5.2 The Definite Integral

In the preceding section we defined the area under a curve in terms of Riemann sums:

However, this definition came with restrictions. We required f ( x ) f ( x ) to be continuous and nonnegative. Unfortunately, real-world problems don’t always meet these restrictions. In this section, we look at how to apply the concept of the area under the curve to a broader set of functions through the use of the definite integral.

Definition and Notation

The definite integral generalizes the concept of the area under a curve. We lift the requirements that f ( x ) f ( x ) be continuous and nonnegative, and define the definite integral as follows.

Definition

If f ( x ) f ( x ) is a function defined on an interval [ a , b ] , [ a , b ] , the definite integral of f from a to b is given by

∫ a b f ( x ) d x = lim n → ∞ ∑ i = 1 n f ( x i * ) Δ x , ∫ a b f ( x ) d x = lim n → ∞ ∑ i = 1 n f ( x i * ) Δ x ,

provided the limit exists. If this limit exists, the function f ( x ) f ( x ) is said to be integrable on [ a , b ] , [ a , b ] , or is an integrable function .

The integral symbol in the previous definition should look familiar. We have seen similar notation in the chapter on Applications of Derivatives, where we used the indefinite integral symbol (without the a and b above and below) to represent an antiderivative. Although the notation for indefinite integrals may look similar to the notation for a definite integral, they are not the same. A definite integral is a number. An indefinite integral is a family of functions. Later in this chapter we examine how these concepts are related. However, close attention should always be paid to notation so we know whether we’re working with a definite integral or an indefinite integral.

Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm Leibniz , who is often considered to be the codiscoverer of calculus, along with Isaac Newton. The integration symbol ∫ is an elongated S, suggesting sigma or summation. On a definite integral, above and below the summation symbol are the boundaries of the interval, [ a , b ] . [ a , b ] . The numbers a and b are x-values and are called the limits of integration ; specifically, a is the lower limit and b is the upper limit. To clarify, we are using the word limit in two different ways in the context of the definite integral. First, we talk about the limit of a sum as n → ∞ . n → ∞ . Second, the boundaries of the region are called the limits of integration.

We call the function f ( x ) f ( x ) the integrand , and the dx indicates that f ( x ) f ( x ) is a function with respect to x, called the variable of integration . Note that, like the index in a sum, the variable of integration is a dummy variable , and has no impact on the computation of the integral. We could use any variable we like as the variable of integration:

∫ a b f ( x ) d x = ∫ a b f ( t ) d t = ∫ a b f ( u ) d u ∫ a b f ( x ) d x = ∫ a b f ( t ) d t = ∫ a b f ( u ) d u

Previously, we discussed the fact that if f ( x ) f ( x ) is continuous on [ a , b ] , [ a , b ] , then the limit lim n → ∞ ∑ i = 1 n f ( x i * ) Δ x lim n → ∞ ∑ i = 1 n f ( x i * ) Δ x exists and is unique. This leads to the following theorem, which we state without proof.

Theorem 5.1

Continuous Functions Are Integrable

If f ( x ) f ( x ) is continuous on [ a , b ] , [ a , b ] , then f is integrable on [ a , b ] . [ a , b ] .

Functions that are not continuous on [ a , b ] [ a , b ] may still be integrable, depending on the nature of the discontinuities. For example, functions continuous on a closed interval, apart from a finite number of jump discontinuities, are integrable.

It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not strictly necessary. Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero. This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums.

Example 5.7

Evaluating an Integral Using the Definition

Use the definition of the definite integral to evaluate ∫ 0 2 x 2 d x . ∫ 0 2 x 2 d x . Use a right-endpoint approximation to generate the Riemann sum.

Solution

We first want to set up a Riemann sum. Based on the limits of integration, we have a = 0 a = 0 and b = 2 . b = 2 . For i = 0 , 1 , 2 ,…, n , i = 0 , 1 , 2 ,…, n , let P = < x i >P = < x i >be a regular partition of [ 0 , 2 ] . [ 0 , 2 ] . Then

Since we are using a right-endpoint approximation to generate Riemann sums, for each i, we need to calculate the function value at the right endpoint of the interval [ x i − 1 , x i ] . [ x i − 1 , x i ] . The right endpoint of the interval is x i , x i , and since P is a regular partition,

Thus, the function value at the right endpoint of the interval is

Then the Riemann sum takes the form

∑ i = 1 n f ( x i ) Δ x = ∑ i = 1 n ( 4 i 2 n 2 ) 2 n = ∑ i = 1 n 8 i 2 n 3 = 8 n 3 ∑ i = 1 n i 2 . ∑ i = 1 n f ( x i ) Δ x = ∑ i = 1 n ( 4 i 2 n 2 ) 2 n = ∑ i = 1 n 8 i 2 n 3 = 8 n 3 ∑ i = 1 n i 2 .

