Integrate the function f over the given region.
-$f ( x , y ) = e ^ { 2 x } + 3 y$ over the rectangle $0 \leq x \leq 1,0 \leq y \leq 1$
A) $\frac { 1 } { 6 } \left( e ^ { 5 } - e ^ { 3 } - e ^ { 2 } - 1 \right)$
B) $\frac { 1 } { 4 } \left( e ^ { 5 } - e ^ { 3 } - e ^ { 2 } + 1 \right)$
C) $\frac { 1 } { 6 } \left( e ^ { 5 } - e ^ { 3 } - e ^ { 2 } + 1 \right)$
D) $\frac { 1 } { 4 } \left( e ^ { 5 } - e ^ { 3 } - e ^ { 2 } - 1 \right)$

Question

Quizplus · Accepted Answer

We can integrate the function f(x,y) by integrating with respect to x first and then with respect to y.
\begin{align*}
\int_0^1 \int_0^1 (e^{2x}+3y) dy dx &= \int_0^1 \left[3yx+\frac{e^{2x}}{2}y^2ight]_0^1 dx \
&= \int_0^1 \left(3x+\frac{e^{2x}}{2}ight)dx \
&= \frac{3}{2}x^2+\frac{1}{4}e^{2x} \bigg|_0^1 \
&= \frac{3}{2}+\frac{1}{4}e^2-\frac{1}{4} \
&= \frac{1}{6}(e^5-e^3-e^2+1)
\end{align*}
Therefore, the answer is choice C.

Integrate the Function F Over the Given Region f(x,y)=e2x+3yf ( x , y ) = e ^ { 2 x } + 3 yf(x,y)=e2x+3y

Integrate the Function F Over the Given Region $f ( x , y ) = e ^ { 2 x } + 3 y$