 \over 2}{\sqrt {0,25+0^{2}}}+{\frac {0,25}{2}}\ln \left|0+{\sqrt {0^{2}+0,25}}\right|)=}
=
2
(
2
16
,
25
+
0
,
125
ln
|
4
+
16
,
25
|
)
−
2
(
0
+
0
,
125
ln
|
0
+
0
,
25
|
)
=
{\displaystyle =2(2{\sqrt {16,25}}+0,125\ln \left|4+{\sqrt {16,25}}\right|)-2(0+0,125\ln \left|0+{\sqrt {0,25}}\right|)=}
≈
2
(
2
⋅
4
,
031128874
+
0
,
125
ln
|
8
,
031128874
|
)
−
2
(
0
,
125
⋅
ln
|
0
,
5
|
)
≈
{\displaystyle \approx 2(2\cdot 4,031128874+0,125\ln |8,031128874|)-2(0,125\cdot \ln |0,5|)\approx }
≈
2
(
8
,
062257748
+
0
,
125
⋅
2
,
0833251
)
−
2
(
0
,
125
⋅
(
−
0
,
69314718
)
)
=
{\displaystyle \approx 2(8,062257748+0,125\cdot 2,0833251)-2(0,125\cdot (-0,69314718))=}
=
2
(
8
,
062257748
+
0
,
260415637
)
−
2
(
−
0
,
086643397
)
=
16
,
64534677
+
0
,
173286795
=
16
,
81863357.
{\displaystyle =2(8,062257748+0,260415637)-2(-0,086643397)=16,64534677+0,173286795=16,81863357.}
Palyginimui, linijos ilgis nuo taško (0; 0) iki taško (4; 16) yra
c
=
a
2
+
b
2
=
4
2
+
16
2
=
272
=
16
,
4924225.
{\displaystyle c={\sqrt {a^{2}+b^{2}}}={\sqrt {4^{2}+16^{2}}}={\sqrt {272}}=16,4924225.}
Apskaičiuosime parabolės
y
=
x
2
{\displaystyle y=x^{2}}
0
≤
x
≤
5.
{\displaystyle 0\leq x\leq 5.}
Randame
y
′
=
(
x
2
)
′
=
2
x
{\displaystyle y'=(x^{2})'=2x}
L
=
∫
a
b
1
+
f
′
2
(
x
)
d
x
=
∫
0
5
1
+
y
′
2
d
x
=
∫
0
5
1
+
(
2
x
)
2
d
x
=
∫
0
5
1
+
4
x
2
d
x
=
{\displaystyle L=\int _{a}^{b}{\sqrt {1+f'^{2}(x)}}dx=\int _{0}^{5}{\sqrt {1+y'^{2}}}dx=\int _{0}^{5}{\sqrt {1+(2x)^{2}}}dx=\int _{0}^{5}{\sqrt {1+4x^{2}}}dx=}
=
1
4
(
2
x
4
x
2
+
1
+
sinh
−
1
(
2
x
)
)
|
0
5
=
1
4
(
2
x
4
x
2
+
1
+
(
e
2
x
−
e
−
2
x
2
)
−
1
)
|
0
5
=
{\displaystyle ={1 \over 4}(2x{\sqrt {4x^{2}+1}}+\sinh ^{-1}(2x))|_{0}^{5}={1 \over 4}(2x{\sqrt {4x^{2}+1}}+({e^{2x}-e^{-2x} \over 2})^{-1})|_{0}^{5}=}
=
1
4
(
2
x
4
x
2
+
1
+
2
e
2
x
−
e
−
2
x
)
|
0
5
=
1
4
(
10
100
+
1
+
2
e
10
−
e
−
10
)
−
1
4
(
0
0
+
1
+
2
e
0
−
e
0
)
=
{\displaystyle ={1 \over 4}(2x{\sqrt {4x^{2}+1}}+{2 \over e^{2x}-e^{-2x}})|_{0}^{5}={1 \over 4}(10{\sqrt {100+1}}+{2 \over e^{10}-e^{-10}})-{1 \over 4}(0{\sqrt {0+1}}+{2 \over e^{0}-e^{0}})=}
=
1
4
(
10
100
+
1
+
2
e
10
−
e
−
10
)
=
1
4
(
100
,
4987562
+
2
22026
,
46579
−
0
,
000045399
≈
{\displaystyle ={1 \over 4}(10{\sqrt {100+1}}+{2 \over e^{10}-e^{-10}})={1 \over 4}(100,4987562+{2 \over 22026,46579-0,000045399}\approx }
≈
1
4
(
100
,
4987562
+
2
22026
,
46574
)
≈
25
,
12471175.
{\displaystyle \approx {1 \over 4}(100,4987562+{2 \over 22026,46574})\approx 25,12471175.}
Palyginimui, linijos ilgis nuo taško (0; 0) iki taško (5; 25) yra:
c
=
a
2
+
b
2
=
5
2
+
25
2
=
25
+
625
=
650
=
25.49509757.
{\displaystyle c={\sqrt {a^{2}+b^{2}}}={\sqrt {5^{2}+25^{2}}}={\sqrt {25+625}}={\sqrt {650}}=25.49509757.}
Apskaičiuosime parabolės
y
=
x
2
{\displaystyle y=x^{2}}
0
≤
x
≤
5.
{\displaystyle 0\leq x\leq 5.}
Randame
y
′
=
(
x
2
)
′
=
2
x
{\displaystyle y'=(x^{2})'=2x}
L
=
∫
0
5
1
+
(
y
′
)
2
d
x
=
∫
0
5
1
+
(
2
x
)
2
d
x
=
∫
0
5
1
+
4
x
2
d
x
=
4
∫
0
5
1
4
+
x
2
d
x
=
{\displaystyle L=\int _{0}^{5}{\sqrt {1+(y')^{2}}}dx=\int _{0}^{5}{\sqrt {1+(2x)^{2}}}dx=\int _{0}^{5}{\sqrt {1+4x^{2}}}dx={\sqrt {4}}\int _{0}^{5}{\sqrt {{1 \over 4}+x^{2}}}dx=}
=
2
(
x
2
1
4
+
x
2
+
1
4
2
ln
|
x
+
x
2
+
1
4
|
)
|
0
5
=
{\displaystyle =2({x \over 2}{\sqrt {{1 \over 4}+x^{2}}}+{\frac {1 \over 4}{2}}\ln \left|x+{\sqrt {x^{2}+{1 \over 4}}}\right|)|_{0}^{5}=}
=
2
(
5
2
1
4
+
5
2
+
1
4
2
ln
|
5
+
5
2
+
1
4
|
)
−
2
(
0
2
⋅
1
4
+
0
2
+
1
4
2
ln
|
0
+
0
2
+
1
4
|
)
=
{\displaystyle =2({5 \over 2}{\sqrt {{\frac {1}{4}}+5^{2}}}+{\frac {\frac {1}{4}}{2}}\ln \left|5+{\sqrt {5^{2}+{\frac {1}{4}}}}\right|)-2({0 \over 2}\cdot {\sqrt {{\frac {1}{4}}+0^{2}}}+{\frac {\frac {1}{4}}{2}}\ln \left|0+{\sqrt {0^{2}+{\frac {1}{4}}}}\right|)=}
=
2
(
5
2
⋅
1
4
+
25
+
1
8
⋅
ln
|
5
+
25
+
1
4
|
)
−
2
⋅
1
8
⋅
ln
|
1
2
|
=
{\displaystyle =2({\frac {5}{2}}\cdot {\sqrt {{\frac {1}{4}}+25}}+{\frac {1}{8}}\cdot \ln \left|5+{\sqrt {25+{\frac {1}{4}}}}\right|)-2\cdot {\frac {1}{8}}\cdot \ln \left|{\frac {1}{2}}\right|=}
=
2
(
5
2
⋅
5.024937811
+
1
8
⋅
ln
|
5
+
5.024937811
|
)
−
1
4
⋅
(
−
0.69314718
)
=
{\displaystyle =2({\frac {5}{2}}\cdot 5.024937811+{\frac {1}{8}}\cdot \ln |5+5.024937811|)-{1 \over 4}\cdot (-0.69314718)=}
=
2
(
12.56234453
+
1
8
⋅
ln
|
10.02493781
|
)
+
0.173286795
=
{\displaystyle =2(12.56234453+{\frac {1}{8}}\cdot \ln |10.02493781|)+0.173286795=}
=
2
(
12.56234453
+
2.30507577
8
)
+
0.173286795
=
2
⋅
(
12.56234453
+
0.288134471
)
+
0.173286795
=
{\displaystyle =2(12.56234453+{\frac {2.30507577}{8}})+0.173286795=2\cdot (12.56234453+0.288134471)+0.173286795=}
=
2
⋅
12.850479
+
0.173286795
=
25.700958
+
0.173286795
=
25.8742448.
