Reverse the order of integration and then evaluate the integral.
-$\int _ { 0 } ^ { 512 } \int _ { \sqrt [ 3 ] { x } } ^ { 8 } \frac { 1 } { y ^ { 4 } + 1 } d y d x$
A) $\frac { \ln 4097 - 1 } { 4 }$
B) $\ln 8 - 1$
C) $\frac { \ln 4097 } { 4 }$
D) $\ln 8$

Question

Quizplus · Accepted Answer

First, we need to switch the order of integration. The bounds for $x$ are from $0$ to $512$ and the bounds for $y$ start at $\sqrt[3]{x}$ and go up to $8$. Therefore, the new bounds for $x$ are from $y^3$ to $512$ and the bounds for $y$ are from $0$ to $2$.

Thus, the integral becomes:

\begin{align*}
\int_0^2 \int_{y^3}^{512} \frac{1}{y^4+1}\,dx\,dy &= \int_0^2 \left[\arctan(yx)ight]_{y^3}^{512}\,dy \
&= \int_0^2 \left[\arctan(512y) - \arctan(y^4)ight]\,dy \
&= \left[\frac{1}{4}\ln(y^8+1) - \frac{1}{2}\ln(y^2+1)ight]_0^2 \
&= \frac{\ln(4097)}{4}
\end{align*}

Therefore, the answer is $\boxed{	extbf{(C)}}$.

Reverse the Order of Integration and Then Evaluate the Integral ∫0512∫x381y4+1dydx\int _ { 0 } ^ { 512 } \int _ { \sqrt [ 3 ] { x } } ^ { 8 } \frac { 1 } { y ^ { 4 } + 1 } d y d x∫0512​∫3x​8​y4+11​dydx

Reverse the Order of Integration and Then Evaluate the Integral $\int _ { 0 } ^ { 512 } \int _ { \sqrt [ 3 ] { x } } ^ { 8 } \frac { 1 } { y ^ { 4 } + 1 } d y d x$