Question

AP Calculus AB:

1. Evaluate the Riemann sum for (x) = x3 − 6x, for 0 ≤ x ≤ 3 with six subintervals, taking the sample points, xi, to be the right endpoint of each interval. Give three decimal places in your answer.

2. Explain, using a graph of f(x), what the Riemann sum in Question #1 represents.

Answer 1

I really have no idea how to start this

Answer 2

you mean \[x^3 - 6x \]

Answer 3

@tburn25 yes sorry about tha

Answer 4

i havent gone past precalc but i understand this; its basically standard deviation

Answer 5

You use the standard deviation formula?

Answer 6

yeah thats what riemanns sum is lol

Answer 7

oh wow i thought it had something to do with integration

Answer 8

since i havent taken calc yet, it probably has something to do with integrals

so you should wait for someone else to answer

Answer 9

so far i did .5(f(.5) + f(1) + .....+ f(3)) and got -7.875

Answer 10

hello.
Pretend our function y=f(x) look like this on [0,3]

we want to find the area between y=f(x) and y=0 (aka x-axis) between the vertical lines x=0 and x=3
we can approximate this area by taking a finite number of rectangles from x=0 to x=3
Let's say we want six sub-intervals
To find the base length of each rectangle (all rectangles will have equal base length) we are going to divide up the length that is from x=0 to x=3 into 6 subintervals... To do you calculate (3-0)/6... And this will be the length of each sub-interval
And it looks like you already did this part from what you just posted right now.
So now we need to calculate the heights of each rectangle using the right endpoint of each sub-interval
the subintervals by the way are as listed:
(0,1/2)
(1/2,1)
(1,3/2)
(3/2,2)
(2,5/2)
(5/2,3)
This is 6 subintervals all with equal length of 1/2.

now to calculate the heights you will do f(of each right endpoint)
It looks like you figure that part out too...
the sum of the areas of each rectangle is given by:
\[\frac{1}{2} f(\frac{1}{2})+\frac{1}{2}f(1)+\frac{1}{2}f(\frac{3}{2})+\frac{1}{2} f(2)+\frac{1}{2}f(\frac{5}{2})+\frac{1}{2}f(3) \\ \text{ or factoring out the } \frac{1}{2} \\ \frac{1}{2}(f(\frac{1}{2})+f(1)+f(\frac{3}{2})+f(2)+f(\frac{5}{2})+f(3))\]