Calculus 1 Lecture Notes, Section 4.3

Calc 1 Lecture NotesSection 4.3Page 1 of 6

Section 4.3: Area

Big idea: The area under a curve can be numerically approximated by dividing the area up into thin vertical rectangles of uniform width and of heights determined by the y value of the curve, and then summing the areas of all the rectangles. More rectangles of smaller width result in a better approximation. In the limit of an infinite number of rectangles of infinitesimal width, the sum is exactly the area under the curve.

Big skill: You should be able to approximate the area under a curve by dividing up an interval into rectangles of equal width, computing the height of each rectangle using the given function, and then sum the areas of the rectangles to get your numerical approximation.

“Area under a curve:”

When we talk about the “area under a curve,” we mean the area of the shape bounded by the curve itself, the x axis, and two vertical segments.

The area under the curve y = x2 between x = a = 1 and x = b = 2 is shaded in the diagram below.

To approximate the area under this curve, we divide it up into thin vertical rectangles of equal width Δx. This is called a regular partition of the interval [a, b]. The picture below shows n = 4 rectangles, each with a separate area Ai. Note that the x-values between partitions are labeled
x0, x1, …, xn.

The height of the rectangles can be determined in one of three ways:

• using the y value of the curve at the left endpoint of the rectangle,
• using the y value of the curve at the right endpoint of the rectangle,
• or using the y value of the curve at the midpoint of the rectangle.

Once a method for picking the heights is settled, the area of the shaded region can be approximated as:

Note that .

Note that

A (TI graphing) calculator can be used to do these sums quickly (see pages 11-12 and 11-18 of the TI-83 user’s manual) (use 2nd LIST to access these functions): On your TI calculator:

sum(seq(expression, variable, begin, end, increment))

For this left-endpoint sum: sum(seq(X^2*0.25, X, 1, 1.75, 0.25)) = 1.96875

Note that

For this right-endpoint sum: sum(seq(X^2*0.25, X, 1.25, 2, 0.25)) = 2.71875

For this midpoint sum: sum(seq(X^2*0.25, X, 1.125, 1.875, 0.25)) = 2.328125

Note that

The exact answer is .

Why does the Midpoint evaluation method work best in this case?

In all three cases, we used different x-values in each partition to find the height of the rectangle. To express the idea that the height of the rectangle can be calculated by plugging anyx-value from a partition into the function, we say that for a set of x-valuesfrom each partition
ci [xi-1, xi] that the Riemann Sum(definition 3.2) for that partition is .

The exact value for the area can be calculated as (definition 3.1): . This works for any choice of ci!

Practice:

1. Compute the Riemann sum for 8 divisions using all three evaluation schemes (i.e., left endpoint, right endpoint, and midpoint evaluation).

1. Prove that the area in the examples above is exactly 7/3.

1. Complete the following table of Riemann sums for the function on the interval
   to show that all three evaluation schemes converge to the same value as n.