Feb 11, 2014 Thread starter #1 S 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.
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.
Feb 12, 2014 Admin #2 M MarkFL Administrator Staff member Feb 24, 2012 13,777 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: \(\displaystyle I=e-2\left(\left[xe^x \right]_0^1-\int_0^1 e^x\,dx \right)=e-2\left(e-(e-1) \right)=e-2(1)=e-2\)
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: \(\displaystyle I=e-2\left(\left[xe^x \right]_0^1-\int_0^1 e^x\,dx \right)=e-2\left(e-(e-1) \right)=e-2(1)=e-2\)