differentiation by substitution problem

[jonnburton]

Hi,

I have a problem in my textbook and am not sure how to get started with this. Could anybody offer any suggestions as to how to deal with this:

\(\displaystyle \int^0_{-ln2} \frac{2}{e^x+1}dx\)

This is supposed to be solved by substitution, where \(\displaystyle u = e^x +1\)

I wasn't sure how to get started with this. At first glance I thought that it's \(\displaystyle \int^0_{-ln2} \frac{2}{u}dx\) but that doesn't go anywhere as there's no way to substitute for 'dx'.

Then I thought perhaps it can be expressed as \(\displaystyle \int^0_{-ln2} \frac{2}{e^x} + 2 dx\) and as \(\displaystyle dx = \frac{1}{e^x}du\) I think the integral can be re-written as: \(\displaystyle 2\int^0_{-ln2} \frac{1}{e^x} + 2 du\)

I'm not sure if what I've done there makes sense as I calculated the definite integral as follows and did not get the answer provided in the back of the book:

\(\displaystyle 2\int^0_{-ln2} e^{-x} + 2 du\)

Now, integrating:

\(\displaystyle 2 \left[-e^{-x} + 2x\right]_{-ln2}^0\)

\(\displaystyle \left[-2e^{-x} + 4x\right]_{-ln2}^0\)

\(\displaystyle \left[-2\right] - \left[-2e^{ln2}-4ln2\right]\)

\(\displaystyle -6-4ln2\)

As ever, I would be very grateful for any information!

 

[DrPhil]

Hi,

I have a problem in my textbook and am not sure how to get started with this. Could anybody offer any suggestions as to how to deal with this:

\(\displaystyle \int^0_{-ln2} \frac{2}{e^x+1}dx\)

This is supposed to be solved by substitution, where \(\displaystyle u = e^x +1\)

If you make a substitution, you also have to calculate du/dx and also the effect on the limits. Thus

......\(\displaystyle u = e^x +1\),............\(\displaystyle du = e^x\ dx\)

......\(\displaystyle x = \ln(u - 1)\),........\(\displaystyle dx = \dfrac{1}{u-1}\ du\)

Limits: \(\displaystyle x=-\ln2 \implies u=3/2\)
..........\(\displaystyle x=0 \implies u = 2\)

Try again!

 

[jonnburton]

If you make a substitution, you also have to calculate du/dx and also the effect on the limits. Thus

......\(\displaystyle u = e^x +1\),............\(\displaystyle du = e^x\ dx\)

......\(\displaystyle x = \ln(u - 1)\),........\(\displaystyle dx = \dfrac{1}{u-1}\ du\)

Limits: \(\displaystyle x=-\ln2 \implies u=3/2\)
..........\(\displaystyle x=0 \implies u = 2\)

Try again!

Ok, thanks for that DrPhil - I will!

 

[jonnburton]

I do find integration by substitution difficult to get my head around. Having followed up on DrPhil's advice, this is what I got:

\(\displaystyle \int^0_{-ln2} \frac{2}{e^x+1}dx\)

\(\displaystyle u = e^x+1 \)
\(\displaystyle du = e^x dx\)
\(\displaystyle dx = \frac{1}{e^x} du = \frac{1}{u-1} du\)

\(\displaystyle \int^2_\frac{3}{2} \frac{2}{u} \cdot \frac{1}{u-1} du\)

\(\displaystyle \int^2_\frac{3}{2} 2u^{-1} \cdot (u-1)^{-1} du \)

\(\displaystyle \int^2_\frac{3}{2} 2u^{-2}-2u^{-1}du\)

\(\displaystyle \left[-2u^{-1}-2\right]^2_\frac{3}{2}\)

\(\displaystyle \left[-\frac{2}{u}-2\right]^2_\frac{3}{2}\)

\(\displaystyle \left[-\frac{2}{2} -2\right] - \left[\frac{2}{\frac{3}{2}} -2 \right]\)

\(\displaystyle \left[-4\right]-\left[\frac{1}{3}-2\right]\)

\(\displaystyle = -6\frac{1}{3}\)

However, this is different from the answer provided by the book, which is \(\displaystyle 2ln \frac{3}{2}\). I must have gone wrong somewhere in this; could anyone tell me where?

 

[pka]

I have a problem in my textbook and am not sure how to get started with this. Could anybody offer any suggestions as to how to deal with this:

\(\displaystyle \int^0_{-ln2} \frac{2}{e^x+1}dx\)

This is supposed to be solved by substitution, where \(\displaystyle u = e^x +1\)

This problem cannot be done with that substitution. There is no \(\displaystyle e^xdx \) there.

You must change to an equivalent problem \(\displaystyle \displaystyle \int^0_{-ln2} \frac{2e^{-x}}{1+e^{-x}}dx\)

Now use the substitution \(\displaystyle u=1+e^{-x}\) then \(\displaystyle du=-e^{-x}dx\) is there.

