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

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...

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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)

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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.

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