How do you find the antiderivative of #(4x^2 - 3x)e^x#?

Question

Emilia Aguilar · Accepted Answer

#4e^x x^2-11e^x x+11e^x#

The antiderivative of a function is its integral. Here, we need to solve:

#int(4x^2-3x)e^xdx#

According to integration by parts, #intf(x)g(x)dx=f(x)intg(x)dx-intf'(x)(intg(x)dx)dx#.

Here, #f(x)=4x^2-3x# and #g(x)=e^x#.

But since #inte^xdx=e^x#, we can just write:

#e^x(4x^2-3x)-int(d/dx(4x^2-3x))e^xdx#

#e^x(4x^2-3x)-int(8x-3)e^xdx#

Integrating by parts the integral, we get:

#e^x(4x^2-3x)-e^x(8x-3)+int8e^xdx#

#e^x(4x^2-3x)-e^x(8x-3)+8e^x#

#e^x((4x^2-3x)-(8x-3)+8)#

#e^x(4x^2-3x-8x+3+8)#

#e^x(4x^2-11x+11)#

#4e^x x^2-11e^x x+11e^x#

Charles Arocha · Answer

undefined To find the antiderivative of $ (4x^2 - 3x)e^x $, you can use integration by parts. Let $ u = 4x^2 - 3x $ and $ dv = e^x \, dx $. Then, $ du = (8x - 3) \, dx $ and $ v = e^x $.

Now, apply the integration by parts formula:

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

Substitute the values:

$$ \int (4x^2 - 3x)e^x \, dx = (4x^2 - 3x)e^x - \int e^x(8x - 3) \, dx $$

Now, integrate $ e^x(8x - 3) $:

$$ = (4x^2 - 3x)e^x - \int (8x - 3)e^x \, dx $$

$$ = (4x^2 - 3x)e^x - \left( 8 \int xe^x \, dx - 3 \int e^x \, dx \right) $$

$$ = (4x^2 - 3x)e^x - \left( 8(xe^x - \int e^x \, dx) - 3e^x \right) $$

$$ = (4x^2 - 3x)e^x - \left( 8xe^x - 8e^x - 3e^x \right) $$

$$ = (4x^2 - 3x)e^x - 8xe^x + 11e^x $$

$$ = (4x^2 - 3x - 8x + 11)e^x $$

$$ = (4x^2 - 11x + 11)e^x + C $$

So, the antiderivative of $ (4x^2 - 3x)e^x $ is $ (4x^2 - 11x + 11)e^x + C $, where $ C $ is the constant of integration.