# Thread: Exponential Functions and Integrals

1. ## Exponential Functions and Integrals

I have a test tomorrow and I'm not getting the answers on a study guide for a couple and I forgot how I did another and lost all my understanding.

These are the formulas I was given: http://i.imgur.com/6rwL0S6.jpg

This is #4 (which I got close to the answer but not the entire thing): http://i.imgur.com/NaJhF5Y.jpg
My work: http://i.imgur.com/l8OHuCD.jpg

This is #9 (where again I got close to the answer but it's not negative): http://i.imgur.com/irn4fUd.jpg
My work: http://i.imgur.com/VGdxQkx.jpg

This is #11 (which I've done before but looking at it again I'm very confused and have no understanding of my work. It'd be nice if someone could show me this one step by step): http://i.imgur.com/8RKTBfe.jpg
My work: http://i.imgur.com/nexHgNv.jpg

2. ## Re: Exponential Functions and Integrals

Originally Posted by cperzely
I have a test tomorrow and I'm not getting the answers on a study guide for a couple and I forgot how I did another and lost all my understanding.

These are the formulas I was given: http://i.imgur.com/6rwL0S6.jpg

This is #4 (which I got close to the answer but not the entire thing): http://i.imgur.com/NaJhF5Y.jpg
My work: http://i.imgur.com/l8OHuCD.jpg
What you give as the "correct answer", $\frac{x- 2}{(ln(2))x(x-1)}$ is NOT equal to $\frac{2}{ln(2)x}- \frac{1}{ln(2)(x- 1)}$. You appear to have done the "partial fractions" incorrectly.

This is #9 (where again I got close to the answer but it's not negative): http://i.imgur.com/irn4fUd.jpg
My work: http://i.imgur.com/VGdxQkx.jpg
The formula you are given is for $\int a^x dx$ but the problem is $\int 4^{-x} dx$. Either use the substitution u= -x or write the problem as $\int (4^{-1})^x dx= \int \left(\frac{1}{4}\right)^x dx$

This is #11 (which I've done before but looking at it again I'm very confused and have no understanding of my work. It'd be nice if someone could show me this one step by step): http://i.imgur.com/8RKTBfe.jpg
My work: http://i.imgur.com/nexHgNv.jpg
Use the substitution $u= (3- x)^2$

3. ## Re: Exponential Functions and Integrals

Thank you for the help HallsofIvy. I redid #11 again and got the answer but did I solve this in a more difficult way than I needed too? I first substituted for 3-x then substituted again for u^2: http://i.imgur.com/wACU8yG.jpg

Also here's the formula I used for this one (as well as for #4): http://i.imgur.com/5A6nktd.jpg

4. ## Re: Exponential Functions and Integrals

Like HallsofIvy said:

Let $u = (3-x)^2\,\, \therefore \mathrm{d}u = -2(3-x)\,\,\mathrm{d}x$

Now the integral becomes:

$\int (3 -x)7^{(3-x)^2}\,\,\mathrm{d}x = -\frac{1}{2}\int 7^u\,\, \mathrm{d}u$

Now you can finish the rest using the formula you mentioned in your last post.