Consider the function f(x)=x^3. In this problem we will find A=(integral from 0 to b) x^3 dx of the region under the curve y=f(x) and over the x-axis interval [0,b].
Formulas that I came up with:
Delta x =b/N
x-sub k= kb/N
F(x-sub k)=(x-subk)^3
a-subk= (delta x)(f(x-subk)
Find A-subN, the approximation of A using N rectangles. First use sigma notation for your formula, then evaluate the sum by using the sum of cubes formula.
I have attempted this problem but I am unsure whether what i got is right because the next part of the problem is :
Evaluate the limit (integral from 0 to b) x^3 dx = limit as N approaches infinity A-sub N to find the area A.
I get infinity as my answer, however I'm pretty sure the answer should be a number...
you appear to be using a right-hand Riemann Sum. If so, then your formulas are correct. So, your sum is
n
∑ xk^3 * ∆x
k=1
= ∑ (kb/n)^3(b/n) = b^4/n^4 ∑k^3 = b^4/n^4 * n^2(n+1)^2/4
as n -> ∞ that becomes just
b^4/n^4 * n^4/4 = b^4/4