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# Area Under a Curve Help

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By McGraw-Hill Professional
Updated on Sep 9, 2011

## Introduction to Area Under a Curve

Consider the curve shown in Fig. 4.1. The curve is the graph of y = f ( x ). We set for ourselves the task of calculating the area A that is (i) under the curve, (ii) above the x -axis, and (iii) between x = a and x = b . Refer to Fig. 4.2 to see the geometric region we are considering.

Fig. 4.1

Fig. 4.2

We take it for granted that the area of a rectangle of length and width w is × w . Now our strategy is to divide the base interval [ a , b ] into equal subintervals. Fix an integer k > 0. We designate the points

P = { x 0 , x 1 , x 2 , ..., x k },

with x 0 = a and x k = b . We require that ∣ x jx j −1 ∣ = ∣ ba ∣/ k ≡ Δ x for j = 1, ..., k . In other words, the points x 0 , x 1 , ..., x k are equally spaced. We call the set P a partition . Sometimes, to be more specific, we call it a uniform partition (to indicate that all the subintervals have the same length). Refer to Fig. 4.3.

Fig. 4.3

The idea is to build an approximation to the area A by erecting rectangles over the segments determined by the partition. The first rectangle R 1 will have as base the interval [ x 0 , x 1 ] and height chosen so that the rectangle touches the curve at its upper right hand corner; this means that the height of the rectangle is f ( x 1 ). The second rectangle R 2 has as base the interval [ x 1 , x 2 ] and height f ( x 2 ). Refer to Fig. 4.4.

Fig. 4.4

Continuing in this manner, we construct precisely k rectangles, R 1 , R 2 , ..., R k , as shown in Fig. 4.5.

Fig. 4.5

Now the sum of the areas of these rectangles is not exactly equal to the area A that we seek. But it is close. The error is the sum of the little semi-triangular pieces that are shaded in Fig. 4.6.

Fig. 4.6

We can make that error as small as we please by making the partition finer . Figure 4.7 illustrates this idea.

Fig. 4.7

Let us denote by R ( f , P ) the sum of the areas of the rectangles that we created from the partition P. This is called a Riemann sum . Thus

Here the symbol denotes the sum of the expression to its right for each of the instances j = 1 to j = k .  The reasoning just presented suggests that the true area A is given by

We call this limit the integral of f from x = a to x = b and we write it as

Thus we have learned that

It is well to take a moment and comment on the integral notation. First, the integral sign

is an elongated “S,” coming from “summation.” The dx is an historical artifact, coming partly from traditional methods of developing the integral, and partly from a need to know explicitly what the variable is. The numbers a and b are called the limits of integration —the number a is the lower limit and b is the upper limit. The function f is called the integrand .

Before we can present a detailed example, we need to record some important information about sums:

I. We need to calculate the sum . To achieve this goal, we write

S = 1 + 2 + ... + ( N − 1) + N

S = N + ( N − 1) + ... + 2 + 1

Thus

2 S = N · ( N + 1)

or

This is a famous formula that was discovered by Carl Friedrich Gauss (1777-1855) when he was a child.

II. The sum is given by

We shall not provide the details of the proof of this formula, but refer the interested reader to [SCH2].

For our first example, we calculate the area under a parabola.

#### Example

Calculate the area under the curve y = x 2, above the x -axis, and between x = 0 and x = 2.

#### Solution

Refer to Fig. 4.8 as we reason along. Let f ( x ) = x 2.

Fig. 4.8

Consider the partition P of the interval [0, 2] consisting of k + 1 points x 0 , x 1 , ..., x k . The corresponding Riemann sum is

Of course

and

As a result, the Riemann sum for the partition P is

Now formula II above enables us to calculate the last sum explicitly. The result is that

In sum,

We conclude that the desired area is 8/3.

You Try It: Use the method presented in the last example to calculate the area under the graph of y = 2 x and above the x -axis, between x = 1 and x = 2. You should obtain the answer 3, which of course can also be determined by elementary considerations—without taking limits.

Find practice problems and solutions for these concepts at: The Integral Practice Test.

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