{\displaystyle =2\cdot 12.850479+0.173286795=25.700958+0.173286795=25.8742448.}
Palyginimui, linijos ilgis nuo taško (0; 0) iki taško (5; 25) yra:
c
=
a
2
+
b
2
=
5
2
+
25
2
=
25
+
625
=
650
=
25.49509757.
{\displaystyle c={\sqrt {a^{2}+b^{2}}}={\sqrt {5^{2}+25^{2}}}={\sqrt {25+625}}={\sqrt {650}}=25.49509757.}
Hiperbolinis arksinusas nėra 1 dalijimas iš hiperbolinio sinuso. Hiperbolinis arksinusas lygus:
arsinh
x
=
ln
(
x
+
x
2
+
1
)
{\displaystyle \operatorname {arsinh} x=\ln \left(x+{\sqrt {x^{2}+1}}\right)}
https://en.wikipedia.org/wiki/Inverse_hyperbolic_functions
Su kompiuterio skaičiuotuvu paspaudus
sinh
−
1
(
2
⋅
5
)
=
2.99822295.
{\displaystyle \sinh ^{-1}(2\cdot 5)=2.99822295.}
O šitą reikšmę pridėjus prie 100.4987562 ir viską padalinus iš 4 gaunamas toks pat atsakymas kaip ir antru budu (25.8742448):
=
1
4
(
2
x
4
x
2
+
1
+
sinh
−
1
(
2
x
)
)
|
0
5
=
{\displaystyle ={1 \over 4}(2x{\sqrt {4x^{2}+1}}+\sinh ^{-1}(2x))|_{0}^{5}=}
=
1
4
(
10
100
+
1
+
sinh
−
1
(
10
)
)
=
{\displaystyle ={1 \over 4}(10{\sqrt {100+1}}+\sinh ^{-1}(10))=}
=
1
4
(
100.4987562112089027
+
2.9982229502979697388
)
=
25.87424479037671811.
{\displaystyle ={1 \over 4}(100.4987562112089027+2.9982229502979697388)=25.87424479037671811.}
Arba dar tikslesnė reikšmė tiesiai iš Windows kalkuliatoriaus: 25.874244790376718110259811166298.
Be to,
sinh
−
1
0
=
0.
{\displaystyle \sinh ^{-1}0=0.}
Va šitas Free Pascal kodas: var
a:longint;
c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(0.000000005)+sqr(sqr(0.000000005*a)-sqr(0.000000005*(a-1))));
writeln(c);
readln;
end.
duoda parabolės ilgio (kai x kinta nuo 0 iki 5) rezultatą 2.5874244790371570E+001 (tai reiškia 25.874244790371570) po 35 sekundžių su 4 Ghz procesoriumi. Tai yra 13 teisingu skaitmenų, nors prarabolė dalinama tik į milijardą atkarpų (viskas taip ir turi būti, nes parabolė yra beveik kaip tiesė...).
Su 2.6 GHz procesorium va tokia eilute c:=c+sqrt(sqr(0.000000005)+sqr(sqr(0.000000005*a)-sqr(0.000000005*(a-1))))*0.000000005*a;
kuri skaičiuoja paraboles masę ir yra truputi ilgesnė duoda atsakyma po 25 sekundžiu. Pasirodo 4 GHz procesorius nustatytas buvo, kad veiktų tarp 1.4 GHz ir 4 GHz (Task manager'yje sukinėjasi dažnis apie 2 Ghz). Nustačius, kad veiktų tik 4 GHz dažniu, tai apskaičiuoja parabolės ilgį po 15 sekundžių (o parabolės masę po 16 sekundžių (Task Manager'yje dažnį rodo 4,15 GHz)). Padalinus AMD FX 8350 Eight-Core 4.00 GHz procesoriaus dažnį iš kažkokio ten AMD Athlon 64 X2 5000+ 2.6 GHz (2 beanduolių) procesoriaus dažnio, gaunamas toks rezultatas: 4.16/2.6=1.6. Padalinus 25 sekundes, kurias skaičiavo 2.6 GHz procesorius iš 1.6 karto didesnio dažnio, gauname: 25/1.6=15.625 sekundes. Kas beveik lygu 16 sekundžių, per kurias susidoroja užsiturbines 4.16 GHz procesorius. Iš to išeina išvada, kad Free Pascal programoje skaičiavimo greitis parabolės masės/ilgio, priklauso tik nuo procesoriaus dažnio, o ne nuo procesoriaus architektūros. Trumpi kodo paaiškinimai. Funkcija sqr(x) duoda
x
2
;
{\displaystyle x^{2};}
x
;
{\displaystyle {\sqrt {x}};}
Su 4 GHz procesorium buvo naudojamas toks Free Pascal (su 2.6 GHz CPU buvo naudojama 2.6.0 versija ar 2.4 kažkokia):
"
Free Pascal IDE Version 1.0.12 [2020 06 04]
Compiler Version 3.2.0
GDB Version GNU gdb (GDB) 7.2
Using configuration files from: C:\FPC\3.2.0\bin\i386-win32\
"
Adresu C:\FPC\3.2.0\bin\i386-win32\ reikia paleisti fp.exe arba iš shortcut'o ant desktopo, kuris įrašinėjant Free Pascal , sukuriamas.
Kaip supratau, 64 bitų Free Pascal Neturi lango palengvinančios ir pele naudojančios programos, todėl, kad palengvintai per langa ir su pele naudotis Free Pascal reikia atsisiųsti iš čia https://www.freepascal.org/down/x86_64/win64-hungary.html
du .exe failus 39 MB (fpc-3.2.0.i386-win32.cross.x86_64-win64.exe) ir 51 MB (fpc-3.2.0.i386-win32.exe) ir juos abu įrašyti į default direktorija (ir tada sukurs shortcut'ą ir galima bus normaliai paleist Free Pascal ). Kai Free Pascal įrašytas, reikia sukurti "New", paskui "paste from Windows" parabolės ilgio kodą ir paskui "Compile" ir "Run" arba iš kart "Run".
Kreivės lanko ilgis randamas pagal šitas formules:
d
s
=
1
+
y
′
2
d
x
,
{\displaystyle ds={\sqrt {1+y'^{2}}}dx,}
L apibrėžta lygtimi y=y(x), o
a
≤
x
≤
b
.
{\displaystyle a\leq x\leq b.}
∫
L
d
s
=
∫
a
b
1
+
(
f
′
(
x
)
)
2
d
x
=
∫
a
b
1
+
(
y
′
)
2
d
x
.
{\displaystyle \int _{L}ds=\int _{a}^{b}{\sqrt {1+(f'(x))^{2}}}dx=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx.}
Kai kreivė L apibrėžta parametrinėmis lygtimis
x
=
x
(
t
)
,
{\displaystyle x=x(t),}
y
=
y
(
t
)
,
{\displaystyle y=y(t),}
t
∈
[
t
0
;
T
]
,
{\displaystyle t\in [t_{0};T],}
d
s
=
x
t
′
2
+
y
t
′
2
d
t
,
{\displaystyle ds={\sqrt {x_{t}'^{2}+y_{t}'^{2}}}dt,}
∫
L
d
s
=
∫
t
0
T
(
x
t
′
)
2
+
(
y
t
′
)
2
d
t
.
{\displaystyle \int _{L}ds=\int _{t_{0}}^{T}{\sqrt {(x_{t}')^{2}+(y_{t}')^{2}}}dt.}
Elementariausias pavyzdis, kai reikia perrašyti funkcija
y
=
x
2
{\displaystyle y=x^{2}}
x
=
t
;
y
=
t
2
{\displaystyle x=t;\;y=t^{2}}
x
t
′
=
t
′
=
1
;
y
t
′
=
(
t
2
)
′
=
2
t
{\displaystyle x_{t}'=t'=1;\;y_{t}'=(t^{2})'=2t}
∫
t
0
T
(
x
t
′
)
2
+
(
y
t
′
)
2
d
t
=
∫
t
0
T
1
2
+
(
2
t
)
2
d
t
=
∫
t
0
T
1
+
4
⋅
t
2
d
t
.
{\displaystyle \int _{t_{0}}^{T}{\sqrt {(x_{t}')^{2}+(y_{t}')^{2}}}dt=\int _{t_{0}}^{T}{\sqrt {1^{2}+(2t)^{2}}}dt=\int _{t_{0}}^{T}{\sqrt {1+4\cdot t^{2}}}dt.}
Kai prametrinėmis lygtimis
x
=
x
(
t
)
,
{\displaystyle x=x(t),}
y
=
y
(
t
)
,
{\displaystyle y=y(t),}
z
=
z
(
t
)
,
{\displaystyle z=z(t),}
t
∈
[
t
0
;
T
]
{\displaystyle t\in [t_{0};T]}
L , tai
∫
L
d
s
=
∫
t
0
T
(
x
t
′
)
2
+
(
y
t
′
)
2
+
(
z
t
′
)
2
d
t
.
