Question

How to find the area bounded by the given curves?

`y=x^2e^-x` and `y=xe^-x`

Answer 1

The area between the curves is `1.208`

Explanation:

Start by finding the intersection points, by solving the system `{(y =x^2e^-x), (y = xe^-x):}`.

`x^2e^-x = xe^-x`

`x^2e^-x -xe^-x = 0`

`xe^-x(x - 1) = 0`

It becomes clear that `x =0` and `x= 1`. These will be our bounds of integration. We now must determine which curve lies above which. We see that when `x= 0.5`, `x^2e^-x < xe^-x`. Thus our integral expression is as follows:

`A = int_0^1 xe^-x - x^2e^-x`

We're going to need to apply integration by parts for both integrals.

Recall that `int udv = uv - int vdu`. We let `dv = e^-x` and `u = x^2`. Therefore `v= -e^-x` and `du = 2x`.

Thus

`intx^2e^-x dx = -x^2e^-x - int -2xe^-xdx`

`intx^2e^-x dx = -x^2e^-x + int 2xe^-x dx`

We apply integration by parts one more time, this time with `u = 2x` and `dv = e^-x`. Therefore `du = 2` and `v = -e^-x`.

`int 2xe^-x dx = -2xe^-x - int -2e^-xdx`

`int 2xe^-x dx = -2xe^-x + int 2e^-x dx`

`int 2xe^-x dx = -2xe^-x - 2e^-x`

Therefore

`intx^2e^-xdx = -x^2e^-x - 2xe^-x - 2e^-x`

We repeat the same few steps with `xe^-x`. We let `u = x` and `dv = e^-x`. Therefore `du = 1` and `v = -e^-x`.

`intxe^-x dx = -xe^-x - int-e^-xdx`

`int xe^-x dx = -xe^-x + inte^-x dx`

`intxe^-x dx = -xe^-x - e^-x`

Now we have to put all this together in our original integral expression,.

`A = [-xe^-x - e^-x -(-x^2e^-x - 2xe^-x - 2e^-x)]_0^1`

`A = [x^2e^-x + xe^-x + e^-x]_0^1`

`A = (1)^2e^(-1) + 1e^(-1) + e^(-1) - e^0`

`A = 1/e + 1/e + 1/e - 1`

`A = 3/e - 1`

`A = (3 - e)/e ~~ 0.104`

This isn't all. We also must consider the region bounded by the curve from `[1, oo]`. This area expression is going to be `int_1^oo x^2e^-x - xe^-x`

Thus

`A_2 =lim_(t->oo) [-t^2e^-t - 2te^-t - 2e^-t - (-te^-t - e^-t)]_1^t`

`A_2 = lim_(t-> oo) [-t^2e^-t - te^-t - e^-t]_1^t`

`A_2 = 1/e + 1/e + 1/e`

`A_2 = 3/e ~~ 1.104`

Therefore our total bounded area is `1.104 + 0.104 = 1.208`

Hopefully this helps!