Integration by parts

krazydog

New member
Joined
Oct 25, 2011
Messages
14
I have a problem where i am not sure how the solutions manual got the step. I understand they are taking
integration by parts twice but i dont see how they are producing the 6 in the second integration.

x3exdx\displaystyle \int x^3e^x dx

u=x3,dv=ex\displaystyle u = x^3 , dv = e^x

Take integration by parts 3 times..
u=x,v=ex,du=dx\displaystyle u = x , v = e^x, du =dx

Now, lets put it in the form udv=uvvdu\displaystyle \int u dv = uv - \int v du

We have
x3exdx=x3ex3x2exdx\displaystyle \int x^3 e^x dx = x^3e^x -3 \int x^2e^x dx


Now from here we need to do integration by parts on the integral again because it doesnt fit any rules?
But when i do

u=x2,dv=ex\displaystyle u = x^2 , dv = e^x
So, then i would have du=2xdx,v=ex\displaystyle du = 2xdx , v = e^x

now, this is where i am not sure how they get 6?

=x3ex3x2ex+6xexdx\displaystyle = x^3e^x - 3x^2e^x + 6 \int xe^x dx

Do i need to distribute the -3 to the integral and times it by the 2xdx?
 
Hello, krazydog!

If you write out all the steps, you shouldn't get lost . . .


I  =  x3ex  ⁣dx\displaystyle \displaystyle I \;=\;\int x^3e^x\,\!dx

By parts: .{u=x3dv=ex  ⁣dxdu=3x2  ⁣dxv=ex}\displaystyle \begin{Bmatrix}u &=& x^3 && dv &=& e^x\,\!dx \\ du &=& 3x^2\,\!dx && v &=& e^x \end{Bmatrix}

Then: .I  =  x3ex3 ⁣ ⁣ ⁣ ⁣x2ex  ⁣dx\displaystyle \displaystyle I \;=\;x^3e^x - 3\!\!\int\!\! x^2e^x\,\!dx

By parts: .{u=x2dv=ex  ⁣dxdu=2xdxv=ex}\displaystyle \begin{Bmatrix}u &=& x^2 && dv &=& e^x\,\!dx \\ du &=& 2x\,dx && v &=& e^x \end{Bmatrix}

Then: . I  =  x3ex3[x2ex2 ⁣ ⁣ ⁣ ⁣xexdx]\displaystyle \displaystyle I \;=\;x^3e^x - 3\left[x^2e^x\,-\,2\!\!\int\!\! xe^x\,dx\right]

. . . . . .I  =  x3ex3x2ex+6 ⁣ ⁣ ⁣ ⁣xexdx\displaystyle \displaystyle I \;=\;x^3e^x - 3x^2e^x + 6\!\!\int\!\! xe^x\,dx

By parts: .{u=xdv=exdxdu=dxv=ex}\displaystyle \begin{Bmatrix}u &=& x && dv &=& e^x\,dx \\ du &=& dx && v &=& e^x \end{Bmatrix}

Then: .I  =  x3ex3x2ex+6[xex ⁣ ⁣exdx]\displaystyle \displaystyle I \;=\;x^3e^x - 3x^2e^x + 6\left[xe^x - \int\!\!e^x\,dx\right]

. . . . . I  =  x3ex3x2ex+6xex6 ⁣ ⁣ ⁣ ⁣exdx\displaystyle \displaystyle I \;=\;x^3e^x - 3x^2e^x + 6xe^x - 6\!\!\int\!\! e^x\,dx

. . . . . I  =  x3ex3x2ex+6xex6ex+C\displaystyle I \;=\;x^3e^x - 3x^2e^x + 6xe^x - 6e^x + C


. . . . . I  =  ex(x33x2+6x6)+C\displaystyle I \;=\;e^x(x^3 - 3x^2 + 6x - 6) + C
 
Yes, it comes from the 3 from before being multiplied by the 2 in the last step.

But, here is a shorter way to do this problem.

Since there is a cubic term in the integrand, the solution will have the form

(ax3+bx2+cx+d)ex\displaystyle (ax^{3}+bx^{2}+cx+d)e^{x}


Differentiate using the product rule:

(ax3+bx2+cx+d)ex+ex(3ax2+2bx+c)ex=x3ex\displaystyle (ax^{3}+bx^{2}+cx+d)e^{x}+e^{x}(3ax^{2}+2bx+c)e^{x}=x^{3}e^{x}

Equate coefficients:

a=1,   3a+b=0,   2b+c=0,   c+d=0\displaystyle a=1, \;\ 3a+b=0, \;\ 2b+c=0, \;\ c+d=0

a is already solved, and this leads to a=1,   b=3,   c=6,   d=6\displaystyle a=1, \;\ b=-3, \;\ c=6, \;\ d=-6

(x33x2+6x6)ex\displaystyle (x^{3}-3x^{2}+6x-6)e^{x}
 
Top