Area Under The Graph using Rectangles

Estimate the area under the graph of f(x)=1+x^2 from the points x=-1 and x=2 using three rectangles. THEN improve your estimate by using six rectangles. What can you conclude about the area under the curve as the number of rectangles increase?

im not sure how to do this... i know it inovles intergals and maybe Riemann summation? I am not sure...

Thank You So Much!

Re: Area Under The Graph using Rectangles

Quote:

Originally Posted by

**biddum1** im not sure how to do this... i know it inovles intergals and maybe Riemann summation? I am not sure...

Thank You So Much!

Don't worry what it's about, just follow the instructions.

Here's the first pic, for you to label.

http://www.ballooncalculus.org/draw/graph/two.png

Then double the resolution, as they suggested.

Re: Area Under The Graph using Rectangles

I do not how to find the answer with rectangles is what I am saying. Once I can do that i should be set..

Re: Area Under The Graph using Rectangles

Whoops, I'll change those trapezia to rectangles! Does it not specify lower or upper? Do both...

Re: Area Under The Graph using Rectangles

Re: Area Under The Graph using Rectangles

In the first picture in post #2, tom@ballooncalculus used rectangle with base vertices at x= -1, x= 0, x= 1 x= 2 and the height of the rectangle the highest value of the function in each interval. For the first two rectangles that height is $\displaystyle (-1)^2+ 1= 1^2+ 1= 2$ and for the third it is $\displaystyle 2^2+ 1= 5$. Find the areas of the three rectangles and add.

In the first picture, he used rectangles with each height the smallest value of the function in each interval. For the first two rectangles, that is $\displaystyle 0^2+ 1= 1$ and for the third it is $\displaystyle 1^2+ 1= 2$. Find the areas of the three rectangles and add.

(If you were to average those two answer, the result is the same as using the "trapezoid rule".)

In the last post, he used rectangle with height given by the midpoint of each interval: for the first two rectangles, $\displaystyle \left(\frac{1}{2}\right)^2+ 1= \left(\frac{1}{2}\right)^2+ 1= \frac{5}{4}$ while the height of the third rectangle is $\displaystyle \left(\frac{3}{2}\right)^2+ 1= \frac{13}{4}$