Calculus Problem Definite Integral from 0 to 1 of (x^2+1)e^-x dx?
Hey guys doing some practice problems for homework could use a little help on this one to check my answer, steps would help a ton as im having trouble simplifying.
The problem is
Definite Integral from 0 to 1 of (x^2+1)e^-x dx
Comments
e^(-x) * dx + x^2 * e^(-x) * dx
-e^(-x) + int(x^2 * e^(-x) * dx)
u = x^2
du = 2x * dx
dv = e^(-x) * dx
v = -e^(-x)
-e^(-x) + (x^2 * (-e^(-x)) - (-2) * int(x * e^(-x) * dx)) =>
-e^(-x) - e^(-x) * x^2 + 2 * int(x * e^(-x) * dx)
u = x
du = dx
dv = e^(-x) * dx
v = -e^(-x) * dx
-e^(-x) - x^2 * e^(-x) + 2 * (-x * e^(-x) + int(e^(-x) * dx)) =>
-e^(-x) - x^2 * e^(-x) - 2x * e^(-x) - 2 * e^(-x) + C =>
-e^(-x) * (1 + x^2 + 2x + 2) + C =>
-e^(-x) * (x^2 + 2x + 3) + C
From 0 to 1
-e^(-1) * (1 + 2 + 3) + e^(0) * (0 + 0 + 3) =>
(-1/e) * (6) + 1 * 3 =>
3 - 6/e =>
(3e - 6) / e =>
(3/e) * (e - 2)
Do a u du substitution. enable u = lnx. Then du = (a million/x)dx So, with changing the bounds into words of u, the quintessential is now i nt(u^2du) from ln12 to ln16. You get (u^3)/3 from ln12 to ln16. So your answer is [(ln16)^3]/3-[(ln12)^3]/3