Using the summation formula for ∑ i = 1 n i 2 , ∑ i = 1 n i 2 , we have

∑ i = 1 n f ( x i ) Δ x = 8 n 3 ∑ i = 1 n i 2 = 8 n 3 [ n ( n + 1 ) ( 2 n + 1 ) 6 ] = 8 n 3 [ 2 n 3 + 3 n 2 + n 6 ] = 16 n 3 + 24 n 2 + 8 n 6 n 3 = 8 3 + 4 n + 8 6 n 2 . ∑ i = 1 n f ( x i ) Δ x = 8 n 3 ∑ i = 1 n i 2 = 8 n 3 [ n ( n + 1 ) ( 2 n + 1 ) 6 ] = 8 n 3 [ 2 n 3 + 3 n 2 + n 6 ] = 16 n 3 + 24 n 2 + 8 n 6 n 3 = 8 3 + 4 n + 8 6 n 2 .

Now, to calculate the definite integral, we need to take the limit as n → ∞ . n → ∞ . We get

∫ 0 2 x 2 d x = lim n → ∞ ∑ i = 1 n f ( x i ) Δ x = lim n → ∞ ( 8 3 + 4 n + 8 6 n 2 ) = lim n → ∞ ( 8 3 ) + lim n → ∞ ( 4 n ) + lim n → ∞ ( 8 6 n 2 ) = 8 3 + 0 + 0 = 8 3 . ∫ 0 2 x 2 d x = lim n → ∞ ∑ i = 1 n f ( x i ) Δ x = lim n → ∞ ( 8 3 + 4 n + 8 6 n 2 ) = lim n → ∞ ( 8 3 ) + lim n → ∞ ( 4 n ) + lim n → ∞ ( 8 6 n 2 ) = 8 3 + 0 + 0 = 8 3 .

Checkpoint 5.7

Use the definition of the definite integral to evaluate ∫ 0 3 ( 2 x − 1 ) d x . ∫ 0 3 ( 2 x − 1 ) d x . Use a right-endpoint approximation to generate the Riemann sum.

Evaluating Definite Integrals

Evaluating definite integrals this way can be quite tedious because of the complexity of the calculations. Later in this chapter we develop techniques for evaluating definite integrals without taking limits of Riemann sums. However, for now, we can rely on the fact that definite integrals represent the area under the curve, and we can evaluate definite integrals by using geometric formulas to calculate that area. We do this to confirm that definite integrals do, indeed, represent areas, so we can then discuss what to do in the case of a curve of a function dropping below the x-axis.

Example 5.8

Using Geometric Formulas to Calculate Definite Integrals

Use the formula for the area of a circle to evaluate ∫ 3 6 9 − ( x − 3 ) 2 d x . ∫ 3 6 9 − ( x − 3 ) 2 d x .

Solution

The function describes a semicircle with radius 3. To find

we want to find the area under the curve over the interval [ 3 , 6 ] . [ 3 , 6 ] . The formula for the area of a circle is A = π r 2 . A = π r 2 . The area of a semicircle is just one-half the area of a circle, or A = ( 1 2 ) π r 2 . A = ( 1 2 ) π r 2 . The shaded area in Figure 5.16 covers one-half of the semicircle, or A = ( 1 4 ) π r 2 . A = ( 1 4 ) π r 2 . Thus,

∫ 3 6 9 − ( x − 3 ) 2 = 1 4 π ( 3 ) 2 = 9 4 π ≈ 7.069. ∫ 3 6 9 − ( x − 3 ) 2 = 1 4 π ( 3 ) 2 = 9 4 π ≈ 7.069.

Figure 5.16 The value of the integral of the function f ( x ) f ( x ) over the interval [ 3 , 6 ] [ 3 , 6 ] is the area of the shaded region.

Checkpoint 5.8

Use the formula for the area of a trapezoid to evaluate ∫ 2 4 ( 2 x + 3 ) d x . ∫ 2 4 ( 2 x + 3 ) d x .

Area and the Definite Integral

When we defined the definite integral, we lifted the requirement that f ( x ) f ( x ) be nonnegative. But how do we interpret “the area under the curve” when f ( x ) f ( x ) is negative?

Net Signed Area

Let us return to the Riemann sum. Consider, for example, the function f ( x ) = 2 − 2 x 2 f ( x ) = 2 − 2 x 2 (shown in Figure 5.17) on the interval [ 0 , 2 ] . [ 0 , 2 ] . Use n = 8 n = 8 and choose < x i * > < x i * >as the left endpoint of each interval. Construct a rectangle on each subinterval of height f ( x i * ) f ( x i * ) and width Δx. When f ( x i * ) f ( x i * ) is positive, the product f ( x i * ) Δ x f ( x i * ) Δ x represents the area of the rectangle, as before. When f ( x i * ) f ( x i * ) is negative, however, the product f ( x i * ) Δ x f ( x i * ) Δ x represents the negative of the area of the rectangle. The Riemann sum then becomes

∑ i = 1 8 f ( x i * ) Δ x = ( Area of rectangles above the x -axis ) − ( Area of rectangles below the x -axis ) ∑ i = 1 8 f ( x i * ) Δ x = ( Area of rectangles above the x -axis ) − ( Area of rectangles below the x -axis )

Eight rectangles are drawn evenly over [0,2] with heights determined by the value of the function at the left endpoints of each." width="325" height="387" />

Figure 5.17 For a function that is partly negative, the Riemann sum is the area of the rectangles above the x-axis less the area of the rectangles below the x-axis.

