How do you integrate #int x^2 e^(-x)dx# using integration by parts?

Question

Scarlett Allison · Accepted Answer

#intx^2e^(-x)dx=-e^(-x)(x^2+2x+2)+C# Integration by parts says that: #intv(du)/(dx)=uv-intu(dv)/(dx)# #u=x^2;(du)/(dx)=2x# #(dv)/(dx)=e^(-x);v=-e^(-x)# #intx^2e^(-x)dx=-x^2e^(-x)-int-2xe^(-2x)dx# Now we do this: #int-2xe^(-2x)dx# #u=2x;(du)/(dx)=2# #(dv)/(dx)=-e^(-x);v=e^(-x)# #int-2xe^(-x)dx=2xe^(-x)-int2e^(-x)dx=2xe^(-x)+2e^(-x)# #intx^2e^(-x)dx=-x^2e^(-x)-(2xe^(-x)+2e^(-x))=-x^2e^(-x)-2xe^(-x)-2e^(-x)+C=-e^(-x)(x^2+2x+2)+C#

Axel Alvarez · Answer

undefined To integrate ∫x^2e^(-x)dx using integration by parts, you can apply the integration by parts formula:

∫u dv = uv - ∫v du

Let u = x^2 and dv = e^(-x)dx. Then, differentiate u to find du and integrate dv to find v:

du = 2x dx
v = -e^(-x)

Now, apply the integration by parts formula:

∫x^2e^(-x)dx = x^2 * (-e^(-x)) - ∫(-e^(-x)) * (2x dx)

= -x^2e^(-x) + 2∫xe^(-x)dx

Now, we have a simpler integral to work with, which can be integrated by parts again. Let u = x and dv = e^(-x)dx:

du = dx
v = -e^(-x)

Applying integration by parts:

2∫xe^(-x)dx = 2(x * (-e^(-x)) - ∫(-e^(-x)) dx)

= -2xe^(-x) + 2∫e^(-x)dx

= -2xe^(-x) - 2e^(-x) + C

Now, substituting back into the original integral:

∫x^2e^(-x)dx = -x^2e^(-x) - 2xe^(-x) - 2e^(-x) + C

Therefore, the integral of x^2e^(-x)dx using integration by parts is:

-x^2e^(-x) - 2xe^(-x) - 2e^(-x) + C