# Thread: Evaluating limit of an area under the graph of a continuous function.

1. ## Evaluating limit of an area under the graph of a continuous function.

The area A of the region S that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles:
(a) Use this definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit

A= lim n-->infinity ( f(x1)1/n +f(x2)1/n + ...+f(xn)1/n )

x1= x1+deltax = 1*deltax
x2= 2deltax
x3= 3deltax

xi=ideltax= i(1/n)

(b) Use the following formula for the sum of the cubes of the first integers to evaluate the limit in part (a).

I am not sure at all how to solve this part.

Any help greatly appreciated!!

2. ## Re: Evaluating limit of an area under the graph of a continuous function.

Starting from your expression $\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \left(\frac{i}{n}\right)^3 \frac{1}{n}$, you can factor $\displaystyle \frac{1}{n^4}$ out of the sum, since it doesn't depend on i:

$\displaystyle \lim_{n\to\infty}\frac{1}{n^4}\sum_{i=1}^ni^3$

and now the sum is exactly $\displaystyle 1^3+2^3+\dots+n^3$, so you can use the formula you were given.

- Hollywood

3. ## Re: Evaluating limit of an area under the graph of a continuous function.

Thanks Hollywood! Since the numbers are infinite, how do I know where to stop adding? It's asking for an exact numerical answer some I'm unsure of the exact amount.

4. ## Re: Evaluating limit of an area under the graph of a continuous function.

You first evaluate the *finite* sum to get a function of n, then take the limit of this function as n goes to infinity.

- Hollywood

5. ## Re: Evaluating limit of an area under the graph of a continuous function.

Ok, after plugging in 1^3 and 2^3 into ( n(n+1)/2 )^2

I got 1+1296 which equals 1297.

Now I use the 1/n^4 summation i^3 as n-->infinity ?

Would the answer be 0 because 1/large = approaches zero?

6. ## Re: Evaluating limit of an area under the graph of a continuous function.

0 wasn't right; scratch that thought

7. ## Re: Evaluating limit of an area under the graph of a continuous function.

$\displaystyle \lim_{n\to\infty}\frac{1}{n^4}\sum_{i=1}^ni^3 =$

$\displaystyle \lim_{n\to\infty}\frac{1}{n^4}\left(\frac{n(n+1)}{ 2}\right)^2 =$

$\displaystyle \lim_{n\to\infty}\frac{1}{n^4}\frac{n^4+2n^3+n^2}{ 4}$

Can you take it from there?

- Hollywood

8. ## Re: Evaluating limit of an area under the graph of a continuous function.

Thanks for your help hollywood; this calculus is new to me so I'm slow at learning it. thanks for the patience. Do we have to plug in 1^3 and 2^3 into that formula?

Or if we are solving for n here is my work:

1/n^4 is zero since a small # over an infinitely large number = 0. However, the n^4+2n^3 +n^2/4 part would be infinity since a very large # over a small number goes to infinity. 0*infinity would be zero though? Am I overlooking something? Sorry for my naivety.

9. ## Re: Evaluating limit of an area under the graph of a continuous function.

Yes, both the numerator and denominator go to infinity as n goes to infinity. But you can simplify the expression to something that you can take the limit of:

$\displaystyle \lim_{n\to\infty}\frac{1}{n^4}\frac{n^4+2n^3+n^2}{ 4}=$

$\displaystyle \lim_{n\to\infty}\frac{n^4+2n^3+n^2}{ 4n^4}=$

$\displaystyle \lim_{n\to\infty}\frac{1}{4}+\frac{1}{2n}+\frac{1} {4n^2}$

and now you can see that $\displaystyle \frac{1}{2n}$ and $\displaystyle \frac{1}{4n^2}$ go to zero as n goes to infinity, and of course $\displaystyle \frac{1}{4}$ stays $\displaystyle \frac{1}{4}$, so the answer is $\displaystyle \frac{1}{4}$.

- Hollywood

,

,

,

### the following formula for the sum of the cubes of the first n integers is provided in appendix E. use it to evaluate the limit in part a 1^3 2^3 3^3....n^3=[n(n 1)/2]^2

Click on a term to search for related topics.