How do you integrate #int x^2 e^(4-x) dx # using integration by parts?

Question

How do you integrate #int x^2  e^(4-x) dx  # using integration by parts?

Isabella Ackerman · Accepted Answer

#-x^2e^(4-x)-2xe^(4-x)+2e^(4-x)+const.#

For integration by parts it's used this scheme:

#int udv=uv-int vdu#

Making #dv=e^(4-x)dx# => #v=-e^(4-x)# #u=x^2# => #du=2xdx# Then the original expression becomes #=x^2(-e^(4-x))-int -e^(4-x)2xdx=-x^2e^(4-x)+2int xe^(4-x)dx#

Applying integration by parts to #int xe^(4-x)dx# Making #dv=e^(4-x)dx# => #v=-e^(4-x)# #u=x# => #du=dx# We get #=x(-e^(4-x))-int e^(4-x)dx=-xe^(4-x)+e^(4-x)#

Using the partial results in the main expression, we get #=-x^2e^(4-x)+2(-xe^(4-x)+e^(4-x))# #=-x^2e^(4-x)-2xe^(4-x)+2e^(4-x)+const.#

Emilia Aguilar · Answer

undefined To integrate $ \int x^2 e^{4-x} \, dx $ using integration by parts:

Let $ u = x^2 $ and $ dv = e^{4-x} \, dx $.

Then, $ du = 2x \, dx $ and $ v = -e^{4-x} $.

Using the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substitute the values:

$$ = x^2 (-e^{4-x}) - \int (-e^{4-x}) (2x \, dx) $$

$$ = -x^2 e^{4-x} + 2 \int x e^{4-x} \, dx $$

Now, we perform integration by parts again:

Let $ u = x $ and $ dv = e^{4-x} \, dx $.

Then, $ du = dx $ and $ v = -e^{4-x} $.

$$ = -x^2 e^{4-x} + 2 \left( -x e^{4-x} - \int -e^{4-x} \, dx \right) $$

$$ = -x^2 e^{4-x} - 2x e^{4-x} + 2 \int e^{4-x} \, dx $$

$$ = -x^2 e^{4-x} - 2x e^{4-x} + 2(-e^{4-x}) + C $$

$$ = -x^2 e^{4-x} - 2x e^{4-x} - 2e^{4-x} + C $$

Thus, the integration of $ \int x^2 e^{4-x} \, dx $ using integration by parts is $ -x^2 e^{4-x} - 2x e^{4-x} - 2e^{4-x} + C $, where $ C $ is the constant of integration.