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f(x)=(25-15sin(x)^29sin(x)^3)9/2−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

∫02π(25−15sin⁡(x)2+9sin⁡(x)3)92dx\int_0^{2\pi}\left(25-15\sin\left(x\right)^2+9\sin\left(x\right)^3\right)\frac{9}{2}dx

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Final answer to the exercise

225π+1358sin⁡(4π)+π⋅−1352+27+(−272⋅sin⁡(2π)2−27)cos⁡(2π)2225\pi +\frac{135}{8}\sin\left(4\pi \right)+\frac{\pi \cdot -135}{2}+27+\left(-\frac{27}{2}\cdot \sin\left(2\pi \right)^2-27\right)\cos\left(2\pi \right)^2

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∫02π(25−15sin(x)2+9sin(x)3)92 dx
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