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I am trying to evaluate the following integral:

$$ \int_{-2}^2 \left(x^3 \cos \left(\frac{x}{2}\right)+\frac{1}{2}\right) \sqrt{4-x^2} \, dx. $$

The answer is pi based on online calculations. But I want to share this to my students that we can express this as pi. I found out that integrating the 2nd term from the left, based on its boundaries, is equivalent to pi. Meaning, I need to show that the integration of the first term from the left is zero based on its boundaries. I need to know how to manually solve the intragral solution to show that the whole expression is equivalent to pi:

Mark McClure
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PRD
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  • Numeric value for integral is: 1.7771715963131930221617993675 and it's not a Pi ? – Mariusz Iwaniuk Nov 14 '19 at 17:12
  • Clearly, you mean $\sqrt{4-x^2}$, rather than just $4-x$. In that case, you can distribute the $\sqrt{4-x^2}$ through. Then, the integral of the first portion is zero by symmetry and the integral of the second portion represents half the area of a semi-circle of radius 2, which gives you $\pi$. – Mark McClure Nov 14 '19 at 17:16
  • @Mariusz. Sorry, It should be $\sqrt{4-x^2}$ – PRD Nov 14 '19 at 17:23
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    Set $x := -x$ to see that the first term is an odd function. – Azlif Nov 14 '19 at 17:23
  • @Mark how can I show that the integral of the first portion is zero by symmetry through solution? – PRD Nov 14 '19 at 17:26
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    @MRA The function is odd. – Mark McClure Nov 14 '19 at 17:27
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    To everyone, thank you for your help. Just to share, I am planning to show the solution through graph first and connect the abstract concepts of the integral of an odd function (which is zero based on the provided boundaries) plus the area of a semicircle with radius 2 units = pi – PRD Nov 14 '19 at 17:50

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