/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 19

Evaluate the following integrals as they are written. $$\int_{0}^{2} \int_{x^{2}}^{2 x} x y d y d x$$

Short Answer

Expert verified
Question: Evaluate the double integral $$\int_{0}^{2} \int_{x^{2}}^{2 x} x y d y d x$$ as it is written, without changing the order of integration. Solution: The final answer, after evaluating the double integral, is \(\frac{32}{3}\).

Step by step solution

01

Understanding the given double integral

The given integral is written in the following form: $$\int_{0}^{2} \int_{x^{2}}^{2 x} x y d y d x$$ Here, we have an inner integral with respect to \(y\) and an outer integral with respect to \(x\). We will first evaluate the inner integral, which is over \(y\), then we will evaluate the outer integral, which is over \(x\).
02

Evaluating the inner integral

To evaluate the inner integral, we need to find the antiderivative of \(xy\) with respect to \(y\). The inner integral is: $$\int_{x^2}^{2x} xy dy$$ To find the antiderivative, we treat \(x\) as a constant since we are integrating with respect to \(y\). The antiderivative of \(xy\) with respect to \(y\) is: $$\frac{1}{2}xy^2$$ Now we need to find the difference between the antiderivative evaluated at the upper and lower limits of \(y\). Thus, we have: $$\frac{1}{2}x(2x)^2 - \frac{1}{2}x(x^2)^2 = \frac{1}{2}x(8x^2) - \frac{1}{2}x(x^4) = 4x^3 - \frac{1}{2}x^5$$ Now that we have evaluated the inner integral, we can move on to the outer integral.
03

Evaluating the outer integral

Now, we evaluate the outer integral with respect to \(x\) using the result from the inner integral: $$\int_{0}^{2} \left(4x^3 - \frac{1}{2}x^5\right)dx$$ To evaluate this integral, we need to find the antiderivative of \(\left(4x^3 - \frac{1}{2}x^5\right)\) with respect to \(x\). The antiderivative is: $$\int \left(4x^3 - \frac{1}{2}x^5\right)dx = x^4 - \frac{1}{12}x^6$$ Now we need to find the difference between the antiderivative evaluated at the upper and lower limits of \(x\). Thus, we have: $$\left[x^4 - \frac{1}{12}x^6\right]_0^2 = \left(2^4 - \frac{1}{12}(2^6)\right) - \left(0^4 - \frac{1}{12}(0^6)\right) = 16 - \frac{64}{12}$$
04

Simplifying and finding the final answer

Now, we just need to simplify our result to find the final answer: $$\int_{0}^{2} \int_{x^2}^{2x} xy dy dx = 16 - \frac{64}{12} = 16 - \frac{32}{6} = 16 - \frac{16}{3}=\frac{32}{3}$$ So, the final answer is \(\frac{32}{3}\).

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Integrals in strips Consider the integral $$I=\iint_{R} \frac{d A}{\left(1+x^{2}+y^{2}\right)^{2}}$$ where \(R=\\{(x, y): 0 \leq x \leq 1,0 \leq y \leq a\\}\) a. Evaluate \(I\) for \(a=1 .\) (Hint: Use polar coordinates.) b. Evaluate \(I\) for arbitrary \(a > 0\) c. Let \(a \rightarrow \infty\) in part (b) to find \(I\) over the infinite strip \(R=\\{(x, y): 0 \leq x \leq 1,0 \leq y < \infty\\}\)
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