Question

[Calculus 2] Having trouble with "The Logarithm as a Definite Integral."

Answer 1

\[\int\limits_{\ln 4}^{\ln 9} 2y/(y^2-25)\]

Answer 2

Have you tried making the substitution \(\large u=y^2-25\)?

Answer 3

Im having trouble understanding it in general... Why neglect 2y?

Answer 4

The objective behind substitution is to pick a function who's derivative appears somewhere in your integrand. You'll see why we don't choose 2y as the substitution once you differentiate \(u=y^2-25\). :-)

Answer 5

Ahhhhh I see, because 2 becomes a constant which then defeats the purpose of finding the area under the curve?

Answer 6

If you pick \(u=2y\), it doesn't change the problem much.

Answer 7

the goal is to turn this into something familiar, like \(1/u\)

Answer 8

and then you're left with integrating something like \(\displaystyle\int\frac{1}{z^2-a^2}\,dz\), which isn't doable if you're not familiar with partial fractions.

Answer 9

Hmmmmm. Okay, so then I have \[\int\limits_{\ln 4}^{\ln 9} 1/(y^2-25)\]. My book tells me the answer is ln|y^2-25| so this is when I take the antiderivative?

Answer 10

Wait, what substitution did you make?

Answer 11

I made the substitution of u = y^2-25. Like you said.

Answer 12

Its an indefinite, >.> I accidently thought of a different problem and made it definite.

Answer 13

Ok. That means then that \(du=2y\,dy\). So, ignoring limits of integration for now, we should see that \[\int \frac{2y}{y^2-25}\,dy \xrightarrow{x=y^2-25}\int\frac{1}{u}\,du.\]

Answer 14

Okay.

Answer 15

Now that that's clear, what do you get when you evaluate \(\displaystyle\int\frac{1}{u}\,du\)?

Answer 16

(Note that you can't use power rule here.)

Answer 17

Substitute for u and take the anti derivative to make it ln|y^2-25|?

Answer 18

+ some constant C

Answer 19

You substitute once you do the integration. So \(\displaystyle\int\frac{1}{u}\,du = \ln|u|+C\), and then you can back substitute \(u=y^2-25\) to get that \(\displaystyle\int\frac{2y}{y^2-25}\,dy = \ln|y^2-25|+C\).

I hope this clarifies things! :-)

Answer 20

Ahhhh I see X3. One more problem if you have time? Youre really helpful ^_^

Answer 21

Sure. Ask away. :-)

Answer 22

\[\int\limits_{0}^{1} (2^-\theta) d \theta\]

Answer 23

Well, I would first make the substitution \(u=-\theta\). Then \(\,du = -\,d\theta\). Also, do you know how to change your limits of integration so we don't need to back substitute later on?

Answer 24

Honestly, No. xD

Answer 25

Ok; when you make a substitution in a definite integral, you can also change the limits of integration. To do this, we use the substitution we made; We said that \(u=-\theta\), but the current limits of integration (0 and 1) are associated with the variable \(\theta\). Thus, when \(\theta=0\), we see that \(u=-0 = 0\) and when \(\theta=1\), we see that \(u=-1\). Thus, 0 and -1 are the new limits of integration for u. Therefore, when you make this substitution, we see that\[\large \int_0^12^{-\theta}\,d\theta \xrightarrow{u=-\theta}-\int_0^{-1}2^u\,du = \int_{-1}^02^u\,du.\]

Are you with me so far? :-)

Answer 26

Yea, you had to make the value negative if you were to switch the interval.

Answer 27

I just realized there a simpler way to do this; note that the work we've done so far will change slightly. First note that we can rewrite this function as e to some power. In particular, \(\large 2^{-\theta} =e^{\ln(2^{-\theta})} = e^{-\theta\ln 2}\). So we see that \(\large \displaystyle \int_0^1 2^{-\theta}\,d\theta = \int_0^1 e^{-\theta\ln 2}\,d\theta\). Here, we instead make the substitution \(\large \displaystyle u=-\theta\ln 2\implies \,du=-\ln 2\,d\theta\implies d\theta= -\frac{\,du}{\ln 2}\). Also, at this time we can change the limits of integration. Hence, when \(\theta=0\), we have \(u=-0\ln 2=0\) and when \(\theta=1\), we have \(u=-1\ln2=-\ln2\). Therefore, when me make our substitution, we see that\[\large\begin{aligned} \int_0^1 e^{-\theta\ln 2}\,d\theta \xrightarrow{u=-\theta\ln 2}{} \int_0^{-\ln 2}e^u\cdot\frac{-\,du}{\ln 2} &= -\frac{1}{\ln 2}\int_0^{-\ln 2}e^u\,du \\ &= \frac{1}{\ln 2}\int_{-\ln 2}^0 e^u\,du\end{aligned}\]If we kept doing things as I originally was doing, we'd have to make another substitution. Sorry that I didn't realize that sooner.

I hope this makes sense as well. The integration require to finish the problem should be straight forward from here. :-)

Answer 28

THANKS!!!! :)