How do I find the average of a section of a curve?

Question

I've done some research already and discovered that the formula I need to do this is

$$\frac{1}{b-a}\int_a^b f(x)\;dx$$

With $a$ and $b$ being the start and end points of the section of curve I want to get the average of.

What I need to know is how I translate a maths $y = ???$ (featuring $x$) function into $f(x)\;dx$, so I can actually work out what the formula I need is.

I gather that $f(x)$ means "a formula of a curve" and that $f(x)\;dx$ means a formula for figuring out the area beneath the curve of the given $f(x)$, but I haven't been able to find a manual for how to translate a given $f(x)$ into the $f(x)\;dx$ that gives it's area.

Can somebody help me with this?

Answer 1

No offense, but your math seems to be far behind of what you wish to achieve with it. If you can live with a value near the average I suggest you take some values between $a$ and $b$, like say

$x_i = a + i\frac{b-a}{n}$

for some fixed numer $n$, calculate the function at these values ($f(x_i)$), sum them all up and divide by $n$. The higher your $n$ the more accurate your average will be.

The big-stretched-out-"S" as you call it (integral) does exactly that, but with $n$ going to infinity. If you want to really understand this - and it is a really beautiful theory with some unimaginable results, everybody should see it - you should read some of the standard literature on calculus. This might also be a promising start:

http://en.wikipedia.org/wiki/Integral_calculus

To interpret the integral as the area between the curve and the x axis is only one way to see it. In my view this way does not help a lot in this case.

Answer 2

Well, if you have an equation $y=x^2$ then take $f(x)=x^2$, meaning that just take $y=f(x)$. If you are not familiar with calculus and insist on finding the average this way, then there isn't a good way for me to explain the dx part. Just write the dx in the integral. Also do you know how to compute the integral $\int_a^b f(x)dx$?

So if you want to find the average value of $y=x^2$ between $x=7$ and $x=9$, then it is

Average = $\frac{1}{9-7}\int_7^9 x^2dx = \frac{1}{2}\cdot\frac{x^3}{3}|_7^9 = \frac{729}{6}-\frac{343}{6}=\frac{386}{6}=64 \frac{1}{3}.$

Also taking y=f(x) is not always possible. The equation must be y = a function of x and x alone. There shouldn't be any y's on the right hand side and the right hand side must pass the vertical line test.