I'm having a really hard time grasping the concept of an integral/area of a region bounded a function.

Let's use $x^3$ as our sample function.

I understand the concept is to create an infinite number of infinitely small rectangles, calculate and sum their area. Using the formula

$$\text{Area}=\lim_{n\to\infty}\sum_{i=1}^n f(C_i)\Delta x=\lim_{n\to\infty}\sum_{i=1}^n\left(\frac{i}n\right)^3\left(\frac1n\right)$$

I understand that $n$ is to represent the number of rectangles, $\Delta x$ is the change in the $x$ values, and that we are summing the series, but I still don't understand what $i$ and $f(C_i)$ do. Is $f(C_i)$ just the value of the function at that point, giving us area?

Sorry to bother you with a homework question. I know how annoying that can be.

P.S. Is there a correct way to enter formulas?

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    $\begingroup$ See here for a quick reference for entering math. $\endgroup$ – Daryl Dec 4 '12 at 4:10
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    $\begingroup$ I took the liberty of changing your $dx$ to the more correct $\Delta x$. $\endgroup$ – Brian M. Scott Dec 4 '12 at 4:12
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    $\begingroup$ No need to apologize about a homework question: you are clearly upfront about that, and you took the time to clarify what you get and specify what you don't get. So your question is perfectly appropriate! $\endgroup$ – amWhy Dec 4 '12 at 4:12

The index of summation $i$ is just a dummy variable that runs through the values $1,2,\dots,n$ and lets us summarize the expression


as the single expression


The points $C_i=\dfrac{i}n$ for $i=1,2,\dots,n$ are the right endpoints of the intervals $\left[0,\frac1n\right],\left[\frac1n,\frac2n\right],\dots\,$, $\left[\frac{n-1}n,\frac{n}n\right]$ into which we’ve subdivided the interval $[0,1]$, and the numbers $f(C_i)$ are indeed just the values of the function $f(x)=x^3$ at those points. The function value $f\left(\frac{i}n\right)$ gives us the height of the rectangle whose base is the interval $\left[\frac{i-1}n,\frac{i}n\right]$ and whose height is the height of the function at the right endpoint of that interval. This area approximates the area under the curve between $x=\frac{i-1}n$ and $x=\frac{i}n$, so the summation, which adds up similar approximations for all of the small subintervals, approximates the total area under $y=x^3$ between $x=0$ and $x=1$.

The final step is taking the limit of better and better approximations over finer and finer subdivisions of $[0,1]$.

The points $C_i$ were found by dividing the interval $[0,1]$ into $n$ equal pieces and finding the right endpoints of those pieces. Since the length of $[0,1]$ is $1$, each piece must have length $\frac1n$. The first begins at $x=0$, so its right endpoint must be at $\frac1n$. The next begins at $\frac1n$, so its right endpoint must be at $\frac1n+\frac1n=\frac2n$. The third begins at $\frac2n$, so its right endpoint must be at $\frac2n+\frac1n=\frac3n$. Continuing in this fashion, you can see that the right endpoint of the $i$-th subinterval must be at $\frac{i}n$.

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  • $\begingroup$ OK. So just out of curiosity, does this method return a approximation of area, or an exact value? $\endgroup$ – OneChillDude Dec 4 '12 at 4:25
  • $\begingroup$ @bwheeler96: The individual summations are approximations; their limit is the exact value. In fact, the area of regions that cannot be chopped up and rearranged into rectangles can be defined as that limit. $\endgroup$ – Brian M. Scott Dec 4 '12 at 4:31

So, $f(C_i)$ is the value of $f$ at $C_i$, but more importantly it is the height of the specific rectangle being used in the approximation. Then $i$ is just the interval which is the base of the rectangle. As $|C_{i+1}-C_i|\rightarrow 0$, this sum becomes the area under the curve.

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  • $\begingroup$ OK. What I meant to say was how do I find $(C_i)$ ? Is it a constant? $\endgroup$ – OneChillDude Dec 4 '12 at 4:20

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