Area between $y = e^x - xe^x$ I am trying to find the area between  $y = e^x - xe^x$ and x= 0
I get stuck on trying to find the antiderivative of $xe^x$ I don't know how to do a complex number like that. I know the curves intersect at 1,1 so I am finding it from 0 to 1.
 A: $$\int_0^1e^x-xe^xdx=\int_0^1e^xdx-\int_0^1xe^xdx=e-1-\int_0^1xe^xdx$$
$$\int_0^1xe^xdx=xe^x| _0^1-\int_0^1e^xdx\ \ \ \ \ \ \ \text{(by parts)}$$
$$=e-(e-1)=1$$
$$\int_0^1e^x-xe^xdx=e-2$$
A: First remember the formula for derivatives by parts. You can remember it by integrating the product formula for derivatives
$$(fg)' = f(g') + (f')g \quad\xrightarrow{\int_0^1}\quad fg|_{x=0}^{1} = \int_0^1 f(g')dx + \int_0^1 (f')g dx $$
Or rephrased: $\int_0^1 (f')g dx = fg|_{x=0}^{1} - \int_0^1 f(g')dx$.
If we chose $f'=e^x \Rightarrow f=e^x$ and $g=x$ we get
$$ \int_0^1 e^x x dx = [e^x x]_{x=0}^{1} - \int_0^1 e^x(1)dx. $$
Now this, we can solve...
$$ \int_0^1 e^x dx - \int_0^1 e^x x dx = - [e^x x]_{x=0}^{1} + 2 \int_0^1 e^x(1)dx $$
$$ = -e + 2(e-1) = e-2 $$
A: You may not know about integration by parts, but you might be expected to use some intelligent guesswork. You know that the derivative of $e^x$ is $e^x$ so how about ...
Take the derivative of $xe^x$ and find that it is $e^x+xe^x$.
So the derivative of $xe^x-e^x$ is ... and go from there.
