View Full Version : Integrals Question


Eli Z
11-22-03, 03:10 AM
How do I integrate a multiplication of two functions?

I need to integrate : |(-at/m)*(e^(bt/m)) dt

Crisp
11-22-03, 05:43 AM
This is an integral that you would typically solve with partial integration.

Bye!

Crisp

HallsofIvy
11-22-03, 10:32 AM
If you are trying to look that up in a text, the technical term is "integration by parts". I say that, not to criticize Crisp, but because "partial integration" might be confused with "partial differentiation" or "partial fractions".

Integration by parts works with products of functions because it is the reverse of the product rule for differentiation. the derivative of the product uv is d(uv)/dx= u dv/dx+ v du/dx which, in differential form, is d(uv)= u dv+ v du. We can rewrite that as
u dv= d(uv)- v du and integrate:

integral u dv= uv- integral v du (integral d(uv)= uv of course).

In this particular case, integral (-at/m)e^(bt/m)dt, let u= -at/m and dv= e^(bt/m)dt. Then du/dt= (-a/m) so du= (-a/m)dt and
v= (m/b)e^(bt/m) (anti-derivative of e^(bt/m)).

uv= {(-a/m)t}{(m/b)e^(bt/m)}= (-a/b)te^(bt/m) and

vdu= {(m/b)e^(bt/m)}{(-a/m)dt}= (-a/b)e^(bt/m)dt.

That's easy to integrate: integral vdu= (-am/b^2)e^(bt/m)

Putting all that together,
integral (-at/m)e^(bt/m)dt= (a/b)te^(bt/m)+ (am/b^2)e^(bt/m)
= (a/b)(t+ m/b)e^(bt/m)+ C

Eli Z
11-22-03, 02:11 PM
Wow. I didnt expect such detailed response. Thanks HallsofIvy. :)