Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
See here for a quick reference for entering math. –  Daryl Dec 4 '12 at 4:10
2  
I took the liberty of changing your $dx$ to the more correct $\Delta x$. –  Brian M. Scott Dec 4 '12 at 4:12
2  
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! –  amWhy Dec 4 '12 at 4:12

2 Answers 2

up vote 2 down vote accepted

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

$$\left(\frac1n\right)^3\left(\frac1n\right)+\left(\frac2n\right)^3\left(\frac1n\right)+\left(\frac3n\right)^3\left(\frac1n\right)+\ldots+\left(\frac{n}n\right)^3\left(\frac1n\right)$$

as the single expression

$$\sum_{k=1}^n\left(\frac{i}n\right)^3\left(\frac1n\right)\;.$$

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 $\frac{i}n$.

share|improve this answer
    
OK. So just out of curiosity, does this method return a approximation of area, or an exact value? –  Brian Wheeler Dec 4 '12 at 4:25
    
@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. –  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.

share|improve this answer
    
OK. What I meant to say was how do I find $(C_i)$ ? Is it a constant? –  Brian Wheeler Dec 4 '12 at 4:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.