Using a Riemman Sum to find the area under a curve (n intervals, left endpoint)

[tesla93]

Hello, I'm having a bit of trouble calculating the area under the curve of x^2 on the interval x=-3 to x=1. The question says that there I have to use n subintervals and left endpoints.

Relevant Equations

Δx=b-a/n

x_i=a+(Δx)(i-1) -its i-1 because we're using a left endpoint, otherwise it would be just i.

f(x_i)= (a+(Δx)(i-1))^2(Δx)

The Attempt


Δx=b-a/n
= 1-(-3)/n
=4/n

x_i = -3+4/n(i-1)

f(x_i)= (-3+4/n(i-1))^2(4/n)

=(9-(24i/n)+(24/n)+(16i^2/n^2)-(32i/n^2)-(16/n^2))*(4/n)

I got stuck at this point. I have no idea how to move on from here. Any advice?

 

[lippyka]

The area under the curve is given by the summation of all the f(xi) values. Using the equation you have for f(xi), you can calculate the value for each xi and then sum them up. To do this, start by setting up the summation for the first n terms:A = 4/n*[f(xi1) + f(xi2) + ... + f(xin)]Then, plug in the expression for f(xi) that you derived earlier into the summation. This should give you an expression for A in terms of n. You can then solve this expression for A to find the total area under the curve.