Evaluate the double integral $\int _ { 0 } ^ { 6 } \int _ { 0 } ^ { x } \frac { 2 } { x ^ { 2 } + 8 } d y d x$ .
A) $\ln 43$
B) $\ln 43 - \ln 8$
C) $\ln 44$
D) $\ln 44 + \ln 8$
E) $\ln 44 - \ln 8$

Question

Quizplus · Accepted Answer

The inner integral with respect to $y$ is straightforward because the integrand does not depend on $y$, so the integral becomes $\int_0^x \frac{2}{x^2+8} dy = \frac{2y}{x^2+8}\Big|_0^x = \frac{2x}{x^2+8}$. Then, integrating with respect to $x$ from 0 to 6 gives $\int_0^6 \frac{2x}{x^2+8} dx$, which can be solved by a substitution method (let $u = x^2 + 8$, then $du = 2x dx$). The integral becomes $\int \frac{1}{u} du$ evaluated from $x=0$ to $x=6$, which is $\ln|u|\Big|_{8}^{44} = \ln 44 - \ln 8$.

Evaluate the Double Integral ∫06∫0x2x2+8dydx\int _ { 0 } ^ { 6 } \int _ { 0 } ^ { x } \frac { 2 } { x ^ { 2 } + 8 } d y d x∫06​∫0x​x2+82​dydx

Evaluate the Double Integral $\int _ { 0 } ^ { 6 } \int _ { 0 } ^ { x } \frac { 2 } { x ^ { 2 } + 8 } d y d x$