Question

How do you find `\int _ { 0} ^ { 2} \frac { e ^ { \sqrt { x + 2} - 3} } { \sqrt { x + 2} } d x`?

Answer 1

`(2(e^2-e^sqrt2))/e^3`

Explanation:

`int_0^2e^(sqrt(x+2)-3)/sqrt(x+2)dx`

Use the substitution `u=sqrt(x+2)`. This implies that `du=1/2(x+2)^(-1/2)dx`, or `du=1/(2sqrt(x+2))dx`. Don't forget to transform the bounds:

We see that:

`=2int_0^2e^(sqrt(x+2)-3)/(2sqrt(x+2))dx=2int_sqrt2^2e^(u-3)du`

Now let `v=u-3`. Then `du=dv`:

`=2int_(sqrt2-3)^1e^vdv=2e^v]_(sqrt2-3)^1=2(e-e^(sqrt2-3))=(2(e^2-e^sqrt2))/e^3`