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:http://www.webassign.net/cgi-perl/sy..._(n)) Delta x)*

(a) Use this definition to find an expression for the area under the curve *y* = *x*^{3} 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 = http://www.webassign.net/cgi-perl/sy...%20%281%29%2Fn

(b) Use the following formula for the sum of the cubes of the first integers to evaluate the limit in part (a). http://www.webassign.net/cgi-perl/sy...%2F2%29%2A%2A2

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

Any help greatly appreciated!!

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

Starting from your expression , you can factor out of the sum, since it doesn't depend on i:

and now the sum is exactly , so you can use the formula you were given.

- Hollywood

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.

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

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?

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

0 wasn't right; scratch that thought

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

Can you take it from there?

- Hollywood

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.

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:

and now you can see that and go to zero as n goes to infinity, and of course stays , so the answer is .

- Hollywood