4.2.5 AP Calculus Exam IBP

[karush]

$\tiny{4.2.5}$

$\displaystyle\int^1_0{xe^x\ dx}$

is equal to
$A.\ \ {1}\quad B. \ \ {-1}\quad C. \ \ {2-e}\quad D.\ \ {\dfrac{e^2}{2}}\quad E.\ \ {e-1}$

Spoiler: IBP

$\begin{array}{lll}
\textit{IBP}
&uv-\displaystyle\int v' dv
&\tiny{(1)}\\ \\
\textit{substitution}
&u=x,\ v'=e^x
&\tiny{(2)}\\ \\
\textit{calculate}
&I=xe^x-\displaystyle\int \ e^xdx\biggr|^1_0
=xe^x-e^x\biggr|^1_0=1
&\tiny{(3)}
\end{array}$

ok I think this is ok possible typos
but curious if this could be solve not using IBP since the only variable is x

 

[skeeter]

$t=e^x \implies x = \ln{t} \implies dx = \dfrac{dt}{t}$

$\displaystyle \int xe^x \, dx = \int \ln{t} \, dt$

If you happen to know the antiderivative of $\ln{t}$ off the top of your head, great ... if not, you would need to use integration by parts anyway.