# Sum of points in an exponential equation

• May 6th 2011, 01:29 AM
qformat
Sum of points in an exponential equation
I'm not sure which sub-board this goes in, so I'm sorry if I chose wrong.

How can you find the sum of all points of an exponential within a certain range? That is:

f(x) = a * x ^ y

I figure the result would be same as the area of the region underneath the curve, above the x-axis, and between the two points (start and end), as highlighted pink in the graph (not included due to errors uploading), but I'm not finding a formula for that either.
• May 6th 2011, 02:21 AM
Ackbeet
It looks to me like you've posted in the correct subforum.

I have a question: what is y in your f(x)?
• May 6th 2011, 02:40 AM
qformat
y is an arbitrary real number, and 'a' is a non-negative integer. Only non-negative results are significant. I think I wrote the function wrong. How's:

f(a, x, y) = a * x ^ y
• May 6th 2011, 02:50 AM
Ackbeet
So is x the only independent parameter you are varying? That is, do you just want to fix a and y, and find the area under the curve for some interval p < x < q?
• May 6th 2011, 02:53 AM
qformat
Yes, that's it! Sorry.

EDIT: Now, I wish I spent more time learning proper notation.
• May 6th 2011, 03:19 AM
Ackbeet
So that means you wish to compute

$\int_{p}^{q}a x^{y}\,dx=a\int_{p}^{q}x^{y}\,dx.$

You can use the power rule here:

$\int x^{n}\,dx=\frac{x^{n+1}}{n+1}$

to get the antiderivative, and then plug in your limits. What do you get?
• May 6th 2011, 03:53 AM
qformat
Hang on while I learn Calculus.
• May 6th 2011, 04:08 AM
Ackbeet
Quote:

Originally Posted by qformat
Hang on while I learn Calculus.

Oh. Well, in that case, let me just compute that for you. I get

$a\int_{p}^{q}x^{y}\,dx=\frac{a}{y+1}\left(q^{y}-p^{y}\right).$

This is a problem you would learn how to solve pretty early on in the second semester of calculus (so, after your review of functions, after limits, continuity, and differentiation).

[EDIT]: See below for a correction.
• May 6th 2011, 04:30 AM
qformat
Wow, I was getting something completely different. I want to go back to school. I better go through the rest of these online tutorials before my next problem. I feel like a primitive aborigine with the privilege of being graced by a modern caretaker. Thank you so much.
• May 6th 2011, 04:34 AM
Ackbeet
Correction: it should be

$\frac{a}{y+1}(q^{y+1}-p^{y+1}).$

Try that out.