# Thread: Trouble with Integration by Parts

1. ## Trouble with Integration by Parts

"Determine the area between the graph of $y=xe^{-x^{2}}$ and the x axis over the interval [0,1]."

my work:

Area = $\int_0^1 \! xe^{-x^{2}} \, dx$
using the integral by parts rule " $\int \! udv = uv - \int \! vdu$ "

u = x | v = $\frac{e^{-x^{2}}}{-2x}$

du = 1 | dv= $e^{-x^{2}}$

thus..
$
\int_0^1 \! xe^{-x^{2}} \, dx = x(\frac{e^{-x^{2}}}{-2x}) - \int \! \frac{e^{-x^{2}}}{-2x} (1)$

This is where I'm like, am I even on the right track?

and if so, how do I integrate $\int \! \frac{e^{-x^{2}}}{-2x}$?

Thanks for the help.

2. You don't need to use an integration by parts. What is the derivative of $x\mapsto \exp (-x^2)$ ?

3. Originally Posted by Savior_Self

Area = $\int_0^1 \! xe^{-x^{2}} \, dx$
No need to use integration by parts; just substitute $y=x^2$

4. Alright, thanks guys, but can we just assume I have to use integration of parts? It's kind of the point of the lesson.

5. Originally Posted by Savior_Self
Alright, thanks guys, but can we just assume I have to use integration of parts? It's kind of the point of the lesson.
That is an old trick of most seasoned calculus teachers.
In a section on integration by parts always throw in some that do not require parts.
It is part of standard teaching practices.

6. Originally Posted by Savior_Self
"Determine the area between the graph of $y=xe^{-x^{2}}$ and the x axis over the interval [0,1]."

my work:

Area = $\int_0^1 \! xe^{-x^{2}} \, dx$
using the integral by parts rule " $\int \! udv = uv - \int \! vdu$ "

u = x | v = $\frac{e^{-x^{2}}}{-2x}$
This is incorrect. With $dv= e^{-x^2}$, v is NOT $\frac{e^{-x^2}}{2x}$
In fact, the anti-derivative of $e^{-x^2}$ is not an elementary function.

du = 1 | dv= $e^{-x^{2}}$

thus..
$
\int_0^1 \! xe^{-x^{2}} \, dx = x(\frac{e^{-x^{2}}}{-2x}) - \int \! \frac{e^{-x^{2}}}{-2x} (1)$

This is where I'm like, am I even on the right track?

and if so, how do I integrate $\int \! \frac{e^{-x^{2}}}{-2x}$?

Thanks for the help.