Question

Use integration by parts to find the integrals in Exercise.
∫^1_0 2x+1/e^x dx.

Answer 1

`\int^1_0 (2x+1)/(e^x) dx=(-5)/(e)+3`

Explanation:

we have the integral

`\int (2x+1)/(e^x) dx`

to integrate by parts we use:

`\int udv=uv-\int vdu`

where u will be
`u=2x+1`, and
`dv=(1)/(e^x)dx=e^(-x) dx`

we need du and v, so:

`du=d(2x+1)\\du=2dx`

and

`v=\int e^(-x)dx\\v=-e^(-x)`

the integral by parts is:

`\int (2x+1)/(e^x) dx=(2x+1)(-e^(-x))-\int(-e^(-x)(2dx))\\=-2xe^(-x)-e^(-x)+2\int e^(-x)dx\\=-2xe^(-x)-e^(-x)+2(-e^(-x))\\=-2xe^(-x)-e^(-x)-2e^(-x)\\=-2xe^(-x)-3e^(-x)\\=(-2x-3)e^(-x)`

now, evaluating with the limits:

`\int^1_0 (2x+1)/(e^x) dx=[(-2x-3)e^(-x)]^1_0`

`=[(-2(1)-3)e^(-(1))]-[(-2(0)-3)e^(-(0))]\\=[(-2-3)e^(-1)]-[(0-3)(1)]\\=-5e^(-1)+3\\=(-5)/(e)+3`