1. ## Find the integral

Hi there,

$\displaystyle \int (x^2-1)e^x dx$

I chose to integrate by parts. This is what I did.

$\displaystyle \int x^2 e^x dx-\int e^x dx$
1st integration $\displaystyle u=x^2 ,du=2x dx ,dv=e^x dx, v=e^x$
$\displaystyle (\int x^2 e^x dx = x^2 e^x -\int e^x 2xdx)-e^x +C$
2nd integration $\displaystyle u=2x, du=2dx, dv=e^x dx, v= e^x$
$\displaystyle \int 2x e^x dx=2xe^x -\int e^x 2dx$
My answer $\displaystyle x^2 e^x +2xe^x -2e^x -e^x +C$
$\displaystyle x^2 e^x +2xe^x -3e^x +C$

Is this correct?

2. Originally Posted by ugkwan
Hi there,

$\displaystyle \int (x^2-1)e^x dx$

I chose to integrate by parts. This is what I did.

$\displaystyle \int x^2 e^x dx-\int e^x dx$
1st integration $\displaystyle u=x^2 ,du=2x dx ,dv=e^x dx, v=e^x$
$\displaystyle (\int x^2 e^x dx = x^2 e^x -\int e^x 2xdx)-e^x +C$
2nd integration $\displaystyle u=2x, du=2dx, dv=e^x dx, v= e^x$
$\displaystyle \int 2x e^x dx=2xe^x -\int e^x 2dx$
My answer $\displaystyle x^2 e^x +2xe^x -2e^x -e^x +C$
$\displaystyle x^2 e^x +2xe^x -3e^x +C$

Is this correct?

Why do you ask? Derivate and check whether you obtain the original function in the integral.

And no, it isn't correct: you have a sign error.

Tonio

3. I see the sign error now. I needed a second opinion.

4. Originally Posted by ugkwan
I see the sign error now. I needed a second opinion.

What for? To remind you that in indefinite integrals you have the tools to check whether you

reached the correct solution or not?

I use to say to my students: any question that can be checked in a reasonable time (the solution to a linear

system of equations, the solution of an indefinite integral, etc.) and STILL is incorrect will be