Question

Evaluate the double integral. $$ \int_{0}^{4} \int_{0}^{x} \frac{2}{x^{2}+1} d y d x $$

Short answer

The value of the double integral is \( \ln 17 \).

Step-by-step solution

Simplify the inner integral

Since the integrand doesn't depend on y, you can perform the integration with respect to it directly. This procedure results in a simple multiplication of the integrated function by the difference of the limits of integration: \( (x - 0) \cdot \frac{2}{x^{2}+1} = \frac{2x}{x^{2}+1} \)

Compute the outer integral

Now, integrate the resulting function, \( \frac{2x}{x^{2}+1} \), with respect to x from 0 to 4. This integral is a basic one requiring substitution. If you let \( u = x^{2}+1 \), \( du = 2x dx \), the integral becomes \( \int u^{-1} du \), which integrates to \( \ln |u| \). Substituting back in for u gives \( \ln |x^{2} + 1| \). Evaluate this at the limits 0 to 4 to get the integral value.

Evaluate limits

Substitute the upper and lower limits to calculate the integral: \( \ln |4^{2} + 1| - \ln |0^{2} + 1| = \ln 17 - \ln 1 \). Since \( \ln 1 = 0 \), the answer simplifies to \( \ln 17 \).