{\displaystyle \int _{L}ds=\int _{t_{0}}^{T}{\sqrt {(x_{t}')^{2}+(y_{t}')^{2}+(z_{t}')^{2}}}dt.}
Kai kreivė L polinėje koordinačių sistemoje apibrėžta lygtimi
ρ
=
ρ
(
ϕ
)
,
{\displaystyle \rho =\rho (\phi ),}
ϕ
∈
[
α
;
β
]
{\displaystyle \phi \in [\alpha ;\beta ]}
d
s
=
ρ
2
+
ρ
′
2
d
ϕ
{\displaystyle ds={\sqrt {\rho ^{2}+\rho '^{2}}}d\phi }
∫
L
d
s
=
∫
α
β
ρ
2
+
(
ρ
′
)
2
d
ϕ
.
{\displaystyle \int _{L}ds=\int _{\alpha }^{\beta }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\phi .}
Polinėse koordinatėse taikoma ta pati pitagoro teorema. Geresniam išaiškiniui, tegu
ρ
=
θ
2
{\displaystyle \rho =\theta ^{2}}
θ
{\displaystyle \theta }
2
π
.
{\displaystyle 2\pi .}
x kinta pastoviu greičiu, o y kitimo greitis kinta pagal tam tikrą funkciją, tada turime formulę
l
=
∫
a
b
1
+
(
y
′
)
2
d
x
.
{\displaystyle l=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx.}
y
′
=
d
y
d
x
{\displaystyle y'={\frac {dy}{dx}}}
ρ
′
=
d
ρ
d
θ
.
{\displaystyle \rho '={\frac {d\rho }{d\theta }}.}
x pastovų pakitimą su y nepastoviu pakitimu (y') pakeltu kvadratu ir paskui traukiant šaknį. Polinėse koordinatėse apskritimo ilgis
c
=
2
π
R
=
2
p
i
ρ
.
{\displaystyle c=2\pi R=2pi\rho .}
2
π
(
R
i
−
R
i
−
1
)
=
2
π
ρ
.
{\displaystyle 2\pi (R_{i}-R_{i-1})=2\pi \rho .}
2
π
(
R
i
−
R
i
−
1
)
{\displaystyle 2\pi (R_{i}-R_{i-1})}
(
R
i
−
R
i
−
1
)
,
{\displaystyle (R_{i}-R_{i-1}),}
(
2
π
(
R
i
−
R
i
−
1
)
)
2
{\displaystyle (2\pi (R_{i}-R_{i-1}))^{2}}
(
R
i
−
R
i
−
1
)
2
{\displaystyle (R_{i}-R_{i-1})^{2}}
2
π
{\displaystyle 2\pi }
u
=
2
π
(
R
i
−
R
i
−
1
)
{\displaystyle u=2\pi (R_{i}-R_{i-1})}
R
=
ρ
.
{\displaystyle R=\rho .}
∫
α
β
ρ
2
+
(
ρ
′
)
2
d
θ
{\displaystyle \int _{\alpha }^{\beta }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta }
ρ
,
{\displaystyle \rho ,}
θ
,
{\displaystyle \theta ,}
ρ
{\displaystyle \rho }
θ
2
,
{\displaystyle \theta ^{2},}
θ
{\displaystyle \theta }
2
π
{\displaystyle 2\pi }
u
i
2
{\displaystyle u_{i}^{2}}
u
i
2
=
(
2
π
n
i
)
4
.
{\displaystyle u_{i}^{2}=(2\pi ni)^{4}.}
n (pavyzdžiui,
n
=
0.01
{\displaystyle n=0.01}
u
,
{\displaystyle u,}
∫
L
d
s
=
∫
α
β
ρ
2
+
(
ρ
′
)
2
d
θ
=
n
∑
1
m
u
i
2
+
v
i
2
=
1
m
∑
1
m
u
i
2
+
v
i
2
.
{\displaystyle \int _{L}ds=\int _{\alpha }^{\beta }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta =n\sum _{1}^{m}{\sqrt {u_{i}^{2}+v_{i}^{2}}}={\frac {1}{m}}\sum _{1}^{m}{\sqrt {u_{i}^{2}+v_{i}^{2}}}.}
l
=
∫
a
b
1
+
(
y
′
)
2
d
x
,
{\displaystyle l=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx,}
x arba y kistu pastoviu greičiu (didėtų linijiniai). Taigi, x didėja linijiniai, o y lyginamas su x greičio kitimu ir y didėja arba mažėja ne linijiniai, kai x didėja linijiniai. Taigi,
v
=
ρ
′
.
{\displaystyle v=\rho '.}
v
i
=
R
i
−
R
i
−
1
2
π
n
(
i
+
1
)
−
2
π
n
i
=
R
i
−
R
i
−
1
2
π
n
.
{\displaystyle v_{i}={\frac {R_{i}-R_{i-1}}{2\pi n(i+1)-2\pi ni}}={\frac {R_{i}-R_{i-1}}{2\pi n}}.}
R
i
−
R
i
−
1
=
d
ρ
{\displaystyle R_{i}-R_{i-1}=d\rho }
2
π
n
=
d
θ
.
{\displaystyle 2\pi n=d\theta .}
v
=
d
ρ
d
θ
=
ρ
′
.
{\displaystyle v={\frac {d\rho }{d\theta }}=\rho '.}
θ
{\displaystyle \theta }
v nelinijiniai, bet pagal tam tikrą funkciją. Baigdami išspręsime pavyzdį, kai
ρ
=
θ
2
,
{\displaystyle \rho =\theta ^{2},}
l
=
∫
L
d
s
=
∫
0
2
π
ρ
2
+
(
ρ
′
)
2
d
θ
=
∫
0
2
π
(
θ
2
)
2
+
(
2
θ
)
2
d
θ
=
∫
0
2
π
θ
4
+
4
θ
2
d
θ
=
{\displaystyle l=\int _{L}ds=\int _{0}^{2\pi }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {(\theta ^{2})^{2}+(2\theta )^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {\theta ^{4}+4\theta ^{2}}}d\theta =}
=
∫
0
2
π
θ
θ
2
+
4
d
θ
=
1
3
(
θ
2
+
4
)
3
2
|
0
2
π
=
1
3
(
(
2
π
)
2
+
4
)
3
2
−
1
3
(
0
2
+
4
)
3
2
=
{\displaystyle =\int _{0}^{2\pi }\theta {\sqrt {\theta ^{2}+4}}\;d\theta ={\frac {1}{3}}\left(\theta ^{2}+4\right)^{3 \over 2}|_{0}^{2\pi }={\frac {1}{3}}\left((2\pi )^{2}+4\right)^{3 \over 2}-{\frac {1}{3}}\left(0^{2}+4\right)^{3 \over 2}=}
=
1
3
(
4
(
π
2
+
1
)
)
3
2
−
8
3
=
(
286.6887126
−
8
)
/
3
=
92.896237521771263212813630524448.
