MHB 4.1.306 AP Calculus Exam Area under Curve

[karush]

$\textsf{What is the area of the region in the first quadrant bounded by the graph of}$
$$y=e^{x/2} \textit{ and the line } x=2$$
a. 2e-2 b. 2e c. $\dfrac{e}{2}-1$ d. $\dfrac{e-1}{2}$ e. e-1Integrate
$\displaystyle \int e^{x/2}=2e^{x/2}$
take the limits
$2e^{x/2}\Biggr|_0^2=2e-2$which is a.any suggestions ?

 

[Prove It]

$\textsf{What is the area of the region in the first quadrant bounded by the graph of}$
$$y=e^{x/2} \textit{ and the line } x=2$$
a. 2e-2 b. 2e c. $\dfrac{e}{2}-1$ d. $\dfrac{e-1}{2}$ e. e-1Integrate
$\displaystyle \int e^{x/2}=2e^{x/2}$
take the limits
$2e^{x/2}\Biggr|_0^2=2e-2$which is a.any suggestions ?

Put a +C at the end of your indefinite integral, and then you are correct.

 

[topsquark]

Integrate
[math]\int e^{x/2}[/math] dx [math]=2e^{x/2}[/math] + C

Tsk tsk.

-Dan

 

[karush]

\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[axis lines=middle, grid=both]
\addplot[
draw = red,
domain=-.5:3,
range=-2:8,
samples=50
] {exp(e^(x/2)};
\draw[solid] (2,0)- -(2,6);
\end{axis}
\end{tikzpicture}ok I tried to tikx $e^{x/2}$ and $x=2$ but kinda ! don't see any vertical line
also be nice for a light shade in the region

 

[Prove It]

\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[axis lines=middle, grid=both]
\addplot[
draw = red,
domain=-.5:3,
range=-2:8,
samples=50
] {exp(e^(x/2)};
\draw[solid] (2,0)- -(2,6);
\end{axis}
\end{tikzpicture}ok I tried to tikx $e^{x/2}$ and $x=2$ but kinda ! don't see any vertical line
also be nice for a light shade in the region

Wait, what? Are you trying to graph the area?