Problem 31 - Calculus Early Transcendentals

Question

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. $$\iint_{R} \frac{x}{(1+x y)^{2}} d A ; R=\\{(x, y): 0 \leq x \leq 4,1 \leq y \leq 2\\}$$

Step-by-Step Solution

Answer

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 1: 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: $$\int_{1}^{2} \int_{0}^{4} \frac{x}{(1+x y)^{2}} dx\, dy$$

Step 2: 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: $$\int_{1}^{2} \left[ -\frac{1}{y(1+4y)} + \frac{1}{y} \right] dy$$

Step 3: Integrate the remaining integral with respect to y

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

Step 4: 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: $$\iint_{R} \frac{x}{(1+x y)^{2}} d A = \ln \left(\frac{2(5^{\frac{1}{4}})}{3^{\frac{1}{2}}}\right)$$