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Missing something within integration

shamieh

Active member
Sep 13, 2013
539
Can someone show me step by step how they are getting e - 2????

I have gone through through this integration 10000 times and can not come to this conclusion!

\(\displaystyle e^x x^2 - \int^1_0 e^x 2x\)

Scratch that . I see it now. (Dull)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I am assuming you are given to evaluate:

\(\displaystyle I=\int_0^1 x^2e^x\,dx\)

Using IBP, let:

\(\displaystyle u=x^2\,\therefore\,du=2x\,dx\)

\(\displaystyle dv=e^x\,\therefore\,v=e^x\)

Hence:

\(\displaystyle I=\left[x^2e^x \right]_0^1-2\int_0^1\,xe^x\,dx=e-2\int_0^1\,xe^x\,dx\)

Using IBP again, let:

\(\displaystyle u=x\,\therefore\,du=dx\)

\(\displaystyle dv=e^x\,\therefore\,v=e^x\)

Hence:

\(\displaystyle I=e-2\left(\left[xe^x \right]_0^1-\int_0^1 e^x\,dx \right)=e-2\left(e-(e-1) \right)=e-2(1)=e-2\)