1. ## Midpoint Rule

Use the Midpoint Rule to approximate the integral
S(with a= -8 and b= -2)(8x–3x^2)dx
with n=3.

This seems really easy and I've done problems like these before, however I seem to have forgotten.
Can someone please start me out?
Thanks, any input is appreciated.

2. Originally Posted by daneeyah
Use the Midpoint Rule to approximate the integral
S(with a= -8 and b= -2)(8x–3x^2)dx
with n=3.

This seems really easy and I've done problems like these before, however I seem to have forgotten.
Can someone please start me out?
Thanks, any input is appreciated.
So first we find $\Delta x$, that is the width of each subinterval we divide the interval into

recall that $\Delta x = \frac {b - a}n$

now, choose the midpoints in each of the subintervals, and call them $x_1,~x_2,~x_3, ..., ~x_n$

the midpoint rule says: $\int_a^b f(x)~dx \approx \Delta x [f(x_1) + f(x_2) + \cdots + f(x_n)]$

can you continue?

3. Thanks, that clarifies a lot.

So would my intervals be:

-8,-5,-2

That doesn't seem right, but once I get my intervals then it'll be simple from there.

4. Originally Posted by daneeyah
Thanks, that clarifies a lot.

So would my intervals be:

-8,-5,-2

That doesn't seem right, but once I get my intervals then it'll be simple from there.
$\Delta x = \frac {b - a}n = \frac {-2 - (-8)}3 = 2$

thus we have the partitions are $a = -8, -6, -4, -2 = b$

the midpoints, therefore, are: $x_1 = -7, ~x_2 = -5, ~x_3 = -3$

now continue

(did you draw a diagram of this?)