{\displaystyle ={\frac {1}{3}}(4(\pi ^{2}+1))^{3 \over 2}-{\frac {8}{3}}=(286.6887126-8)/3=92.896237521771263212813630524448.}
l
=
∫
L
d
s
=
∫
0
2
π
ρ
2
+
(
ρ
′
)
2
d
θ
=
∫
0
2
π
(
θ
2
)
2
+
(
2
θ
)
2
d
θ
=
∫
0
2
π
θ
4
+
4
θ
2
d
θ
=
{\displaystyle l=\int _{L}ds=\int _{0}^{2\pi }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {(\theta ^{2})^{2}+(2\theta )^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {\theta ^{4}+4\theta ^{2}}}d\theta =}
=
∫
0
2
π
θ
θ
2
+
4
d
θ
=
1
2
∫
0
2
π
θ
2
+
4
d
(
θ
2
)
=
1
2
(
θ
2
θ
2
+
4
+
4
2
ln
|
θ
+
θ
2
+
4
|
)
|
0
2
π
=
{\displaystyle =\int _{0}^{2\pi }\theta {\sqrt {\theta ^{2}+4}}\;d\theta ={\frac {1}{2}}\int _{0}^{2\pi }{\sqrt {\theta ^{2}+4}}\;d(\theta ^{2})={\frac {1}{2}}\left({\frac {\theta }{2}}{\sqrt {\theta ^{2}+4}}+{\frac {4}{2}}\ln |\theta +{\sqrt {\theta ^{2}+4}}|\right)|_{0}^{2\pi }=}
=
1
2
(
2
π
2
(
2
π
)
2
+
4
+
4
2
ln
|
2
π
+
(
2
π
)
2
+
4
|
)
−
1
2
(
0
2
0
2
+
4
+
4
2
ln
|
0
+
0
2
+
4
|
)
=
{\displaystyle ={\frac {1}{2}}\left({\frac {2\pi }{2}}{\sqrt {(2\pi )^{2}+4}}+{\frac {4}{2}}\ln |2\pi +{\sqrt {(2\pi )^{2}+4}}|\right)-{\frac {1}{2}}\left({\frac {0}{2}}{\sqrt {0^{2}+4}}+{\frac {4}{2}}\ln |0+{\sqrt {0^{2}+4}}|\right)=}
=
1
2
(
2
π
4
π
2
+
4
+
2
ln
|
2
π
+
4
π
2
+
4
|
)
−
1
2
(
2
ln
|
2
|
)
=
{\displaystyle ={\frac {1}{2}}\left(2\pi {\sqrt {4\pi ^{2}+4}}+2\ln |2\pi +{\sqrt {4\pi ^{2}+4}}|\right)-{\frac {1}{2}}\left(2\ln |2|\right)=}
=
1
2
(
4
π
π
2
+
1
+
2
ln
|
2
π
+
2
π
2
+
1
|
)
−
ln
|
2
|
=
{\displaystyle ={\frac {1}{2}}\left(4\pi {\sqrt {\pi ^{2}+1}}+2\ln |2\pi +2{\sqrt {\pi ^{2}+1}}|\right)-\ln |2|=}
=
2
π
π
2
+
1
+
ln
|
2
π
+
2
π
2
+
1
|
−
ln
|
2
|
=
{\displaystyle =2\pi {\sqrt {\pi ^{2}+1}}+\ln |2\pi +2{\sqrt {\pi ^{2}+1}}|-\ln |2|=}
=20,71508584921547418835543503203+2,5554429238707935293055934466408-0,69314718055994530941723212145818=22,577381592526322408243796357213.
čia
d
(
θ
2
)
=
2
θ
d
θ
,
d
θ
=
d
(
θ
2
)
2
θ
{\displaystyle d(\theta ^{2})=2\theta \;d\theta ,\;d\theta ={\frac {d(\theta ^{2})}{2\theta }}}
∫
x
x
2
±
a
2
d
x
=
1
3
(
x
2
±
a
2
)
3
2
{\displaystyle \int x{\sqrt {x^{2}\pm a^{2}}}dx={\frac {1}{3}}\left(x^{2}\pm a^{2}\right)^{3 \over 2}}
∫
x
2
±
a
2
d
x
=
x
2
x
2
±
a
2
+
a
2
2
ln
|
x
+
x
2
±
a
2
|
.
{\displaystyle \int {\sqrt {x^{2}\pm a^{2}}}dx={\frac {x}{2}}{\sqrt {x^{2}\pm a^{2}}}+{\frac {a^{2}}{2}}\ln |x+{\sqrt {x^{2}\pm a^{2}}}|.}
Antru budu neteisingai išintegruota, nes
1
2
∫
0
2
π
θ
2
+
4
d
(
θ
2
)
{\displaystyle {\frac {1}{2}}\int _{0}^{2\pi }{\sqrt {\theta ^{2}+4}}\;d(\theta ^{2})}
∫
x
2
±
a
2
d
x
=
x
2
x
2
±
a
2
+
a
2
2
ln
|
x
+
x
2
±
a
2
|
,
{\displaystyle \int {\sqrt {x^{2}\pm a^{2}}}dx={\frac {x}{2}}{\sqrt {x^{2}\pm a^{2}}}+{\frac {a^{2}}{2}}\ln |x+{\sqrt {x^{2}\pm a^{2}}}|,}
o kaip
∫
x
+
a
d
x
=
∫
x
+
a
d
(
x
+
a
)
=
(
x
+
a
)
3
2
3
2
,
{\displaystyle \int {\sqrt {x+a}}dx=\int {\sqrt {x+a}}d(x+a)={\frac {(x+a)^{3 \over 2}}{3 \over 2}},}
Kai kreivė L polinėje koordinačių sistemoje apibrėžta lygtimi
ρ
=
ρ
(
ϕ
)
,
{\displaystyle \rho =\rho (\phi ),}
ϕ
∈
[
α
;
β
]
{\displaystyle \phi \in [\alpha ;\beta ]}
d
s
=
ρ
2
+
ρ
′
2
d
ϕ
{\displaystyle ds={\sqrt {\rho ^{2}+\rho '^{2}}}d\phi }
∫
L
d
s
=
∫
α
β
ρ
2
+
(
ρ
′
)
2
d
ϕ
.
{\displaystyle \int _{L}ds=\int _{\alpha }^{\beta }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\phi .}
Polinėse koordinatėse taikoma ta pati pitagoro teorema. Geresniam išaiškiniui, tegu
ρ
=
θ
2
{\displaystyle \rho =\theta ^{2}}
θ
{\displaystyle \theta }
2
π
.
{\displaystyle 2\pi .}
x kinta pastoviu greičiu, o y kitimo greitis kinta pagal tam tikrą funkciją, tada turime formulę
l
=
∫
a
b
1
+
(
y
′
)
2
d
x
.
{\displaystyle l=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx.}
y
′
=
d
y
d
x
{\displaystyle y'={\frac {dy}{dx}}}
ρ
′
=
d
ρ
d
θ
.
{\displaystyle \rho '={\frac {d\rho }{d\theta }}.}
x pastovų pakitimą su y nepastoviu pakitimu (y') pakeltu kvadratu ir paskui traukiame šaknį. Polinėse koordinatėse apskritimo ilgis
c
=
2
π
R
=
2
π
ρ
.
{\displaystyle c=2\pi R=2\pi \rho .}
2
π
(
R
i
−
R
i
−
1
)
=
2
π
d
ρ
.
{\displaystyle 2\pi (R_{i}-R_{i-1})=2\pi \;{\text{d}}\rho .}
2
π
(
R
i
−
R
i
−
1
)
{\displaystyle 2\pi (R_{i}-R_{i-1})}
(
R
i
−
R
i
−
1
)
,
{\displaystyle (R_{i}-R_{i-1}),}
(
2
π
(
R
i
−
R
i
−
1
)
)
2
{\displaystyle (2\pi (R_{i}-R_{i-1}))^{2}}
(
R
i
−
R
i
−
1
)
2
{\displaystyle (R_{i}-R_{i-1})^{2}}
2
π
{\displaystyle 2\pi }
u
=
2
π
(
R
i
−
R
i
−
1
)
{\displaystyle u=2\pi (R_{i}-R_{i-1})}
R
=
ρ
.
{\displaystyle R=\rho .}
∫
α
β
ρ
2
+
(
ρ
′
)
2
d
θ
{\displaystyle \int _{\alpha }^{\beta }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta }
ρ
,
{\displaystyle \rho ,}
θ
,
{\displaystyle \theta ,}
ρ
{\displaystyle \rho }
θ
2
,
{\displaystyle \theta ^{2},}
θ
{\displaystyle \theta }
2
π
{\displaystyle 2\pi }
u
i
{\displaystyle u_{i}}
u
i
2
=
(
(
2
π
n
i
)
2
−
(
2
π
n
(
i
−
1
)
)
2
)
2
,
u
i
=
θ
i
2
−
θ
i
−
1
2
=
(
2
π
n
i
)
2
−
(
2
π
n
(
i
−
1
)
)
2
.
{\displaystyle u_{i}^{2}=((2\pi ni)^{2}-(2\pi n(i-1))^{2})^{2},\;u_{i}=\theta _{i}^{2}-\theta _{i-1}^{2}=(2\pi ni)^{2}-(2\pi n(i-1))^{2}.}
n (pavyzdžiui,
n
=
0.01
{\displaystyle n=0.01}
u
,
{\displaystyle u,}
∫
L
d
s
=
∫
α
β
ρ
2
+
(
ρ
′
)
2
d
θ
=
n
∑
1
m
u
i
2
+
v
i
2
=
1
m
∑
1
m
u
i
2
+
v
i
2
.
{\displaystyle \int _{L}ds=\int _{\alpha }^{\beta }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta =n\sum _{1}^{m}{\sqrt {u_{i}^{2}+v_{i}^{2}}}={\frac {1}{m}}\sum _{1}^{m}{\sqrt {u_{i}^{2}+v_{i}^{2}}}.}
l
=
∫
a
b
1
+
(
y
′
)
2
d
x
,
{\displaystyle l=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx,}
x arba y kistu pastoviu greičiu (didėtų linijiniai). Taigi, x didėja linijiniai, o y lyginamas su x greičio kitimu ir y didėja arba mažėja ne linijiniai, kai x didėja linijiniai. Taigi,
v
=
ρ
′
.