Taking the limit as n → ∞ , n → ∞ , the Riemann sum approaches the area between the curve above the x-axis and the x-axis, less the area between the curve below the x-axis and the x-axis, as shown in Figure 5.18. Then,

∫ 0 2 f ( x ) d x = lim n → ∞ ∑ i = 1 n f ( c i ) Δ x = A 1 − A 2 . ∫ 0 2 f ( x ) d x = lim n → ∞ ∑ i = 1 n f ( c i ) Δ x = A 1 − A 2 .

The quantity A 1 − A 2 A 1 − A 2 is called the net signed area .

Notice that net signed area can be positive, negative, or zero. If the area above the x-axis is larger, the net signed area is positive. If the area below the x-axis is larger, the net signed area is negative. If the areas above and below the x-axis are equal, the net signed area is zero.

Example 5.9

Finding the Net Signed Area

Find the net signed area between the curve of the function f ( x ) = 2 x f ( x ) = 2 x and the x-axis over the interval [ −3 , 3 ] . [ −3 , 3 ] .

Solution

The function produces a straight line that forms two triangles: one from x = −3 x = −3 to x = 0 x = 0 and the other from x = 0 x = 0 to x = 3 x = 3 (Figure 5.19). Using the geometric formula for the area of a triangle, A = 1 2 b h , A = 1 2 b h , the area of triangle A₁, above the axis, is

where 3 is the base and 2 ( 3 ) = 6 2 ( 3 ) = 6 is the height. The area of triangle A₂, below the axis, is

where 3 is the base and 6 is the height. Thus, the net area is

Figure 5.19 The area above the curve and below the x-axis equals the area below the curve and above the x-axis.

Analysis

If A₁ is the area above the x-axis and A₂ is the area below the x-axis, then the net area is A 1 − A 2 . A 1 − A 2 . Since the areas of the two triangles are equal, the net area is zero.

Checkpoint 5.9

Find the net signed area of f ( x ) = x − 2 f ( x ) = x − 2 over the interval [ 0 , 6 ] , [ 0 , 6 ] , illustrated in the following image.

Total Area

One application of the definite integral is finding displacement when given a velocity function. If v ( t ) v ( t ) represents the velocity of an object as a function of time, then the area under the curve tells us how far the object is from its original position. This is a very important application of the definite integral, and we examine it in more detail later in the chapter. For now, we’re just going to look at some basics to get a feel for how this works by studying constant velocities.

When velocity is a constant, the area under the curve is just velocity times time. This idea is already very familiar. If a car travels away from its starting position in a straight line at a speed of 70 mph for 2 hours, then it is 140 mi away from its original position (Figure 5.20). Using integral notation, we have

Figure 5.20 The area under the curve v ( t ) = 70 v ( t ) = 70 tells us how far the car is from its starting point at a given time.

In the context of displacement, net signed area allows us to take direction into account. If a car travels straight north at a speed of 60 mph for 2 hours, it is 120 mi north of its starting position. If the car then turns around and travels south at a speed of 40 mph for 3 hours, it will be back at it starting position (Figure 5.21). Again, using integral notation, we have

In this case the displacement is zero.

Figure 5.21 The area above the axis and the area below the axis are equal, so the net signed area is zero.

Suppose we want to know how far the car travels overall, regardless of direction. In this case, we want to know the area between the curve and the x-axis, regardless of whether that area is above or below the axis. This is called the total area .

Graphically, it is easiest to think of calculating total area by adding the areas above the axis and the areas below the axis (rather than subtracting the areas below the axis, as we did with net signed area). To accomplish this mathematically, we use the absolute value function. Thus, the total distance traveled by the car is

∫ 0 2 | 60 | d t + ∫ 2 5 | −40 | d t = ∫ 0 2 60 d t + ∫ 2 5 40 d t = 120 + 120 = 240. ∫ 0 2 | 60 | d t + ∫ 2 5 | −40 | d t = ∫ 0 2 60 d t + ∫ 2 5 40 d t = 120 + 120 = 240.

Bringing these ideas together formally, we state the following definitions.

Definition

Let f ( x ) f ( x ) be an integrable function defined on an interval [ a , b ] . [ a , b ] . Let A₁ represent the area between f ( x ) f ( x ) and the x-axis that lies above the axis and let A₂ represent the area between f ( x ) f ( x ) and the x-axis that lies below the axis. Then, the net signed area between f ( x ) f ( x ) and the x-axis is given by

The total area between f ( x ) f ( x ) and the x-axis is given by