Reverse the order of integration and then evaluate the integral.
-$$\int _ { 0 } ^ { 6 \sqrt { \ln 10 } } \int _ { y / 6 } ^ { \sqrt { \ln 10 } } e ^ { x ^ { 2 } } d x d y$$
A) 25
B) 35
C) 33
D) 27

Question

Quizplus · Accepted Answer

We begin by reversing the order of integration:

$$\int _ { 0 } ^ { \sqrt { \ln 10 } } \int _ { 6x } ^ { 6 \sqrt { \ln 10 } } e ^ { x ^ { 2 } } d y d x$$

Now we integrate first with respect to $y$:

\begin{align*}
&\int _ { 0 } ^ { \sqrt { \ln 10 } } \int _ { 6x } ^ { 6 \sqrt { \ln 10 } } e ^ { x ^ { 2 } } d y d x \
&= \int _ { 0 } ^ { \sqrt { \ln 10 } } \left[ y e^{x^2} \right] _ { 6x } ^ { 6 \sqrt { \ln 10 } } d x \
&= \int _ { 0 } ^ { \sqrt { \ln 10 } } \left( 6 \sqrt{\ln 10} - 6x \right) e^{x^2} d x \
&= 6\sqrt{\ln 10} \int _ { 0 } ^ { \sqrt { \ln 10 } } e^{x^2} d x - 6\int _ { 0 } ^ { \sqrt { \ln 10 } } x e^{x^2} d x \
&= 3\sqrt{\ln 10} \left( e^{\ln 10} - 1 \right) - 3\int _ { 0 } ^ { \left(\sqrt { \ln 10 }\right)^2 } e^{u} d u \
&=3\sqrt{\ln 10} \left( e^{\ln 10} - 1 \right) - 3\left[ e^u \right] _ { 0 } ^ { \left(\sqrt{\ln 10}\right)^2 }  \
&= 3\sqrt{\ln 10} \left( e^{\ln 10} - 1 \right) - 3\left( e^{\ln 10} - 1 \right) \
&= \boxed{27}
\end{align*}

Reverse the Order of Integration and Then Evaluate the Integral ∫06ln⁡10∫y/6ln⁡10ex2dxdy\int _ { 0 } ^ { 6 \sqrt { \ln 10 } } \int _ { y / 6 } ^ { \sqrt { \ln 10 } } e ^ { x ^ { 2 } } d x d y∫06ln10​​∫y/6ln10​​ex2dxdy

Reverse the Order of Integration and Then Evaluate the Integral $\int _ { 0 } ^ { 6 \sqrt { \ln 10 } } \int _ { y / 6 } ^ { \sqrt { \ln 10 } } e ^ { x ^ { 2 } } d x d y$