{\displaystyle v=\rho '.}
v
i
=
R
i
−
R
i
−
1
2
π
n
i
−
2
π
n
(
i
−
1
)
=
R
i
−
R
i
−
1
2
π
n
=
θ
i
2
−
θ
i
−
1
2
2
π
n
=
(
2
π
n
i
)
2
−
(
2
π
n
(
i
−
1
)
)
2
2
π
n
.
{\displaystyle v_{i}={\frac {R_{i}-R_{i-1}}{2\pi ni-2\pi n(i-1)}}={\frac {R_{i}-R_{i-1}}{2\pi n}}={\frac {\theta _{i}^{2}-\theta _{i-1}^{2}}{2\pi n}}={\frac {(2\pi ni)^{2}-(2\pi n(i-1))^{2}}{2\pi n}}.}
R
i
−
R
i
−
1
=
d
ρ
{\displaystyle R_{i}-R_{i-1}=d\rho }
2
π
n
=
d
θ
.
{\displaystyle 2\pi n=d\theta .}
v
=
d
ρ
d
θ
=
ρ
′
.
{\displaystyle v={\frac {d\rho }{d\theta }}=\rho '.}
θ
{\displaystyle \theta }
v nelinijiniai, bet pagal tam tikrą funkciją. Kad viskas būtų palyginta per
θ
{\displaystyle \theta }
i (i naturalus skaičius), mes turime, kad po kiekvieno i padidėjimo apėjimas vyksta greičiu
2
π
n
i
.
{\displaystyle 2\pi ni.}
(
R
i
−
R
i
−
1
)
2
{\displaystyle (R_{i}-R_{i-1})^{2}}
(
2
π
(
R
i
−
R
i
−
1
)
)
2
{\displaystyle (2\pi (R_{i}-R_{i-1}))^{2}}
R
i
−
R
i
−
1
{\displaystyle R_{i}-R_{i-1}}
2
π
n
{\displaystyle 2\pi n}
R
i
−
R
i
−
1
2
π
n
=
R
i
−
R
i
−
1
2
π
1
m
=
v
i
=
ρ
′
.
{\displaystyle {\frac {R_{i}-R_{i-1}}{2\pi n}}={\frac {R_{i}-R_{i-1}}{2\pi {\frac {1}{m}}}}=v_{i}=\rho '.}
l
=
1
m
∑
1
m
(
2
π
(
R
i
−
R
i
−
1
)
)
2
+
(
R
i
−
R
i
−
1
2
π
⋅
1
m
)
2
=
1
m
∑
1
m
u
i
2
+
v
i
2
.
{\displaystyle l={\frac {1}{m}}\sum _{1}^{m}{\sqrt {(2\pi (R_{i}-R_{i-1}))^{2}+\left({\frac {R_{i}-R_{i-1}}{2\pi \cdot {\frac {1}{m}}}}\right)^{2}}}={\frac {1}{m}}\sum _{1}^{m}{\sqrt {u_{i}^{2}+v_{i}^{2}}}.}
R
i
−
R
i
−
1
{\displaystyle R_{i}-R_{i-1}}
2
π
/
m
{\displaystyle 2\pi /m}
C
=
2
π
{\displaystyle C=2\pi }
R
{\displaystyle R}
Baigdami išspręsime pavyzdį, kai
ρ
=
θ
2
,
{\displaystyle \rho =\theta ^{2},}
l
=
∫
L
d
s
=
∫
0
2
π
ρ
2
+
(
ρ
′
)
2
d
θ
=
∫
0
2
π
(
θ
2
)
2
+
(
2
θ
)
2
d
θ
=
∫
0
2
π
θ
4
+
4
θ
2
d
θ
=
{\displaystyle l=\int _{L}ds=\int _{0}^{2\pi }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {(\theta ^{2})^{2}+(2\theta )^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {\theta ^{4}+4\theta ^{2}}}d\theta =}
=
∫
0
2
π
θ
θ
2
+
4
d
θ
=
1
3
(
θ
2
+
4
)
3
2
|
0
2
π
=
1
3
(
(
2
π
)
2
+
4
)
3
2
−
1
3
(
0
2
+
4
)
3
2
=
{\displaystyle =\int _{0}^{2\pi }\theta {\sqrt {\theta ^{2}+4}}\;d\theta ={\frac {1}{3}}\left(\theta ^{2}+4\right)^{3 \over 2}|_{0}^{2\pi }={\frac {1}{3}}\left((2\pi )^{2}+4\right)^{3 \over 2}-{\frac {1}{3}}\left(0^{2}+4\right)^{3 \over 2}=}
=
1
3
(
4
(
π
2
+
1
)
)
3
2
−
8
3
=
(
286.6887126
−
8
)
/
3
=
92.896237521771263212813630524448
;
{\displaystyle ={\frac {1}{3}}(4(\pi ^{2}+1))^{3 \over 2}-{\frac {8}{3}}=(286.6887126-8)/3=92.896237521771263212813630524448;}
l
=
∫
L
d
s
=
∫
0
2
π
ρ
2
+
(
ρ
′
)
2
d
θ
=
∫
0
2
π
(
θ
2
)
2
+
(
2
θ
)
2
d
θ
=
∫
0
2
π
θ
4
+
4
θ
2
d
θ
=
{\displaystyle l=\int _{L}ds=\int _{0}^{2\pi }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {(\theta ^{2})^{2}+(2\theta )^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {\theta ^{4}+4\theta ^{2}}}d\theta =}
=
∫
0
2
π
θ
θ
2
+
4
d
θ
=
1
2
∫
0
2
π
θ
2
+
4
d
(
θ
2
+
4
)
=
1
2
⋅
(
θ
2
+
4
)
3
/
2
3
2
|
0
2
π
=
(
θ
2
+
4
)
3
/
2
3
|
0
2
π
=
92.89623752
;
{\displaystyle =\int _{0}^{2\pi }\theta {\sqrt {\theta ^{2}+4}}\;d\theta ={\frac {1}{2}}\int _{0}^{2\pi }{\sqrt {\theta ^{2}+4}}\;d(\theta ^{2}+4)={\frac {1}{2}}\cdot {\frac {(\theta ^{2}+4)^{3/2}}{\frac {3}{2}}}|_{0}^{2\pi }={\frac {(\theta ^{2}+4)^{3/2}}{3}}|_{0}^{2\pi }=92.89623752;}
l
=
∫
L
d
s
=
∫
0
2
π
ρ
2
+
(
ρ
′
)
2
d
θ
=
∫
0
2
π
(
θ
2
)
2
+
(
2
θ
)
2
d
θ
=
∫
0
2
π
θ
4
+
4
θ
2
d
θ
=
{\displaystyle l=\int _{L}ds=\int _{0}^{2\pi }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {(\theta ^{2})^{2}+(2\theta )^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {\theta ^{4}+4\theta ^{2}}}d\theta =}
=
∫
0
2
π
θ
θ
2
+
4
d
θ
=
1
2
∫
0
2
π
θ
2
+
4
d
(
θ
2
)
=
1
2
(
2
⋅
4
3
+
2
θ
2
3
)
θ
2
+
4
|
0
2
π
=
{\displaystyle =\int _{0}^{2\pi }\theta {\sqrt {\theta ^{2}+4}}\;d\theta ={\frac {1}{2}}\int _{0}^{2\pi }{\sqrt {\theta ^{2}+4}}\;d(\theta ^{2})={\frac {1}{2}}\left({\frac {2\cdot 4}{3}}+{\frac {2\theta ^{2}}{3}}\right){\sqrt {\theta ^{2}+4}}|_{0}^{2\pi }=}
=
1
2
(
8
3
+
2
⋅
(
2
π
)
2
3
)
(
2
π
)
2
+
4
−
1
2
(
8
3
+
2
⋅
0
2
3
)
0
2
+
4
=
(
4
3
+
4
π
2
3
)
4
π
2
+
4
−
8
3
=
{\displaystyle ={\frac {1}{2}}\left({\frac {8}{3}}+{\frac {2\cdot (2\pi )^{2}}{3}}\right){\sqrt {(2\pi )^{2}+4}}-{\frac {1}{2}}\left({\frac {8}{3}}+{\frac {2\cdot 0^{2}}{3}}\right){\sqrt {0^{2}+4}}=\left({\frac {4}{3}}+{\frac {4\pi ^{2}}{3}}\right){\sqrt {4\pi ^{2}+4}}-{\frac {8}{3}}=}
=
8
+
8
π
2
3
π
2
+
1
−
8
3
=
8
(
π
2
+
1
)
3
π
2
+
1
−
8
3
=
8
(
π
2
+
1
)
3
2
3
−
8
3
=
{\displaystyle ={\frac {8+8\pi ^{2}}{3}}{\sqrt {\pi ^{2}+1}}-{\frac {8}{3}}={\frac {8(\pi ^{2}+1)}{3}}{\sqrt {\pi ^{2}+1}}-{\frac {8}{3}}={\frac {8(\pi ^{2}+1)^{3 \over 2}}{3}}-{\frac {8}{3}}=}
=95,562904188437929879480297191115-2,6(6)=92,896237521771263212813630524448;
čia
d
(
θ
2
+
4
)
=
2
θ
d
θ
,
d
θ
=
d
(
θ
2
+
4
)
2
θ
{\displaystyle d(\theta ^{2}+4)=2\theta \;d\theta ,\;d\theta ={\frac {d(\theta ^{2}+4)}{2\theta }}}
∫
x
x
2
±
a
2
d
x
=
1
3
(
x
2
±
a
2
)
3
2
{\displaystyle \int x{\sqrt {x^{2}\pm a^{2}}}dx={\frac {1}{3}}\left(x^{2}\pm a^{2}\right)^{3 \over 2}}
∫
a
x
+
b
d
x
=
(
2
b
3
a
+
2
x
3
)
a
x
+
b
.
