Problem 23 - Calculus Early Transcendentals

Question

Evaluate the following integrals as they are written. $$\int_{-2}^{2} \int_{x^{2}}^{8-x^{2}} x d y d x$$

Step-by-Step Solution

Answer

In this problem, we have computed the value of the given double integral. The step-by-step solution was provided by first integrating with respect to y, followed by integrating with respect to x. The final result of the integral is 0.

Step 1: Integrate with respect to y

We are given the integral $$\int_{-2}^{2} \int_{x^{2}}^{8-x^{2}} x d y d x$$, so we start by integrating with respect to y: $$\int_{-2}^{2} [ \int_{x^{2}}^{8-x^{2}} x d y ] d x$$ The antiderivative of x with respect to y is xy: $$\int_{-2}^{2} [ xy\Big|_{x^2}^{8-x^2} ] d x$$

Step 2: Find values of the antiderivative

Evaluate the antiderivative using the given bounds, and subtract in the order \(8-x^2\) (the higher bound) and then \(x^2\) (the lower bound): $$\int_{-2}^{2} [ (x(8-x^2)) - (x(x^2)) ] d x$$ Simplify the expression inside the integral: $$\int_{-2}^{2} [ 8x-x^3 - x^3] dx$$ Combine like terms: $$\int_{-2}^{2} [ 8x-2x^3 ] dx$$

Step 3: Integrate with respect to x

Now we integrate the expression with respect to x: $$\int_{-2}^{2} [ 8x-2x^3 ] dx$$ The antiderivative of \(8x-2x^3\) with respect to x is \(4x^2-\frac{1}{2}x^4\): $$[4x^2 - \frac{1}{2}x^4 \Big|_{-2}^{2}]$$

Step 4: Evaluate the integral

Finally, evaluate the integral using the given bounds and subtract the values to find the result: $$\Big(4(2)^2 - \frac{1}{2}(2)^4\Big) - \Big(4(-2)^2 - \frac{1}{2}(-2)^4\Big)$$ Calculate the values: $$(16-16)-(16-16)=0$$ So, the result of the given integral is 0: $$\int_{-2}^{2} \int_{x^{2}}^{8-x^{2}} x d y d x = 0$$