Matematika/Apibrėžtinis integralas: Skirtumas tarp puslapio versijų

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:<math>V=8\pi a b^2\int_0^{\pi\over 2} \cos^2 (t) dt=8\pi\cdot 4\cdot 3^2\int_0^{\pi\over 2}\frac{\cos(2t)+1}{2} dt=16\pi\cdot 9\int_0^{\pi\over 2}(\cos(2t)+1) dt=144\pi[\int_0^{\pi\over 2}\cos(2t)dt +\int_0^{\pi\over 2} dt]=</math>
:<math>=144\pi[\int_0^{\pi\over 2}\cos(2t)\frac{d(2t)}{2} +{\pi\over 2}]=144\pi[\frac{1}{2}\cdot \sin(2t)|_0^{\pi\over 2} +{\pi\over 2}]=72\pi[ \sin(2t)|_0^{\pi\over 2} +\pi]=72\pi[ \sin(2\cdot\frac{\pi}{2})-\sin(2\cdot 0)+\pi]=72\pi^2=710.6115169.</math>
 
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