Thread: Integration By Parts Problem

1. Integration By Parts Problem

Problem:

During a surge in the demand for electricity, the rate, r, at which energy is used can be approximated by the following function where t is the time in hours and b is a positive constant.

$r(t) = te^{-bt}$

a.)
(a) Find the total energy, E, used in the first T hours. Give your answer as a function of b.

My Work:

So I know that I want the integral of r(t) from 0 to T. I tried to find the integral and got:

$\frac{1}{b} e^{-bt}*(1-t)$

Is this right?

And also, I have NO IDEA what it means when the question says that it wants the answer in terms of b. "b" is a constant! How do you do that?! Is that like E(b), and if so, what would that look like?

2. Originally Posted by lysserloo
Problem:

During a surge in the demand for electricity, the rate, r, at which energy is used can be approximated by the following function where t is the time in hours and b is a positive constant.

$r(t) = te^{-bt}$

a.)
(a) Find the total energy, E, used in the first T hours. Give your answer as a function of b.

My Work:

So I know that I want the integral of r(t) from 0 to T. I tried to find the integral and got:

$\frac{1}{b} e^{-bt}*(1-t)$

Is this right?

And also, I have NO IDEA what it means when the question says that it wants the answer in terms of b. "b" is a constant! How do you do that?! Is that like E(b), and if so, what would that look like?

Nope, that's not the integral. try it again, you want to use integration by parts with $u = t$ and $dv = e^{-bt}~dt$

3. I redid the problem and got this:

$(\frac{1}{b}e^{-bt}) (\frac{1}{b}-t)$

4. Originally Posted by lysserloo
I redid the problem and got this:

$(\frac{1}{b}e^{-bt}) (\frac{1}{b}-t)$
no. it is $- \frac {e^{-bt}}b \left( t + \frac 1b \right)$

5. It looks to me like I got it right, but distributed a negative sign wrong somewhere, or something.

I'll redo it.

However, my main concern with this problem was the second part. The "solving it in terms of b" thing. Could somebody explain how you would do that? The rest I'm sure I can figure out.

6. It just means write the equation as $b = \dots$.