# Does the improper integral ( x^2 e^-x dx from 1 to infintiy) converge?

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Using integration by parts

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

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

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

`int^(oo)_1 x^2e^(-x) dx = (-x^2e^(-x) - 2xe^(-x) -2e^(-x))|_1^(oo)`

`int^(oo)_1 x^2e^(-x)dx = lim_(n->oo)-(x^2+2x+2)e^(-x) - (-e^(-x)-2e^(-x)-2e^(-x))`

`int^(oo)_1 x^2e^(-x)dx=5e^(-x)-lim_(x->oo)(x^2+2x+2)/e^(x)`

`lim_(x->oo) (x^2+2x+2)/e^x` is `(oo)/(oo)` so we can use L'Hopital's rule

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

Finally we know `lim_(x->oo) 2e^(-x) = 2lim_(x->oo)e^(-x) = 0`

So `int^(oo)_1 x^2 e^(-x) = 5e^(-x)`

You may define the improper integral such that:

`int_1^oo x^2*e^(-x) dx = lim_(1-gtn) int_1^n x^2*e^(-x) dx`

You need to evaluate the integral `int_1^n x^2*e^(-x) dx` using integration by parts.

`u = x^2 =gt du = 2xdx`

`` `dv = e^(-x) =gt v = -e^(-x)`

Using the formula int udv = uv - int vdu y ields:

`int x^2*e^(-x) dx = -x^2*e^(-x)+ 2intx*e^(-x) dx`

Use the integration by parts for int `x*e^(-x) dx` .

`u = x =gt du = dx`

`dv = e^(-x) =gt v = -e^(-x)`

`` `intx*e^(-x) dx = -xe^(-x) + int e^(-x) dx`

`intx*e^(-x) dx = -xe^(-x) - e^(-x) + c`

`int x^2*e^(-x) dx = -x^2*e^(-x)- 2xe^(-x) - 2e^(-x)+ c`

`` `int x^2*e^(-x) dx = -e^(-x)*(x^2 + 2x + 2) + c`

`int_1^n x^2*e^(-x) dx = -e^(-n)*(n^2 + 2n + 2) + e^(-1)*(1^2 + 2 + 2)`

`` `int_1^n x^2*e^(-x) dx = -e^(-n)*(n^2 + 2n + 2) + 5/e`

`lim_(n-gtoo) -(n^2 + 2n + 2)/(e^n) + lim_(n-gtoo) 5/e`

`n^2 + 2n + 2 lt e^n =gt lim_(n-gtoo) -(n^2 + 2n + 2)/(e^n) = 0`

`` `lim_(n-gtoo) -(n^2 + 2n + 2)/(e^n) + lim_(n-gtoo) 5/e = 5/e`

**Evaluating the improper integral yields 5/e => the integral `int_1^n x^2*e^(-x) dx` is convergent.**