 

[DrPhil]

I do find integration by substitution difficult to get my head around. Having followed up on DrPhil's advice, this is what I got:

\(\displaystyle \int^0_{-ln2} \frac{2}{e^x+1}dx\)

\(\displaystyle u = e^x+1 \)
\(\displaystyle du = e^x dx\)
\(\displaystyle dx = \frac{1}{e^x} du = \frac{1}{u-1} du\)

\(\displaystyle \int^2_\frac{3}{2} \frac{2}{u} \cdot \frac{1}{u-1} du\)......OK to here

\(\displaystyle \int^2_\frac{3}{2} 2u^{-1} \cdot (u-1)^{-1} du \)

You have to expand as partial fractions.

\(\displaystyle \dfrac{2}{u(u-1)} = \dfrac{A}{u} + \dfrac{B}{u-1} \implies A = -2,\;\;\;B = 2 \)

and the integral becomes

\(\displaystyle \displaystyle 2\ \int_{3/2}^2 \left[ \dfrac{1}{u-1} - \dfrac{1}{u} \right] du \)

From there it is a matter of dealing with logarithms to get the book answer (with which I agree).

EDIT: in your previous post, you did have the term du/u, but you did not integrate it correctly. Be careful!

 

[DrPhil]

This problem cannot be done with that substitution. There is no \(\displaystyle e^xdx \) there.

You must change to an equivalent problem \(\displaystyle \displaystyle \int^0_{-ln2} \frac{2e^{-x}}{1+e^{-x}}dx\)

Now use the substitution \(\displaystyle u=1+e^{-x}\) then \(\displaystyle du=-e^{-x}dx\) is there.

Not impossible - and that substitution was specified in the presentation of the problem. The substitution you suggest is quicker - it avoids the partial fractions and integrates a single term.

 

[pka]

Not impossible - and that substitution was specified in the presentation of the problem.

You wrong about the original post. Please look again at the question:

\(\displaystyle \displaystyle\int^0_{-ln2} \frac{2}{e^x+1}dx\)

This is supposed to be solved by substitution, where \(\displaystyle u = e^x +1\)

There is no \(\displaystyle e^xdx\) in that statement. And one cannot be introduced.

There no way to get the \(\displaystyle e^xdx\) into the numerator.

 

[DrPhil]

You wrong about the original post. Please look again at the question:


There is no \(\displaystyle e^xdx\) in that statement. And one cannot be introduced. Multiply by e^x/e^x

There no way to get the \(\displaystyle e^xdx\) into the numerator.

Fortunately, you don't need to have e^x in the numerator! The substitution for dx is.....\(\displaystyle dx = \dfrac{du}{u - 1} \)

The "missing" e^x before the substitution shows up as (u-1) (which is equal to e^x) in the denominator after the substitution. If you check our work, you will find we have accomplished what you say is impossible. Admittedly your substitution is easier, but we didn't have to ignore the last line of the question.

 

[jonnburton]

thanks for your posts. I see that there are a couple of ways of doing this, although in this case I do have to use \(\displaystyle u = e^{x+1}\)

I think I can see how to solve this question now though...

 

[Deleted member 4993]

Hi,

I have a problem in my textbook and am not sure how to get started with this. Could anybody offer any suggestions as to how to deal with this:

\(\displaystyle \int^0_{-ln2} \frac{2}{e^x+1}dx\)

!

u = e^x +1

du = e^x dx → dx = du/e^x → dx = du/(u-1)

when x = - ln(2) → e^x = e^-ln(2) = e^ln(1/2) = 1/2 → u = e^x + 1 = 1.5

when x = 0 → e^x = e⁰⁾ = 1 → u = e^x + 1 = 2

\(\displaystyle \displaystyle \int^0_{-ln2} \frac{2}{e^x+1}dx\)

\(\displaystyle = \ \displaystyle \int^2_{1.5} \frac{2}{u}\frac{du}{u-1}\)

\(\displaystyle = \ \ 2 * \displaystyle \left [ \int^2_{1.5} \frac{du}{u-1} \ - \ \int^2_{1.5} \frac{du}{u} \right ]\)

\(\displaystyle = \ \ 2 * \displaystyle \left [ ln(1) \ - \ ln(0.5) \ - \ ln(2) \ + \ ln(1.5) \right ]\)

= 2 * ln(1.5)

Answer in your book is correct.

 

[HallsofIvy]

thanks for your posts. I see that there are a couple of ways of doing this, although in this case I do have to use \(\displaystyle u = e^{x+1}\)

I think I can see how to solve this question now though...

\(\displaystyle e^{x+1}\)? The problem you first stated had \(\displaystyle e^x+ 1\), a completely different problem.