{\displaystyle \int {\sqrt {ax+b}}dx=\left({\frac {2b}{3a}}+{\frac {2x}{3}}\right){\sqrt {ax+b}}.}
Išspręsime šį pavyzdį kitaip:
l
=
∫
L
d
s
=
∫
0
2
π
ρ
2
+
(
ρ
2
π
)
2
d
θ
=
∫
0
2
π
ρ
2
+
ρ
2
4
π
2
d
θ
=
1
2
π
∫
0
2
π
4
π
2
ρ
2
+
ρ
2
d
θ
=
{\displaystyle l=\int _{L}ds=\int _{0}^{2\pi }{\sqrt {\rho ^{2}+({\frac {\rho }{2\pi }})^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {\rho ^{2}+{\frac {\rho ^{2}}{4\pi ^{2}}}}}d\theta ={\frac {1}{2\pi }}\int _{0}^{2\pi }{\sqrt {4\pi ^{2}\rho ^{2}+\rho ^{2}}}d\theta =}
=
1
2
π
∫
0
2
π
ρ
4
π
2
+
1
d
θ
=
1
2
π
4
π
2
+
1
ρ
2
2
|
0
2
π
=
1
2
1
+
1
4
π
2
ρ
2
|
0
2
π
=
4
π
2
2
1
+
1
4
π
2
=
{\displaystyle ={\frac {1}{2\pi }}\int _{0}^{2\pi }\rho {\sqrt {4\pi ^{2}+1}}d\theta ={\frac {1}{2\pi }}{\sqrt {4\pi ^{2}+1}}{\frac {\rho ^{2}}{2}}|_{0}^{2\pi }={\frac {1}{2}}{\sqrt {1+{\frac {1}{4\pi ^{2}}}}}\rho ^{2}|_{0}^{2\pi }={\frac {4\pi ^{2}}{2}}{\sqrt {1+{\frac {1}{4\pi ^{2}}}}}=}
=
2
π
2
4
π
2
+
1
=
π
4
π
2
+
1
=
19.9876454.
{\displaystyle ={\frac {2\pi }{2}}{\sqrt {4\pi ^{2}+1}}=\pi {\sqrt {4\pi ^{2}+1}}=19.9876454.}
Kreivės lanko ilgis randamas pagal šitas formules:
d
s
=
1
+
y
′
2
d
x
,
{\displaystyle ds={\sqrt {1+y'^{2}}}dx,}
L apibrėžta lygtimi y=y(x), o
a
≤
x
≤
b
.
{\displaystyle a\leq x\leq b.}
∫
L
d
s
=
∫
a
b
1
+
(
f
′
(
x
)
)
2
d
x
=
∫
a
b
1
+
(
y
′
)
2
d
x
.
{\displaystyle \int _{L}ds=\int _{a}^{b}{\sqrt {1+(f'(x))^{2}}}dx=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx.}
Kai kreivė L apibrėžta parametrinėmis lygtimis
x
=
x
(
t
)
,
{\displaystyle x=x(t),}
y
=
y
(
t
)
,
{\displaystyle y=y(t),}
t
∈
[
t
0
;
T
]
,
{\displaystyle t\in [t_{0};T],}
d
s
=
x
t
′
2
+
y
t
′
2
d
t
,
{\displaystyle ds={\sqrt {x_{t}'^{2}+y_{t}'^{2}}}dt,}
∫
L
d
s
=
∫
t
0
T
(
x
t
′
)
2
+
(
y
t
′
)
2
d
t
.
{\displaystyle \int _{L}ds=\int _{t_{0}}^{T}{\sqrt {(x_{t}')^{2}+(y_{t}')^{2}}}dt.}
Elementariausias pavyzdis, kai reikia perrašyti funkcija
y
=
x
2
{\displaystyle y=x^{2}}
x
=
t
;
y
=
t
2
{\displaystyle x=t;\;y=t^{2}}
x
t
′
=
t
′
=
1
;
y
t
′
=
(
t
2
)
′
=
2
t
{\displaystyle x_{t}'=t'=1;\;y_{t}'=(t^{2})'=2t}
∫
t
0
T
(
x
t
′
)
2
+
(
y
t
′
)
2
d
t
=
∫
t
0
T
1
2
+
(
2
t
)
2
d
t
=
∫
t
0
T
1
+
4
⋅
t
2
d
t
.
{\displaystyle \int _{t_{0}}^{T}{\sqrt {(x_{t}')^{2}+(y_{t}')^{2}}}dt=\int _{t_{0}}^{T}{\sqrt {1^{2}+(2t)^{2}}}dt=\int _{t_{0}}^{T}{\sqrt {1+4\cdot t^{2}}}dt.}
Kai prametrinėmis lygtimis
x
=
x
(
t
)
,
{\displaystyle x=x(t),}
y
=
y
(
t
)
,
{\displaystyle y=y(t),}
z
=
z
(
t
)
,
{\displaystyle z=z(t),}
t
∈
[
t
0
;
T
]
{\displaystyle t\in [t_{0};T]}
L , tai
∫
L
d
s
=
∫
t
0
T
(
x
t
′
)
2
+
(
y
t
′
)
2
+
(
z
t
′
)
2
d
t
.
{\displaystyle \int _{L}ds=\int _{t_{0}}^{T}{\sqrt {(x_{t}')^{2}+(y_{t}')^{2}+(z_{t}')^{2}}}dt.}
Kai kreivė L polinėje koordinačių sistemoje apibrėžta lygtimi
ρ
=
ρ
(
ϕ
)
,
{\displaystyle \rho =\rho (\phi ),}
ϕ
∈
[
α
;
β
]
{\displaystyle \phi \in [\alpha ;\beta ]}
d
s
=
ρ
2
+
ρ
′
2
d
ϕ
{\displaystyle ds={\sqrt {\rho ^{2}+\rho '^{2}}}d\phi }
∫
L
d
s
=
∫
α
β
ρ
2
+
(
ρ
′
)
2
d
ϕ
.
{\displaystyle \int _{L}ds=\int _{\alpha }^{\beta }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\phi .}
Polinėse koordinatėse taikoma ta pati pitagoro teorema. Geresniam išaiškiniui, tegu
ρ
=
θ
2
{\displaystyle \rho =\theta ^{2}}
θ
{\displaystyle \theta }
2
π
.
{\displaystyle 2\pi .}
x kinta pastoviu greičiu, o y kitimo greitis kinta pagal tam tikrą funkciją, tada turime formulę
l
=
∫
a
b
1
+
(
y
′
)
2
d
x
.
{\displaystyle l=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx.}
y
′
=
d
y
d
x
{\displaystyle y'={\frac {dy}{dx}}}
ρ
′
=
d
ρ
d
θ
.
{\displaystyle \rho '={\frac {d\rho }{d\theta }}.}
x pastovų pakitimą su y nepastoviu pakitimu (y') pakeltu kvadratu ir paskui traukiame šaknį. Polinėse koordinatėse apskritimo ilgis
c
=
2
π
R
=
2
π
ρ
.
{\displaystyle c=2\pi R=2\pi \rho .}
2
π
(
R
i
−
R
i
−
1
)
=
2
π
d
ρ
.
{\displaystyle 2\pi (R_{i}-R_{i-1})=2\pi \;{\text{d}}\rho .}
2
π
(
R
i
−
R
i
−
1
)
{\displaystyle 2\pi (R_{i}-R_{i-1})}
(
R
i
−
R
i
−
1
)
,
{\displaystyle (R_{i}-R_{i-1}),}
(
2
π
(
R
i
−
R
i
−
1
)
)
2
{\displaystyle (2\pi (R_{i}-R_{i-1}))^{2}}
(
R
i
−
R
i
−
1
)
2
{\displaystyle (R_{i}-R_{i-1})^{2}}
2
π
{\displaystyle 2\pi }
u
=
2
π
(
R
i
−
R
i
−
1
)
{\displaystyle u=2\pi (R_{i}-R_{i-1})}
R
=
ρ
.
