Question

How do you find the definite integral for: `xe^(x^2 + 2)` for the intervals `[0, 2]`?

Answer 1

`I=int_0^2xe^(x^2+2)dx=1/2e^2(e^4-1)`

Explanation:

`I=int_0^2xe^(x^2+2)dx`

Use the substitution `u=x^2+2`. This implies that `du=2xdx`. When we change variables, we will also have to change the bounds. To do this, take the current bounds and plug them into `u=x^2+2`.

`I=1/2int_0^2e^(x^2+2)(2xdx)=1/2int_2^6e^udu`

The integral of `e^u` is itself:

`I=1/2[e^u]_2^6=1/2(e^6-e^2)=1/2e^2(e^4-1)`