Integrate. int_(-2)^(2)(x^(3)cos((x)/(2))+(1)/(2))sqrt(4-x^(2))dx

Given Integral: We are given the integral to evaluate: int_(-2)^(2)(x^(3)cos((x)/(2))+(1)/(2))sqrt(4-x^(2))dx This integral can be split into two separate integrals: int_(-2)^(2)x^(3)cos((x)/(2))dx + int_(-2)^(2)(1)/(2)sqrt(4-x^(2))dxFirst Integral: First, let's focus on the first integral: int_(-2)^(2)x^(3)cos((x)/(2))dx This integral does not have an elementary antiderivative, and it is an odd function over a symmetric interval around zero. Since the cosine function is even and x^3 is odd, their product is an odd function. The integral of an odd function over a symmetric interval around zero is zero. Therefore, int_(-2)^(2)x^(3)cos((x)/(2))dx = 0Second Integral: Now, let's focus on the second integral: int_(-2)^(2)(1)/(2)sqrt(4-x^(2))dx This integral represents the area of a semicircle with radius 2, because the integrand is the equation of a semicircle in the upper half-plane. The area of a semicircle is given by (1/2)πr^2, where r is the radius.Area Calculation: Since the radius r is 2, we can calculate the area of the semicircle: Area = (1/2)π(2)^2 Area = (1/2)π(4) Area = 2π Therefore, int_(-2)^(2)(1)/(2)sqrt(4-x^(2))dx = 2πFinal Result: Adding the results of the two integrals together, we get: 0 (from the first integral) + 2π (from the second integral) = 2π

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Integrate.
int_(-2)^(2)(x^(3)cos((x)/(2))+(1)/(2))sqrt(4-x^(2))dx

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

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Calculus

Evaluate definite integrals using the chain rule

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Q. Integrate.

\newline

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

Given Integral: We are given the integral to evaluate: $\int_{-2}^{2}(x^{3}\cos(\frac{x}{2})+\frac{1}{2})\sqrt{4-x^{2}}\,dx$ This integral can be split into two separate integrals: $\int_{-2}^{2}x^{3}\cos(\frac{x}{2})\,dx + \int_{-2}^{2}\frac{1}{2}\sqrt{4-x^{2}}\,dx$
First Integral: First, let's focus on the first integral: $\newline$ $\int_{-2}^{2}x^{3}\cos\left(\frac{x}{2}\right)dx$ $\newline$ This integral does not have an elementary antiderivative, and it is an odd function over a symmetric interval around zero. Since the cosine function is even and $x^3$ is odd, their product is an odd function. The integral of an odd function over a symmetric interval around zero is zero. $\newline$ Therefore, $\int_{-2}^{2}x^{3}\cos\left(\frac{x}{2}\right)dx = 0$
Second Integral: Now, let's focus on the second integral: $\newline$ $\int_{-2}^{2}\frac{1}{2}\sqrt{4-x^{2}}\,dx$ $\newline$ This integral represents the area of a semicircle with radius $2$ , because the integrand is the equation of a semicircle in the upper half-plane. $\newline$ The area of a semicircle is given by $(1/2)\pi r^{2}$ , where $r$ is the radius.
Area Calculation: Since the radius $r$ is $2$ , we can calculate the area of the semicircle: $\newline$ Area = $(1/2)\pi(2)^2$ $\newline$ Area = $(1/2)\pi(4)$ $\newline$ Area = $2\pi$ $\newline$ Therefore, $\int_{-2}^{2}\frac{1}{2}\sqrt{4-x^{2}}dx = 2\pi$
Final Result: Adding the results of the two integrals together, we get: $0$ (from the first integral) + $2\pi$ (from the second integral) = $2\pi$