Question

find the integral of x^2 e^-xdx with the limits of negative infinity to zero.

so far, I have,
lim as t approaches negative infinity of (-e^-x [x(x+2) + 2] from 0 to t
does this converge or diverge?

Answer 1

if u = x^2 then du = 2 x dx
if dv = e^-x dx then v = -e^-x

u v = -x^2e^-x
v du = -2 x e^-x dx
u v - integral v du
= -x^2 e^-x + 2 integral x e^-x dx
now do integral x e^-x dx
u = x so du = dx
dv = e^-x dx so v = -e^-x
u v - v du = -xe^-x -integral -e^-x dx
= -x e^-x - e^-x
put that back in
-x^2 e^-x - 2 [x e^-x + e^-x]
= -e^-x [ x^2 + 2 x + 2]
so I agree, now the limits
if x = 0, integral = -2
if x = -oo that is my infinity symbol
- (1/e^x) x^2 as x --> oo
but e^big x = 1 + x + x^2/2! + x^3/3! ...
which gets very big indeed for big x
so limit at -oo is 0
so
-2 - 0 = -2