The area we are trying to find is shaded in this graph:

Since* *`n= 5`, the width of each rectangle will be:

`h=Deltax=(b-a)/n=(1-0.5)/5=0.1`

We aim to find the sum of the areas of the following 5 rectangles:

Now the height of each rectangle is given by the function value for that particular *x*-value.

For example, since `y = f(x) = 1 − x^2`, the first rectangle has height given by:

f(0.5) = 1 − (0.5)^{2}= 0.75

It has area given by:

`"Area"_1= 0.75 × 0.1 = 0.075`

The second rectangle has height:

f(0.6) = 1 − (0.6)^{2}= 0.64

The 5th rectangle has height

f(0.9) = 1 − (0.9)^{2}= 0.19

Adding the areas together gives us the following. (We are writing it using summation notation, which just means the sum of the 5 rectangles. Also, we are adding the heights first then multiplying by the width, which is the same for each rectangle.)

`A=sum_(i=1)^5A_i`

`=(0.75+0.64+0.51+` `0.36+` `{:0.19)(0.1)`

`=2.45(0.1)`

`=0.245`

In the above answer, we are finding the area of the "outer" rectangles. To find a better approximation, we could also find the area of the **inner rectangles**, and then average the 2 results. The graph for the inner rectangles is as follows:

And this is the sum of the areas for the inner rectangles (the 5th one has height 0, so area 0):

`A=sum_(i=1)^5A_i`

` = (0.64+0.51+0.36+` `0.19+` `{:0)(0.1)`

`=1.7(0.1)`

`=0.17`

The average of the 2 areas is given by: `(0.245 + 0.17)/2 = 0.2075`.

A **third way** of doing this problem would be to find the **mid-point rectangles**. The diagram for this would be:

This time our area is

`(0.6975 + 0.5775 +` ` 0.4375 +` ` 0.2775 +` ` {:0.0975) × 0.1` `= 0.20875`

(The first one comes from *f*(0.55) = 1 − (0.55)^{2} = 0.6975).

This answer is slightly above the average of the outer and inner rectangles, and less work!