# Integration Help!

1. Aug 2, 2010

### Pete_01

1. The problem statement, all variables and given/known data
It's been a while since I've done some of these less common integrals. This is for a physics problem but I need to integrate....

int from 0 to 2sec: (emf^2/R) (1-e^(-Rt/L))
basically integrating voltage*current for an RL circuit.

How do I do this?

2. Aug 2, 2010

### Staff: Mentor

Is this the integral?
$$\int_0^2 \frac{E^2}{R}(1 - e^{-Rt/L})dt$$

I'm assuming that the only variable is t. Split the integral above into two integrals, the first of which is very easy. For the second integral, use the substitution u = -Rt/L, and du = -R/L dt.

3. Aug 2, 2010

### Pete_01

ok so the first integral $$\frac{E^2}{R}$$ can be pulled out leaving:

$$\frac{E^2}{R}\int_0^2 \frac{-E^2}{R}(e^{-Rt/L})dt$$

correct?

can you pull out the other E^2/R and have just:

$$-\frac{E^4}{R^2}\int_0^2 (e^{-Rt/L})dt$$

and then by u substitution get:

$$-\frac{E^4}{R^2}\int_0^2 (e^{u})du$$ ?

Sorry, haven't done these for a lonng while.

4. Aug 2, 2010

### eumyang

No. You need to split the integrals using
$$\int (f(x) - g(x)) dx = \int f(x) dx - \int g(x) dx$$

Distribute the fraction, then split the integral.

69

5. Aug 2, 2010

### Pete_01

Ok, but in my problem the E and R are constants so can't I take them out?

6. Aug 2, 2010

### Staff: Mentor

If you take out E^2/R first, you will have two terms in the integrand. If you do it that way, I would still recommend splitting the integral into two parts.

7. Aug 2, 2010

### Pete_01

Could someone please go step by step and show me how to do this? It's such a small part of the original problem. Thank you

8. Aug 2, 2010

### rock.freak667

You have this right?

$$\int_0^2 \frac{E^2}{R}(1 - e^{-Rt/L})dt$$

So let's do as Mark44 said and take out the E2/R. You will be left with

$$\frac{E^2}{R} \int_0^2 (1 - e^{-Rt/L})dt$$

Are you able to evaluate this now?