# Homework Help: Integration by parts of a function

1. Feb 2, 2010

### apiwowar

the function is c = 15te-.2t

the goal is to integrate it from t = 0 to t = 3

so to set up the integral i took out the 15 first so i got:

15 * integral from 0 to 3 of t*e-.2t

i set u = t and so du = dt
dv = e-.2tdt so v= -5e-.2t

so following the integration by parts formula i got:

15(-5*e-.2t + 5* integral of e-.2tdt

then after integrating the second part i got

15(-5e-.2t - 25e-.2t) evaluated from 0 to 3

when evaluated i got

(-123/4826181 - 205.8043635)-(-75-375) = 120.7130184

but thats wrong,

anyone have any hints or sees any mistakes that i made?

thanks

2. Feb 2, 2010

### rock.freak667

this should be

15(-5te-0.2t+5∫e-.2tdt

3. Feb 2, 2010

### apiwowar

your right, thats just a typo, i had the t there in my calculations

4. Feb 2, 2010

### rock.freak667

it looks like you did it correct, is your answer supposed to be in terms of e?

5. Feb 2, 2010

6. Feb 2, 2010

### rock.freak667

Then you need to write it in terms of the constant e. So recalculate the answer, just not using a calculator.

7. Feb 2, 2010

### apiwowar

so just leave it as with the base e but calculate the exponents of e?

8. Feb 2, 2010

Yes.

9. Feb 2, 2010

### apiwowar

ok, thanks alot