Question

The value of integral $\int_{-2}^{2} (x^3 \cos(\frac{x}{2}) + \frac{1}{2}) \sqrt{4 - x^2} dx$ equals

• A. the area enclosed by $x^2 + y^2 = 1$
• B. $2 \pi$
• C. the area enclosed by $\frac{x^2}{4} + 4y^2 = 1$
• D. the area enclosed by $\frac{x^2}{4} + 16y^2 = 1$

Answer 1

Answer: (A), (C)

A, C

Answer 2

The given integral is $I = \int_{-2}^{2} (x^3 \cos(\frac{x}{2}) + \frac{1}{2}) \sqrt{4 - x^2} dx$.

We can split the integral into two parts: $I = \int_{-2}^{2} x^3 \cos(\frac{x}{2}) \sqrt{4 - x^2} dx + \int_{-2}^{2} \frac{1}{2} \sqrt{4 - x^2} dx$

Let's evaluate the first part, $I_1 = \int_{-2}^{2} x^3 \cos(\frac{x}{2}) \sqrt{4 - x^2} dx$. Let $f(x) = x^3 \cos(\frac{x}{2}) \sqrt{4 - x^2}$. We check if $f(x)$ is an odd or even function: $f(-x) = (-x)^3 \cos(\frac{-x}{2}) \sqrt{4 - (-x)^2}$ $f(-x) = -x^3 \cos(\frac{x}{2}) \sqrt{4 - x^2}$ $f(-x) = -f(x)$

Since $f(x)$ is an odd function and the limits of integration are symmetric about zero (from -2 to 2), the integral of $f(x)$ over this interval is zero. So, $I_1 = 0$.

Now, let's evaluate the second part, $I_2 = \int_{-2}^{2} \frac{1}{2} \sqrt{4 - x^2} dx$. $I_2 = \frac{1}{2} \int_{-2}^{2} \sqrt{4 - x^2} dx$

The integral $\int_{-2}^{2} \sqrt{4 - x^2} dx$ represents the area of the upper semicircle of radius $r=2$. The equation $y = \sqrt{4 - x^2}$ describes the upper half of the circle $x^2 + y^2 = 4$. The area of a full circle with radius $r$ is $\pi r^2$. The area of a semicircle with radius $r$ is $\frac{1}{2} \pi r^2$. For $r=2$, the area of the semicircle is $\frac{1}{2} \pi (2)^2 = \frac{1}{2} \pi (4) = 2\pi$. Therefore, $I_2 = \frac{1}{2} (2\pi) = \pi$.

Combining the two parts of the integral: $I = I_1 + I_2 = 0 + \pi = \pi$.

Now, we compare this value with the areas given in the options:

A. The area enclosed by $x^2 + y^2 = 1$. This is a circle with radius $r=1$. Area = $\pi r^2 = \pi (1)^2 = \pi$. This matches our calculated value.

B. $2 \pi$. This does not match our calculated value.

C. The area enclosed by $\frac{x^2}{4} + 4y^2 = 1$. This is an ellipse. The standard form of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Comparing, we have $a^2 = 4 \implies a = 2$. And $4y^2 = 1 \implies \frac{y^2}{1/4} = 1 \implies b^2 = 1/4 \implies b = 1/2$. The area of an ellipse is $\pi ab$. Area = $\pi (2)(\frac{1}{2}) = \pi$. This also matches our calculated value.

D. The area enclosed by $\frac{x^2}{4} + 16y^2 = 1$. This is an ellipse. $a^2 = 4 \implies a = 2$. $16y^2 = 1 \implies \frac{y^2}{1/16} = 1 \implies b^2 = 1/16 \implies b = 1/4$. Area = $\pi ab = \pi (2)(\frac{1}{4}) = \frac{\pi}{2}$. This does not match our calculated value.

Both options A and C yield the value $\pi$. Therefore, both are correct descriptions of the value of the integral.