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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?

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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.

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Ah! so it's just as Sebastian's link explained it. To get f(x)dx, you simply reverse the process for getting your formula's primative - you need to figure out what formula your base formula is the primative of, itself. – Cambot Oct 16 '12 at 18:35
A small correction. To get f(x), you just use the function y. To get $\int f(x)dx$, you use the antiderivative. I am assuming "primative" means derivative. – Fixed Point Oct 16 '12 at 19:42
I'm not sure. I remember in Math's class they taught us how to get primatives, which are the formulas for the gradiant of a curve at any give point. i.e. the primative of x-squared is 2x, the primative of sin(x) is cos(x). – Cambot Oct 17 '12 at 5:04

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:

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.

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Yes, I was aware of the theory behind it. What I wanted to know was how to translate a straight curve-drawing equation into it's associated average equation. Figuring out such an equation would be far more accurate and likely far less time consuming then calculating the area of many small sub-divisions beneath the curve and averaging them. Thanks for the wiki link. My understanding of what it says I need to do to get the area formula is just reverse the process I'd use on the base formula to get it's primative. I remember a bit about primatives from highschool, so this should help. Thanks. – Cambot Oct 16 '12 at 17:08

we usually don't sum all the approximating rectangles, but perform that task indirectly by invoking the fundamental theorem of calculus.

we suppose that $$F(t)=\int_a^t f(x)dx$$ represents the area under the curve (or more correctly, the net area between the curve and the x-axis) of $f$ where $a\le x\le t$.

Now consider the derivative of A (the function whose values represent the slope of $A$) $$F'(t)=\lim_{h\to 0} \frac{F(t+h)-F(t)}{h}$$ $$F'(t)=\lim_{h\to 0} \frac{\int_a^{t+h} f(t)dt-\int_a^{t} f(t)dt}{h}$$ $$F'(t)=\lim_{h\to 0} \frac{\int_a^{t} f(t)dt+\int_t^{t+h} f(t)dt-\int_a^{t} f(t)dt}{h}$$ $$F'(t)=\lim_{h\to 0} \frac{\int_t^{t+h} f(t)dt}{h}$$

enter image description here

Now think about what would happen to the shaded region to the right (between $t$ and $t+h$) were $h$ to become very small. The left and right sides of the region would become closer to each other in length and the region would resemble a rectangle with a width of $h$ and a height of $f(t)$. thus $$F'(t)=\lim_{h\to 0} \frac{\int_t^{t+h} f(t)dt}{h}$$ $$F'(t)=\lim_{h\to 0} \frac{hf(t)}{h}$$ $$F'(t)=f(t)$$ Put another way, we say that the area function ($F$) is some anti-derivative of $f$.

For example, if we know that the antidervivative of $x^2$ is $x^3/3+C$ (we have an unknown constant because parallel curve have the same slope), we know that the area between where x ranges between $1$ and $2$ would be $2^3/3+C$, which is a pretty useless result. To fix the problem we note that area of a region is just the sum of the areas of its components.

enter image description here

In the second graphic, the total area under $f$ between $c$ and $b$ is just the sum of the blue and green areas. In the language of integrals, $$\int_c^bf(x)dx = \int_c^af(x)dx + \int_a^bf(x)dx$$ thus $$\int_a^bf(x)dx = \int_c^bf(x)dx - \int_c^af(x)dx$$ So to solve the area problem, we need to evaluate 2 anti-derivatives, not one. In our example $$\int_1^2 x^2dx = (2^3/3+C)-(1^3/3+C)=(8-1)/3=7/3$$ Notice how nicely the $C$s cancel out.

By the way, in single variable calculus, you can think of $dx$ as an infinitely small non-zero positive $\Delta x$. Note that $d$ and $\Delta$ are the Latin and Greek representations for the first sound in the word difference.

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