Integrating xe^{ax}: A Step-by-Step Solution

  • Thread starter Thread starter Radarithm
  • Start date Start date
  • Tags Tags
    Integrating
Radarithm
Gold Member
Messages
158
Reaction score
2

Homework Statement



Evaluate: \int{xe^{ax}}dx

Homework Equations



Integration by substitution

The Attempt at a Solution



I'm on a phone at the moment. My work: http://postimg.org/image/v4hdr5uqx/

The correct answer was:
\frac{xe^{ax}}{a}-\frac{e^{ax}}{a^2}
 
Last edited by a moderator:
Physics news on Phys.org
You should do this integral by parts. Some of your original substitutions don't look OK.
 
http://postimg.org/image/ao3mi4ygz/

I feel like I'm getting closer but I'm still making a dumb mistake. Is it with the derivatives?
 
Last edited by a moderator:
If dv = e^{ax} dx, then v = e^{ax} is wrong. There are also multiple errors on the second line, but you need to fix what I said first.
 
Got it. Thanks for pointing out my mistakes.
\int{xe^{ax}}dx u=x du=dx dv=e^{ax}dx v=\frac{e^{ax}}{a}
\frac{xe^{ax}}{a}-\int{\frac{e^{ax}}{a}}dx
\frac{xe^{ax}}{a}-\int{e^{ax}a^{-1}}dx=\frac{xe^{ax}}{a}-\frac{e^{ax}}{a^2}

Power rule + derivative mistake
Thanks for the help.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top