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Area and the Definite Integral

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

Net Signed Area

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

\[\sum_{i = 1}^{8} f \left(\right. x_{i}^{\star} \left.\right) \Delta x = \left(\right. \text{Area of rectangles above the} x -\text{axis} \left.\right) - \left(\right. \text{Area of rectangles below the} x -\text{axis} \left.\right)\]
A graph of a downward opening parabola over [-1, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). Eight rectangles are drawn evenly over [0,2] with heights determined by the value of the function at the left endpoints of each.
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 \(n \rightarrow \infty ,\) 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,

\[\begin{aligned} \int_{0}^{2} f \left(\right. x \left.\right) d x & = \underset{n \rightarrow \infty}{\text{lim}} \sum_{i = 1}^{n} f \left(\right. c_{i} \left.\right) \Delta x \\ & = A_{1} - A_{2} . \end{aligned}\]

The quantity \(A_{1} - A_{2}\) is called the net signed area.

A graph of a downward opening parabola over [-2, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). The area in quadrant one under the curve is shaded blue and labeled A1. The area in quadrant four above the curve and to the left of x=2 is shaded blue and labeled A2.
Figure 5.18 In the limit, the definite integral equals area A1 less area A2, or 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 \(f \left(\right. x \left.\right) = 2 x\) and the x-axis over the interval \(\left[\right. −3 , 3 \left]\right. .\)

Solution

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

\[A_{1} = \frac{1}{2} 3 \left(\right. 6 \left.\right) = 9 ,\]

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

\[A_{2} = \frac{1}{2} \left(\right. 3 \left.\right) \left(\right. 6 \left.\right) = 9 ,\]

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

\[\int_{−3}^{3} 2 x d x = A_{1} - A_{2} = 9 - 9 = 0 .\]
A graph of an increasing line over [-6, 6] going through the origin and (-3, -6) and (3,6). The area under the line in quadrant one over [0,3] is shaded blue and labeled A1, and the area above the line in quadrant three over [-3,0] is shaded blue and labeled A2.
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 \(A_{1} - A_{2} .\) Since the areas of the two triangles are equal, the net area is zero.

Checkpoint 5.9

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

A graph of an increasing line going through (-2,-4), (0,-2), (2,0), (4,2) and (6,4). The area above the curve in quadrant four is shaded blue and labeled A2, and the area under the curve and to the left of x=6 in quadrant one is shaded and labeled A1.

Total Area

One application of the definite integral is finding displacement when given a velocity function. If \(v \left(\right. t \left.\right)\) 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

\[\int_{0}^{2} 70 d t = 140 .\]
A graph in quadrant 1 with the x-axis labeled as t (hours) and y-axis labeled as v (mi/hr). The area under the line v(t) = 70 is shaded blue over [0,2].
Figure 5.20 The area under the curve \(v \left(\right. t \left.\right) = 70\) 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

\[\begin{aligned} \int_{0}^{2} 60 d t + \int_{2}^{5} −40 d t & = 120 - 120 \\ & = 0. \end{aligned}\]

In this case the displacement is zero.

A graph in quadrants one and four with the x-axis labeled as t (hours) and the y axis labeled as v (mi/hr). The first part of the graph is the line v(t) = 60 over [0,2], and the area under the line in quadrant one is shaded. The second part of the graph is the line v(t) = -40 over [2,5], and the area above the line in quadrant four is shaded.
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

\[\begin{aligned} \int_{0}^{2} \left|\right. 60 \left|\right. d t + \int_{2}^{5} \left|\right. −40 \left|\right. d t & = \int_{0}^{2} 60 d t + \int_{2}^{5} 40 d t \\ & = 120 + 120 \\ & = 240. \end{aligned}\]

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

Definition

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

\[\int_{a}^{b} f \left(\right. x \left.\right) d x = A_{1} - A_{2} .\]

The total area between \(f \left(\right. x \left.\right)\) and the x-axis is given by

\[\int_{a}^{b} \left|\right. f \left(\right. x \left.\right) \left|\right. d x = A_{1} + A_{2} .\]

Example 5.10

Finding the Total Area

Find the total area between \(f \left(\right. x \left.\right) = x - 2\) and the x-axis over the interval \(\left[\right. 0 , 6 \left]\right. .\)

Solution

Calculate the x-intercept as \(\left(\right. 2 , 0 \left.\right)\) (set \(y = 0 ,\) solve for x). To find the total area, take the area below the x-axis over the subinterval \(\left[\right. 0 , 2 \left]\right.\) and add it to the area above the x-axis on the subinterval \(\left[\right. 2 , 6 \left]\right.\) (Figure 5.22).

A graph of a increasing line f(x) = x-2 going through the points (-2,-4), (0,2), (2,0), (4,2), and (6,4). The area under the line in quadrant one and to the left of the line x=6 is shaded and labeled A1. The area above the line in quadrant four is shaded and labeled A2.
Figure 5.22 The total area between the line and the x-axis over \(\left[\right. 0 , 6 \left]\right.\) is A2 plus A1.

We have

\[\int_{0}^{6} \left|\right. \left(\right. x - 2 \left.\right) \left|\right. d x = A_{2} + A_{1} .\]

Then, using the formula for the area of a triangle, we obtain

\[A_{2} = \frac{1}{2} b h = \frac{1}{2} \cdot 2 \cdot 2 = 2\]
\[A_{1} = \frac{1}{2} b h = \frac{1}{2} \cdot 4 \cdot 4 = 8 .\]

The total area, then, is

\[A_{1} + A_{2} = 8 + 2 = 10 .\]

Checkpoint 5.10

Find the total area between the function \(f \left(\right. x \left.\right) = 2 x\) and the x-axis over the interval \(\left[\right. −3 , 3 \left]\right. .\)

This lesson is part of:

Integration

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