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Chapter 13: Problem 31

When converted to an iterated integral, the following double integrals are easier to evaluate in one order than the other. Find the best order and evaluate the integral. $$\iint_{R} \frac{x}{(1+x y)^{2}} d A ; R=\\{(x, y): 0 \leq x \leq 4,1 \leq y \leq 2\\}$$

Short Answer

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Question: Evaluate the double integral $$\iint_{R} \frac{x}{(1+x y)^{2}} d A ; R=\\{(x, y): 0 \leq x \leq 4,1 \leq y \leq 2\\}$$ Answer: $$\iint_{R} \frac{x}{(1+x y)^{2}} d A = \ln \left(\frac{2(5^{\frac{1}{4}})}{3^{\frac{1}{2}}}\right)$$

Step by step solution

01

Setup the iterated integral with the given order of integration (dx dy)

We set up the integral by using the limits for x (0 to 4) and y (1 to 2) and choosing the order of integration dx dy: $$\int_{1}^{2} \int_{0}^{4} \frac{x}{(1+x y)^{2}} dx\, dy$$
02

Integrate the inner integral with respect to x

Now, integrate the inner integral with respect to x: $$\int_{1}^{2} \left[ -\frac{1}{y(1+xy)} \right]_{0}^{4} dy$$ We plug in the x limits to obtain: $$\int_{1}^{2} \left[ -\frac{1}{y(1+4y)} + \frac{1}{y} \right] dy$$
03

Integrate the remaining integral with respect to y

Now, integrate the remaining integral with respect to y: $$\left[ -\frac{1}{4}\ln |1+4y| + \ln |y| \right]_{1}^{2}$$
04

Evaluate the result at the y limits

Finally, plug in the y limits of integration (1 to 2) into the result: $$\left[ -\frac{1}{4}\ln |1+4(2)| + \ln |2| \right] - \left[ -\frac{1}{4}\ln |1+4(1)| + \ln |1| \right]$$ Simplify the expression: $$-\frac{1}{4}\ln 9 + \ln 2 + \frac{1}{4}\ln 5$$ Combine logarithms: $$\ln \left(\frac{2(5^{\frac{1}{4}})}{3^{\frac{1}{2}}}\right)$$ Thus, the final result is: $$\iint_{R} \frac{x}{(1+x y)^{2}} d A = \ln \left(\frac{2(5^{\frac{1}{4}})}{3^{\frac{1}{2}}}\right)$$

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