Question

How do you determine if the improper integral converges or diverges `intx^2 e^-x dx` from 0 to infinity?

Answer 1

`int_0^oox^2e^(-x)dx = 2`

Explanation:

We will make use of integration by parts to find a solution to the indefinite integral, and then apply the limits to evaluate the improper definite integral.

`intx^2e^(-x)dx`

Integration by parts 1:

Let `u = x^2` and `dv = e^-xdx`
Then `du = 2xdx` and `v = -e^-x`

Applying the integration by parts formula `intudv = uv-intvdu`:

`intx^2e^-xdx = -x^2e^-x+2intxe^-xdx`

Integration by parts 2:

Let `u = x` and `dv = e^-xdx`
Then `du = dx` and `v = -e^-x`

Applying the formula to the remaining integral:

`intxe^-xdx = -xe^-x+inte^-xdx`

`= -xe^-x-e^-x+C`

Substituting this back in, we have

`intx^2e^-xdx = -x^2e^-x+2(-xe^-x-e^-x)+C`

`=-e^(-x)(x^2+2x+2)+C`

---

Now we can check the definite integral:

`int_0^oox^2e^-xdx = lim_(M->oo)int_0^Mx^2e^-xdx`

`=lim_(M->oo)[-e^-x(x^2+2x+2)]_0^M`

`=lim_(M->oo)-(M^2+2M+2)/e^M+2`

Intuitively we can say at this point that the first term will go to `0`, as the exponential `e^x` in the denominator will grow much faster than the polynomial in the numerator. This will give us an answer of `2`. Still, we can prove this using L'Hopital's rule to show that the initial term does converge to `0`.

---

`lim_(x->oo)(x^2+2x+2)/e^x = (d/dx(x^2+2x+2))/(d/dxe^x)`

`=lim_(x->oo)(2x+2)/e^x`

`=lim_(x->oo)(d/dx(2x+2))/(d/dxe^x)`

`=lim_(x->oo)(2/e^x)`

`2/oo`

`=0`

---

With that, using that if two functions converge at a limit, then the limit of their sum is equal to the sum of their limits, we have:

`lim_(M->oo)2-(M^2+2M+2)/e^M = lim_(M->oo)2-lim_(M->oo)(M^2+2M+2)/e^M`

`=2+0`

`+2`

Therefore the integral converges to `2`.