midpoint Riemann sum

[catalle]

Problem says:
Sketch a graph of the integrand and then show the setup for using a midpoint Riemann sum to approximate int((-2x+12)dx) on [0, 5] using five rectangles of equal base.

I put it in the calc, then sketched the graph, starting at the point y=12, and ends at y=0 when x=6
now what? each rectangle's base would equal 1
what point do I start to graph midpoint Riemann sums?

Tried drawing it, looks really messy and is going to hinder the calculation process -_-

 

[stapel]

What is the midpoint of the first interval? :wink:

Eliz.

 

[catalle]

Re:

What is the midpoint of the first interval? :wink:

Eliz.

is it 3?
but how will the graph look? :?:
do I go down 2 right 1 cause of the -2x slope
are any part of the rectangles above the curve, any space that isn't covered beneath the curve?

 

[catalle]

nevermind
would it be 2.5?
then the graph would have rectangles at every half?

 

[catalle]

yep I don't know how to solve this, please help!

OK so what #s do I plug into the \(\displaystyle (-2x+12) dx\) ?

thanks to some java grapher I finally graphed the integrand and divided it into 5 rectangles
base width of all rectangles is 1
height of rectangles:
1st: h=11 a=11
2nd: h=9 a=9
3rd: h=7 a=7
4th: h=5 a=5
5th: h=3 a=3

\(\displaystyle 1(11+9+7+5+3)=35\)

however, what to do about the integrand? does it play any role in finding the area?

 

[mmm4444bot]

… thanks to some java grapher I finally graphed the integrand …

Good grief.

(At the risk of having one of my posts censored, again, I state my unsolicited opinion that people who cannot graph lines are simply torturing themselves by trying to learn calculus.)

Do you know?

You're trying to approximate the area enclosed by the line and the positive x- and y-axes by constructing those five rectangles and adding their areas.

You found five rectangles. You added their areas. You're done.

(Well, you're done with the approximation part. I cannot help you with the sketching part unless you first explain why you cannot graph a line.)

Now you're trying to ask whether or not the value of an integral depends upon it's integrand?

I'm confident that you can answer this question on your own by repeatedly changing the slope in the given integrand and recalculating the new approximated value of the resulting integral.

You also asked something about the overages and underages. I think that the areas of the overages match the areas of the underages in this exercise, so it all evens out in the end. Let's see.

The triangular region of the requested area has base 5 and height 10.

The formula Area = (1/2) * Base * Height gives 25 square units.

The remaining rectangular region of the requested area has length 5 and width 2.

The formula Area = Length * Width gives 10 square units.

25 + 10 = 35

Yup, it looks like the approximation is actually exact. This tells us that all of the overages in the mid-point approximation method "fill in" all of the resulting underages, with this integral.

This happens, when finding the area underneath a straight line, because of symmetry.