{\displaystyle R=\rho .}
∫
α
β
ρ
2
+
(
ρ
′
)
2
d
θ
{\displaystyle \int _{\alpha }^{\beta }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta }
ρ
,
{\displaystyle \rho ,}
θ
,
{\displaystyle \theta ,}
ρ
{\displaystyle \rho }
θ
2
,
{\displaystyle \theta ^{2},}
θ
{\displaystyle \theta }
2
π
{\displaystyle 2\pi }
u
i
{\displaystyle u_{i}}
u
i
2
=
(
(
2
π
n
i
)
2
−
(
2
π
n
(
i
−
1
)
)
2
)
2
,
u
i
=
θ
i
2
−
θ
i
−
1
2
=
(
2
π
n
i
)
2
−
(
2
π
n
(
i
−
1
)
)
2
.
{\displaystyle u_{i}^{2}=((2\pi ni)^{2}-(2\pi n(i-1))^{2})^{2},\;u_{i}=\theta _{i}^{2}-\theta _{i-1}^{2}=(2\pi ni)^{2}-(2\pi n(i-1))^{2}.}
n (pavyzdžiui,
n
=
0.01
{\displaystyle n=0.01}
u
,
{\displaystyle u,}
∫
L
d
s
=
∫
α
β
ρ
2
+
(
ρ
′
)
2
d
θ
=
n
∑
1
m
u
i
2
+
v
i
2
=
1
m
∑
1
m
u
i
2
+
v
i
2
.
{\displaystyle \int _{L}ds=\int _{\alpha }^{\beta }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta =n\sum _{1}^{m}{\sqrt {u_{i}^{2}+v_{i}^{2}}}={\frac {1}{m}}\sum _{1}^{m}{\sqrt {u_{i}^{2}+v_{i}^{2}}}.}
l
=
∫
a
b
1
+
(
y
′
)
2
d
x
,
{\displaystyle l=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx,}
x arba y kistu pastoviu greičiu (didėtų linijiniai). Taigi, x didėja linijiniai, o y lyginamas su x greičio kitimu ir y didėja arba mažėja ne linijiniai, kai x didėja linijiniai. Taigi,
v
=
ρ
′
.
{\displaystyle v=\rho '.}
v
i
=
R
i
−
R
i
−
1
2
π
n
i
−
2
π
n
(
i
−
1
)
=
R
i
−
R
i
−
1
2
π
n
=
θ
i
2
−
θ
i
−
1
2
2
π
n
=
(
2
π
n
i
)
2
−
(
2
π
n
(
i
−
1
)
)
2
2
π
n
.
{\displaystyle v_{i}={\frac {R_{i}-R_{i-1}}{2\pi ni-2\pi n(i-1)}}={\frac {R_{i}-R_{i-1}}{2\pi n}}={\frac {\theta _{i}^{2}-\theta _{i-1}^{2}}{2\pi n}}={\frac {(2\pi ni)^{2}-(2\pi n(i-1))^{2}}{2\pi n}}.}
R
i
−
R
i
−
1
=
d
ρ
{\displaystyle R_{i}-R_{i-1}=d\rho }
2
π
n
=
d
θ
.
{\displaystyle 2\pi n=d\theta .}
v
=
d
ρ
d
θ
=
ρ
′
.
{\displaystyle v={\frac {d\rho }{d\theta }}=\rho '.}
θ
{\displaystyle \theta }
v nelinijiniai, bet pagal tam tikrą funkciją. Kad viskas būtų palyginta per
θ
{\displaystyle \theta }
i (i naturalus skaičius), mes turime, kad po kiekvieno i padidėjimo apėjimas vyksta greičiu
2
π
n
i
.
{\displaystyle 2\pi ni.}
(
R
i
−
R
i
−
1
)
2
{\displaystyle (R_{i}-R_{i-1})^{2}}
(
2
π
(
R
i
−
R
i
−
1
)
)
2
{\displaystyle (2\pi (R_{i}-R_{i-1}))^{2}}
R
i
−
R
i
−
1
{\displaystyle R_{i}-R_{i-1}}
2
π
n
{\displaystyle 2\pi n}
R
i
−
R
i
−
1
2
π
n
=
R
i
−
R
i
−
1
2
π
1
m
=
v
i
=
ρ
′
.
{\displaystyle {\frac {R_{i}-R_{i-1}}{2\pi n}}={\frac {R_{i}-R_{i-1}}{2\pi {\frac {1}{m}}}}=v_{i}=\rho '.}
l
=
1
m
∑
1
m
(
2
π
(
R
i
−
R
i
−
1
)
)
2
+
(
R
i
−
R
i
−
1
2
π
⋅
1
m
)
2
=
1
m
∑
1
m
u
i
2
+
v
i
2
.
{\displaystyle l={\frac {1}{m}}\sum _{1}^{m}{\sqrt {(2\pi (R_{i}-R_{i-1}))^{2}+\left({\frac {R_{i}-R_{i-1}}{2\pi \cdot {\frac {1}{m}}}}\right)^{2}}}={\frac {1}{m}}\sum _{1}^{m}{\sqrt {u_{i}^{2}+v_{i}^{2}}}.}
R
i
−
R
i
−
1
{\displaystyle R_{i}-R_{i-1}}
2
π
/
m
{\displaystyle 2\pi /m}
C
=
2
π
{\displaystyle C=2\pi }
R
{\displaystyle R}
Baigdami išspręsime pavyzdį, kai
ρ
=
θ
2
,
{\displaystyle \rho =\theta ^{2},}
l
=
∫
L
d
s
=
∫
0
2
π
ρ
2
+
(
ρ
′
)
2
d
θ
=
∫
0
2
π
(
θ
2
)
2
+
(
2
θ
)
2
d
θ
=
∫
0
2
π
θ
4
+
4
θ
2
d
θ
=
{\displaystyle l=\int _{L}ds=\int _{0}^{2\pi }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {(\theta ^{2})^{2}+(2\theta )^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {\theta ^{4}+4\theta ^{2}}}d\theta =}
=
∫
0
2
π
θ
θ
2
+
4
d
θ
=
1
3
(
θ
2
+
4
)
3
2
|
0
2
π
=
1
3
(
(
2
π
)
2
+
4
)
3
2
−
1
3
(
0
2
+
4
)
3
2
=
{\displaystyle =\int _{0}^{2\pi }\theta {\sqrt {\theta ^{2}+4}}\;d\theta ={\frac {1}{3}}\left(\theta ^{2}+4\right)^{3 \over 2}|_{0}^{2\pi }={\frac {1}{3}}\left((2\pi )^{2}+4\right)^{3 \over 2}-{\frac {1}{3}}\left(0^{2}+4\right)^{3 \over 2}=}
=
1
3
(
4
(
π
2
+
1
)
)
3
2
−
8
3
=
(
286.6887126
−
8
)
/
3
=
92.896237521771263212813630524448
;
{\displaystyle ={\frac {1}{3}}(4(\pi ^{2}+1))^{3 \over 2}-{\frac {8}{3}}=(286.6887126-8)/3=92.896237521771263212813630524448;}
l
=
∫
L
d
s
=
∫
0
2
π
ρ
2
+
(
ρ
′
)
2
d
θ
=
∫
0
2
π
(
θ
2
)
2
+
(
2
θ
)
2
d
θ
=
∫
0
2
π
θ
4
+
4
θ
2
d
θ
=
{\displaystyle l=\int _{L}ds=\int _{0}^{2\pi }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {(\theta ^{2})^{2}+(2\theta )^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {\theta ^{4}+4\theta ^{2}}}d\theta =}
=
∫
0
2
π
θ
θ
2
+
4
d
θ
=
1
2
∫
0
2
π
θ
2
+
4
d
(
θ
2
+
4
)
=
1
2
⋅
(
θ
2
+
4
)
3
/
2
3
2
|
0
2
π
=
(
θ
2
+
4
)
3
/
2
3
|
0
2
π
=
92.89623752
;
{\displaystyle =\int _{0}^{2\pi }\theta {\sqrt {\theta ^{2}+4}}\;d\theta ={\frac {1}{2}}\int _{0}^{2\pi }{\sqrt {\theta ^{2}+4}}\;d(\theta ^{2}+4)={\frac {1}{2}}\cdot {\frac {(\theta ^{2}+4)^{3/2}}{\frac {3}{2}}}|_{0}^{2\pi }={\frac {(\theta ^{2}+4)^{3/2}}{3}}|_{0}^{2\pi }=92.89623752;}
l
=
∫
L
d
s
=
∫
0
2
π
ρ
2
+
(
ρ
′
)
2
d
θ
=
∫
0
2
π
(
θ
2
)
2
+
(
2
θ
)
2
d
θ
=
∫
0
2
π
θ
4
+
4
θ
2
d
θ
=
{\displaystyle l=\int _{L}ds=\int _{0}^{2\pi }{\sqrt {\rho ^{2}+(\rho ')^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {(\theta ^{2})^{2}+(2\theta )^{2}}}d\theta =\int _{0}^{2\pi }{\sqrt {\theta ^{4}+4\theta ^{2}}}d\theta =}
=
∫
0
2
π
θ
θ
2
+
4
d
θ
=
1
2
∫
0
2
π
θ
2
+
4
d
(
θ
2
)
=
1
2
(
2
⋅
4
3
+
2
θ
2
3
)
θ
2
+
4
|
0
2
π
=
