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

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∻Solution by Steps∻ ‖ step 1 ⋮ Evaluate the definite integral of the given function over the interval from -2 to 2. ‖ ‖ step 2 ⋮ The function to integrate is $\sqrt{4 - x^2} \left(\frac{1}{2} + x^3 \cos\left(\frac{x}{2}\right)\right)$. ‖ ‖ step 3 ⋮ The integral is symmetric around the y-axis, and the function $x^3 \cos\left(\frac{x}{2}\right)$ is an odd function, which will integrate to zero over the symmetric interval. ‖ ‖ step 4 ⋮ The remaining function to integrate is $\frac{1}{2} \sqrt{4 - x^2}$, which is an even function. ‖ ‖ step 5 ⋮ The integral of an even function over a symmetric interval can be computed as twice the integral from 0 to the upper limit. ‖ ‖ step 6 ⋮ Compute the integral $\int_{0}^{2} \frac{1}{2} \sqrt{4 - x^2} dx$. ‖ ‖ step 7 ⋮ Recognize that $\sqrt{4 - x^2}$ represents a semicircle with radius 2, and the integral computes the area of a semicircle. ‖ ‖ step 8 ⋮ The area of a semicircle with radius 2 is $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2)^2 = 2\pi$. ‖ ‖ step 9 ⋮ Multiply the area by $\frac{1}{2}$ to account for the coefficient in the integral. ‖ ‖ step 10 ⋮ The final result is $\frac{1}{2} \cdot 2\pi = \pi$. ‖ ∻Answer∻ ⚹ $\pi$ ⚹ ∻Key Concept∻ ⚹Symmetry in Integration⚹ ∻Explanation∻ ⚹The integral of an odd function over a symmetric interval is zero, and the integral of an even function can be simplified by considering only half of the interval and doubling the result. In this case, the odd part of the function integrates to zero, and the even part corresponds to the area of a semicircle.⚹◊What is the definite integral $\int_{-2}^{2}\left(x^{3} \cos \frac{x}{2}+\frac{1}{2}\right) \sqrt{4-x^{2}} dx$?⍭ Generate me a similar question◊

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