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Chapter 13: Problem 14

Evaluate the following iterated integrals. $$\int_{0}^{1} \int_{0}^{1} \frac{y}{1+x^{2}} d x d y$$

Short Answer

Expert verified
Question: Evaluate the iterated integral of the function \(\frac{y}{1+x^2}\) over the square region \([0, 1] \times [0, 1]\) with respect to \(x\) first and then with respect to \(y\). Answer: \(\frac{\pi}{8}\)

Step by step solution

01

Evaluate the inner integral with respect to x

First, we need to find the antiderivative of the function \(\frac{y}{1+x^2}\) with respect to x. $$\int_{0}^{1} \frac{y}{1+x^{2}} d x$$ Since y is a constant with respect to x in this case, the integral becomes: $$y \int_{0}^{1} \frac{1}{1+x^{2}} d x$$ The antiderivative of \(\frac{1}{1+x^2}\) with respect to x is \(\arctan{x}\) Evaluate the antiderivative at the limits, 1 and 0. $$y \left[\arctan(1) - \arctan(0)\right]$$ Result: $$y \left[\frac{\pi}{4} - 0\right]$$ This reduces to: $$\frac{\pi}{4}y$$
02

Evaluate the outer integral with respect to y

Now we take the result from Step 1 and compute the outer integral with respect to y: $$\int_{0}^{1} \frac{\pi}{4}y d y$$ Find the antiderivative of \(\frac{\pi}{4}y\) with respect to y. This is: $$\frac{\pi}{8}y^2$$ Evaluate the antiderivative at the limits, 1 and 0. $$\frac{\pi}{8}(1^2 - 0^2)$$ The final result is: $$\frac{\pi}{8}$$

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