{\displaystyle =\int _{0}^{2\pi }\theta {\sqrt {\theta ^{2}+4}}\;d\theta ={\frac {1}{2}}\int _{0}^{2\pi }{\sqrt {\theta ^{2}+4}}\;d(\theta ^{2})={\frac {1}{2}}\left({\frac {2\cdot 4}{3}}+{\frac {2\theta ^{2}}{3}}\right){\sqrt {\theta ^{2}+4}}|_{0}^{2\pi }=}
=
1
2
(
8
3
+
2
⋅
(
2
π
)
2
3
)
(
2
π
)
2
+
4
−
1
2
(
8
3
+
2
⋅
0
2
3
)
0
2
+
4
=
(
4
3
+
4
π
2
3
)
4
π
2
+
4
−
8
3
=
{\displaystyle ={\frac {1}{2}}\left({\frac {8}{3}}+{\frac {2\cdot (2\pi )^{2}}{3}}\right){\sqrt {(2\pi )^{2}+4}}-{\frac {1}{2}}\left({\frac {8}{3}}+{\frac {2\cdot 0^{2}}{3}}\right){\sqrt {0^{2}+4}}=\left({\frac {4}{3}}+{\frac {4\pi ^{2}}{3}}\right){\sqrt {4\pi ^{2}+4}}-{\frac {8}{3}}=}
=
8
+
8
π
2
3
π
2
+
1
−
8
3
=
8
(
π
2
+
1
)
3
π
2
+
1
−
8
3
=
8
(
π
2
+
1
)
3
2
3
−
8
3
=
{\displaystyle ={\frac {8+8\pi ^{2}}{3}}{\sqrt {\pi ^{2}+1}}-{\frac {8}{3}}={\frac {8(\pi ^{2}+1)}{3}}{\sqrt {\pi ^{2}+1}}-{\frac {8}{3}}={\frac {8(\pi ^{2}+1)^{3 \over 2}}{3}}-{\frac {8}{3}}=}
=95,562904188437929879480297191115-2,6(6)=92,896237521771263212813630524448;
čia
d
(
θ
2
+
4
)
=
2
θ
d
θ
,
d
θ
=
d
(
θ
2
+
4
)
2
θ
{\displaystyle d(\theta ^{2}+4)=2\theta \;d\theta ,\;d\theta ={\frac {d(\theta ^{2}+4)}{2\theta }}}
∫
x
x
2
±
a
2
d
x
=
1
3
(
x
2
±
a
2
)
3
2
{\displaystyle \int x{\sqrt {x^{2}\pm a^{2}}}dx={\frac {1}{3}}\left(x^{2}\pm a^{2}\right)^{3 \over 2}}
∫
a
x
+
b
d
x
=
(
2
b
3
a
+
2
x
3
)
a
x
+
b
.
{\displaystyle \int {\sqrt {ax+b}}dx=\left({\frac {2b}{3a}}+{\frac {2x}{3}}\right){\sqrt {ax+b}}.}
Bandydami apskaičiuoti analogiškai su programa "Free Pascal", panaudojant kodą: var
a:longint;
c:real;
begin
for a:=1 to 100 do
c:=c+sqrt(sqr(sqr(a*0.062831853)-sqr((a-1)*0.062831853))+sqr((sqr(a*0.062831853)-sqr((a-1)*0.062831853))/0.062831853));
writeln(c);
readln;
end.
gauname atsakymą 629,557559394517. Šis atsakymas padalintas iš
2
π
{\displaystyle 2\pi }
Kai panaudojame šį kodą: var
a:longint;
c:real;
begin
for a:=1 to 10000 do
c:=c+sqrt(sqr(sqr(a*0.00062831853)-sqr((a-1)*0.00062831853))+sqr((sqr(a*0.00062831853)-sqr((a-1)*0.00062831853))/0.00062831853));
writeln(c);
readln;
end.
tai gauname atsakymą 62831,8654025094, kas byloje apie bloga teorijos išaiškinimą. Va čia kodas: var
a:longint;
c:real;
begin
for a:=1 to 628318531 do
c:=c+0.00000001*sqrt(sqr(sqr(a*0.00000001))+sqr(a*2.0/100000000));
writeln(c);
readln;
end.
kuris duoda atsakymą 92,8962378457359 po 16 sekundžių su 2,6 GHz procesoriumi. Optimizuotas šio kodo variantas: var
a:longint;
c,b:real;
begin
for a:=1 to 628318531 do
c:=c+sqrt(sqr(sqr(a*0.00000001))+sqr(a*0.00000002));
b:=c*0.00000001;
writeln(b);
readln;
end.
duoda atsakymą 92,8962378457489 po 11 sekundžių su 2,6 GHz procesoriumi. Šis kodas: var
a:longint;
c,b:real;
begin
for a:=1 to 62831853 do
c:=c+sqrt(sqr(sqr(a*0.0000001))+sqr((sqr(a*0.0000001)-sqr((a-1)*0.0000001))/0.0000001));
b:=c*0.0000001;
writeln(b);
readln;
end.
duoda atsakymą 92,8962389233553 po dviejų sekundžių. Tikslesnė (ne daug tikslesnė, nes kaip tik
2
π
⋅
10
8
=
628318530.7179586476925286766559
{\displaystyle 2\pi \cdot 10^{8}=628318530.7179586476925286766559}
var
a:longint;
c,b:real;
begin
for a:=1 to 628318531 do
c:=c+sqrt(sqr(sqr(a*0.00000001))+sqr((sqr(a*0.00000001)-sqr((a-1)*0.00000001))*100000000));
b:=c*0.00000001;
writeln(b);
readln;
end.
duoda atsakymą 92,8962378085099 po 15 sekundžių su 2,6 GHz procesoriumi. Labiausiai teoriją atitinkantis kodas yra šis: var
a:longint;
c,b:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(sqr(a*0.0000000062831853))+sqr((sqr(a*0.0000000062831853)-sqr((a-1)*0.0000000062831853))/0.0000000062831853));
b:=c*0.0000000062831853;
writeln(b);
readln;
end.
duodantis atsakymą 92,8962373310520 po 30 sekundžių su 2,6 GHz procesoriumi. Su patikslinta
2
π
{\displaystyle 2\pi }
var
a:longint;
c,b:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt(sqr(sqr(a*0.000000006283185307179586))+sqr((sqr(a*0.000000006283185307179586)-sqr((a-1)*0.000000006283185307179586))/0.000000006283185307179586));
b:=c*0.000000006283185307179586;
writeln(b);
readln;
end
gauname atsakymą 92,8962376285006 po 31 sekundės su 2,6 GHz procesoriumi.
Suprastinta kodo su šita eilute c:=c+sqrt(sqr(sqr(a*0.00000001))+sqr((sqr(a*0.00000001)-sqr((a-1)*0.00000001))*100000000));
versija yra kodas: var a:longint; c,b:real;
begin
for a:=1 to 628318531 do
c:=c+sqrt(sqr(sqr(a*0.00000001))+sqr((sqr(a*1.0)-sqr((a-1)*1.0))*0.00000001));
b:=c*0.00000001;
writeln(b);
readln;
end.
kuris duoda atsakymą 9.2896237808507621E+001 po 9 sekundžių su 4.16 GHz dažniu veikiančiu procesoriumi. Jeigu vietoje sqr(a*1.0) parašyti sqr(a), tai gaunamas visai ne tokių skaitmenų atsakymas ir 8 eilėmis didesnis (8.2683404394761600E+009). O jeigu vietoje sqr((a-1)*1.0) parašyti sqr(a-1), tai iš vis kažkokia klaida ir negaunamas joks atsakymas. 
![{\displaystyle t\in [t_{0};T],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b874c1b132c7f3e6239aa445b37e2d8d961fd2c)
![{\displaystyle t\in [t_{0};T]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c296cc887890e455320a28f1796e1e443e9be296)
![{\displaystyle \phi \in [\alpha ;\beta ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2007849b5cdd53dbc01143bff5739409c7e1efd9)
