What is f(x) = int x^2e^(2x-1)-x^3e^x dxf(x)=∫x2e2x−1−x3exdx if f(2) = 7 f(2)=7?
1 Answer
I got:
(x^2/2 - x/2 + 1/4)e^(2x - 1) - (x^3 - 3x^2 + 6x - 6)e^x - 5/4e^3 + 2e^2 + 7(x22−x2+14)e2x−1−(x3−3x2+6x−6)ex−54e3+2e2+7
This has several integration by parts. But we do have access to an integral table.
int x^me^(ax)dx = 1/a x^me^(ax) - m/aint x^(m-1)e^(ax)dx∫xmeaxdx=1axmeax−ma∫xm−1eaxdx form >= 2m≥2
int xe^(ax)dx = 1/a^2e^(ax)(ax - 1) + C∫xeaxdx=1a2eax(ax−1)+C
So the first integral should give:
int x^2e^(2x-1)dx∫x2e2x−1dx
= 1/2 x^2e^(2x-1) - int xe^(2x-1)dx=12x2e2x−1−∫xe2x−1dx
= x^2/2e^(2x-1) - 1/4 e^(2x-1)(2x-1)=x22e2x−1−14e2x−1(2x−1)
= x^2/2e^(2x-1) - x/2 e^(2x-1) + 1/4e^(2x - 1)=x22e2x−1−x2e2x−1+14e2x−1
= (x^2/2 - x/2 + 1/4)e^(2x - 1)=(x22−x2+14)e2x−1
and the second integral should give:
int x^3e^xdx∫x3exdx
= x^3e^x - 3intx^2e^xdx=x3ex−3∫x2exdx
= x^3e^x - 3(x^2e^x - 2int xe^xdx)=x3ex−3(x2ex−2∫xexdx)
= x^3e^x - 3[x^2e^x - 2(e^x(x-1))]=x3ex−3[x2ex−2(ex(x−1))]
= x^3e^x - 3x^2e^x + 6xe^x - 6e^x=x3ex−3x2ex+6xex−6ex
= (x^3 - 3x^2 + 6x - 6)e^x=(x3−3x2+6x−6)ex
These two integral solutions combine to give:
color(green)((x^2/2 - x/2 + 1/4)e^(2x - 1) - (x^3 - 3x^2 + 6x - 6)e^x + C)(x22−x2+14)e2x−1−(x3−3x2+6x−6)ex+C
Further, you know that
((2)^2/2 - (2)/2 + 1/4)e^(2(2)-1) - ((2)^3 - 3(2)^2 + 6(2) - 6)e^(2) + C = 7((2)22−22+14)e2(2)−1−((2)3−3(2)2+6(2)−6)e2+C=7
(2 - 1 + 1/4)e^3 - (8 - 12 + 12 - 6)e^2 + C = 7(2−1+14)e3−(8−12+12−6)e2+C=7
5/4e^3 - 2e^2 + C = 754e3−2e2+C=7
Thus, your integration constant is
color(blue)(int x^2e^(2x-1) - x^3e^xdx)∫x2e2x−1−x3exdx
= color(blue)((x^2/2 - x/2 + 1/4)e^(2x - 1) - (x^3 - 3x^2 + 6x - 6)e^x - 5/4e^3 + 2e^2 + 7)=(x22−x2+14)e2x−1−(x3−3x2+6x−6)ex−54e3+2e2+7
