Problem 66 - Calculus Early Transcendentals

Question

The following integrals can be evaluated only by reversing the order of integration. Sketch the region of integration, reverse the order of integration, and evaluate the integral. $$\int_{0}^{4} \int_{\sqrt{x}}^{2} \frac{x}{y^{5}+1} d y d x$$

Step-by-Step Solution

Answer

\(\frac{1}{10}\ln 33\)

Step 1: Identify the region and reverse the order

Original: \(0 \leq x \leq 4\), \(\sqrt{x} \leq y \leq 2\).
Equivalently: \(0 \leq y \leq 2\), \(0 \leq x \leq y^2\).

Reversed integral: \(\int_0^2 \int_0^{y^2} \frac{x}{y^5+1}\,dx\,dy\)

Step 2: Evaluate the inner integral

\(\int_0^{y^2} \frac{x}{y^5+1}\,dx = \frac{1}{y^5+1} \cdot \frac{x^2}{2}\Big|_0^{y^2} = \frac{y^4}{2(y^5+1)}\)

Step 3: Evaluate the outer integral

\(\frac{1}{2}\int_0^2 \frac{y^4}{y^5+1}\,dy\). Let \(u = y^5+1\), \(du = 5y^4\,dy\):
\(= \frac{1}{10}\int_1^{33} \frac{du}{u} = \frac{1}{10}\ln 33\)