MHB Missing something within integration

[shamieh]

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!

$$e^x x^2 - \int^1_0 e^x 2x$$

Scratch that . I see it now. (Dull)

 

[MarkFL]

I am assuming you are given to evaluate:

$$I=\int_0^1 x^2e^x\,dx$$

Using IBP, let:

$$u=x^2\,\therefore\,du=2x\,dx$$

$$dv=e^x\,\therefore\,v=e^x$$

Hence:

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

Using IBP again, let:

$$u=x\,\therefore\,du=dx$$

$$dv=e^x\,\therefore\,v=e^x$$

Hence:

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