Reverse the order of integration and then evaluate the integral.
-$$\int _ { 0 } ^ { 63 } \int _ { y / 9 } ^ { 7 } \frac { \cos x } { x } d x d y$$
A) 9 cos 7
B) 9 sin 7
C) 7 sin 9
D) 7 cos 9

Question

Quizplus · Accepted Answer

To reverse the order of integration, we need to draw the region of integration and switch the order of the limits of integration.

The region of integration is the triangle with vertices at $(0,0)$, $(63,0)$, and $(63,7\cdot9) = (63,63)$.

Changing the order of integration gives:

\begin{align*}
\int_0^7\int_{0}^{9y/7}\frac{\cos x}{x}\,dx\,dy &= \int_0^7\left(\int_{0}^{9y/7}\frac{\cos x}{x}\,dxight)\,dy\
&= \int_0^7\left(\int_{y/9}^{7}\frac{\cos x}{x}\,dyight)\,dx
\end{align*}

We can now evaluate the integral:

\begin{align*}
\int_0^7\left(\int_{y/9}^{7}\frac{\cos x}{x}\,dyight)\,dx &= \int_0^7 \left(\frac{1}{9}\int_{9y}^{63}\frac{\cos x}{x}\,dxight)\,dy\
&= \int_0^7 \left(\frac{\sin 63 - \sin 9y}{9}ight)\,dy\
&= \frac{1}{9}\left(\int_0^7 \sin 63\,dy - \int_0^7 \sin 9y\,dyight)\
&= \frac{1}{9}\left(\sin 63\cdot 7 - \left[-\frac{\cos 9y}{9}ight]_{0}^{7}ight)\
&= \frac{1}{9}\left(\sin 63\cdot 7 + \frac{\cos 63 - 1}{9}ight)\
&= \frac{1}{9}\left(\sin 63\cdot 7 + \frac{\cos 63}{9} - \frac{1}{9}ight)\
&\approx \boxed{0.448}	ext{(rounded to three decimal places)}
\end{align*}

Reverse the Order of Integration and Then Evaluate the Integral ∫063∫y/97cos⁡xxdxdy\int _ { 0 } ^ { 63 } \int _ { y / 9 } ^ { 7 } \frac { \cos x } { x } d x d y∫063​∫y/97​xcosx​dxdy

Reverse the Order of Integration and Then Evaluate the Integral $\int _ { 0 } ^ { 63 } \int _ { y / 9 } ^ { 7 } \frac { \cos x } { x } d x d y$