Reverse the order of integration and then evaluate the integral.
-$$\int _ { 0 } ^ { 3 } \int _ { x / 3 } ^ { 1 } 3 y ^ { 2 } \cos x y d y d x$$
A) $5 ( 1 - \cos 3 )$
B) $\cos 3 - 1$
C) $\frac { 1 } { 2 } ( 1 - \cos 3 )$
D) $3 ( 1 - \cos 3 )$

Question

Quizplus · Accepted Answer

We will reverse the order of integration and write the integral as follows:

$$\int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 3 y } 3 y ^ { 2 } \cos x y d x d y$$

Now we evaluate the integral, keeping in mind that $\cos xy$ is the only term involving $x$:

\begin{align*}
\int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 3 y } 3 y ^ { 2 } \cos x y d x d y & = \int _ { 0 } ^ { 1 } \left[ 3 y ^ { 2 } \frac { \sin xy } { y } \right] _ { x=0 } ^ { x=3y } d y \
& = \int _ { 0 } ^ { 1 } 3 y ( \sin 3y - \sin 0 ) d y \
& = [ - \frac { 3 } { 2 } y \cos 3y ] _ { 0 } ^ { 1 } \
& = \frac { 3 } { 2 } ( 1 - \cos 3 )
\end{align*}

Therefore, the answer is C.

Reverse the Order of Integration and Then Evaluate the Integral ∫03∫x/313y2cos⁡xydydx\int _ { 0 } ^ { 3 } \int _ { x / 3 } ^ { 1 } 3 y ^ { 2 } \cos x y d y d x∫03​∫x/31​3y2cosxydydx

Reverse the Order of Integration and Then Evaluate the Integral $\int _ { 0 } ^ { 3 } \int _ { x / 3 } ^ { 1 } 3 y ^ { 2 } \cos x y d y d x$