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

**Steelers72** · February 9th 2013, 06:57 PM

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)

My answer =

(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!!

**hollywood** · February 9th 2013, 07:44 PM

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

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

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

- Hollywood

**Steelers72** · February 9th 2013, 07:57 PM

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.

**hollywood** · February 9th 2013, 08:38 PM

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

**Steelers72** · February 9th 2013, 09:04 PM

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?

**Steelers72** · February 10th 2013, 12:45 PM

0 wasn't right; scratch that thought

**hollywood** · February 10th 2013, 12:57 PM

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

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

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

Can you take it from there?

- Hollywood

**Steelers72** · February 10th 2013, 01:37 PM

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.

**hollywood** · February 10th 2013, 09:21 PM

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:

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

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

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

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

- Hollywood

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