Sums involving functions of the form y=a^x

• Feb 16th 2011, 05:55 AM
quantoembryo
Sums involving functions of the form y=a^x
I have my calculus 2 midterm today, and after going through the chapters I have found the only reccuring issue I have is with functions of the form $y=a^x$.

If someone here could help explain how to solve these function for areas without using integral calculus I would greatly appreciate it.

I'll give an example and my steps involved:

Solve for the area under the function $y=e^x$ on the interval $[0, b]$.

Let $x_i=0+(b/n)i$, and thus $y_i=e^{(bi)/n$

$\displaystyle\sum_{i=1}^{n}{y_i\Delta{x}$

${\displaystyle\sum_{i=1}^{n}{e^{(bi)/n}(b/n)$

$\displaystyle\sum_{i=1}^{n}{e^{(bi)/n}(b/n)$

$(b/n)\displaystyle\sum_{i=1}^{n}{e^{(bi)/n}$

$(b/n)\frac{e^{b/n}-e^{b+1}}{1-e^{b/n}}$

It's at this point that I become flustered. I understand there is manipulation that must occur, but can someone give me tips or show me how, with ease? Keep in mind I am not a math wizard.
• Feb 16th 2011, 06:49 AM
Ackbeet
I think you performed the sum incorrectly (last line):

$(b/n)\displaystyle\sum_{i=1}^{n}{e^{(bi)/n}=(b/n)\frac{e^{b/n}-e^{b+b/n}}{1-e^{b/n}}.$

You now need to compute the limit:

$\displaystyle\lim_{n\to\infty}(b/n)\frac{e^{b/n}-e^{b+b/n}}{1-e^{b/n}}.$

Can you do that?
• Feb 16th 2011, 07:06 AM
quantoembryo
Sorry, that was a typo. The issue I have is calculating the limit, because it's obvious the answer is not zero. I'm not quite sure how to manipulate it such that I can calculate it.
• Feb 16th 2011, 07:15 AM
Ackbeet
Are you allowed to use L'Hopital's Rule?
• Feb 16th 2011, 07:55 AM
quantoembryo
I would assume so. It doesn't really state. How would you use this rule?
• Feb 16th 2011, 08:02 AM
Ackbeet
Quote:

Originally Posted by quantoembryo
I would assume so. It doesn't really state. How would you use this rule?

Well, you have to get your fraction either into the form $0/0$ or $\infty/\infty.$ Try this:

$\displaystyle\lim_{n\to\infty}(b/n)\frac{e^{b/n}-e^{b+b/n}}{1-e^{b/n}}=b(1-e^{b})\lim_{n\to\infty}(1/n)\frac{e^{b/n}}{1-e^{b/n}}$

$=\displaystyle b(1-e^{b})\left[\lim_{n\to\infty}e^{b/n}\right]\left[\lim_{n\to\infty}\frac{1/n}{1-e^{b/n}}\right].$

The limit on the left is just 1. The limit on the right is ready for L'Hopital's rule, which states that, under certain conditions, you can compute

$\displaystyle\lim_{n\to\infty}\frac{f(n)}{g(n)}=\l im_{n\to\infty}\frac{f'(n)}{g'(n)}.$

So what do you get when you apply the rule?
• Feb 16th 2011, 08:17 AM
quantoembryo
Would the answer be 1-e^b?
• Feb 16th 2011, 08:19 AM
Ackbeet
Nope, not quite. Can you show your work?
• Feb 16th 2011, 08:48 AM
quantoembryo
$\frac{f(n)}{g(n)}=\lim_{n\to\infty}\frac{f'(n)}{g' (n)}$

$\frac{f(n)}{g(n)}=\lim_{n\to\infty}\frac{-1/n^2}{e^{b/n}b/n^2}$

$\frac{f(n)}{g(n)}=\lim_{n\to\infty}\frac{-1/n^2}{e^{b/n}b/n^2}=\frac{-1}{e^{b/n}b}$

and therefore,
$=b(1-e^{b})(1)(\frac{-1}{e^{b/n}b})$

$=\frac{e^b-1}{e^{b/n}}$
• Feb 16th 2011, 08:50 AM
Ackbeet
The way you're writing it is a bit off. Keep the limit sign there! Your final result should not have an n in it, because you're taking the limit as n approaches infinity.