How Is Area Related To A Definite Integral?

You might have seen the marketing campaign for the Super Bass-o-Matic 76.

Let’s look at some options for finding the increased costs when changing production of Super Bass-o-Matic blenders  from 200 blenders to 800 blenders. Suppose the marginal cost is given by

\displaystyle {C}'(x)=30-0.02x {\text{ dollars per blender}}

By finding the area of rectangles, we get values that indicate an increase in cost. For instance, the first term of the left hand sum below,

\displaystyle \left( 26\frac{\$}{\text{blender}} \right)\left( 100\text{ blenders} \right)=\$2600

This means that if we estimate the cost per blender as 26 dollars per blender for all 100 blenders from 200 to 300 blenders, the increased cost is $2600. If we continue this for more rectangles until 800, the sum of the areas give an estimate for the increased cost from increasing production from 200 to 800.

In this case the left hand sum is $11400 and the right hand sum is $12600…the difference between these estimates is $1200.

If we do the the same process for 60 rectangles (each being 10 wide), we get additional estimates.

In this case, the left hand sum is $11940 and the right hand sum is 12060. The difference between the estimate for 60 rectangles is $120. Ten times as many rectangles leads to a tenth of the difference.

We get a similar picture for 600 rectangles.

The left hand sum is $11994 and the right hand sum is $12006. In this case the difference between the estimates is $12. Also notice that as the number of rectangles gets larger, the area in the rectangles looks more and more like the area under the function and above the horizontal axis.

Let’s summarize what happens as the number of rectangles increases.

Number of Rectangles
















You might guess that 6000 rectangles would lead to a difference of $1.2 and you would be correct. The left hand sum is $11999.4 and the right hand sum is $12000.6.

In fact, as the number of rectangles gets larger and larger (each rectangle getting narrower and narrower) each estimate gets closer and closer to $12000.

This is the exact area in the blue shaded region under the function, above the horizontal axis, and between 200 and 800. To convince yourself that this is the case, we can this area by breaking the region up into a rectangle and triangle.

The area is

\displaystyle 14\cdot 600+\frac{1}{2}\cdot 12\cdot 60=12000

The exact area under the function is represented with a definite integral,

\displaystyle \int\limits_{200}^{800}{(30-0.02x)\,dx}=12000