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
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 = 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 =
(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!!
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?
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.
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