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

Sign in to unlock the answer

Answer from Sia

Posted 4 months ago

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.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Upgrade to Asksia Pro

Join a AskSia's Pro Plan, and get 24/7 AI tutoring for your reviews, assignments, quizzes and exam preps.

Unlimited chat query usages

Strong algorithms that better know you

Early access to new release features

Study Other Question