Riemann Sums Problem

Question

Estimate the area under the graph of f(x) = x^2+2x+1 over the interval [−1,1] using ten approximating rectangles and right endpoints.

Rn=   

Repeat the approximation using left endpoints.

Ln=  

Answer 1

The problem is asking us to approximate the area between the graph of f(x) (a parabola) and the x-axis using 10 rectangles.

We'll first divided the interval from -1 to 1 into 10 equal segments each being .2 units wide.

(The spans don't HAVE to be equal, but why make ourselves crazy, right?)

Now we draw 10 rectangles so that their widths are each of our 10 segments, and their heights are determined by where their upper right corners intersect the parabola. (Why, because the problem said to use the right side. The second part of the problem we will use the top left corners.)

Before continuing, look at your drawing.

Notice how tops of the rectangles are above the parabola except for the point of intersection. They include extra area will result in an over approximation.

Can you anticipate it, yet? In the second part of the problem when we use the top-left corners, the tops of the rectangles will be below the parabola resulting in "Uncounted area" and an underestimate.

Can you also see how me could get better and better approximations by drawing more rectangles?

(But, let's get back to finishing the first part of the problem.)

Take note of the x values of the intersections. -.8, -.6, -.4, -.2, 0, .2, .4, .6, .8 & 1.

Calculate the corresponding y's by whatever means you're allowed.

(I'm a grown-up and I'm giving myself permission to use my TI-84.)

The corresponding y values are .04, .16, .36, .64, 1, 1.44, 1.96, 2.56, 3.24 & 4

The areas of the rectangles are found by multiplying the heights (y's) by the corresponding widths (x's) and the total approximation is the sum of the 10 rectangles.

I'll leave the arithmetic to you.

Do you agree, Rieman sums are easy, yet tedious, right?

Have fun.