Net Signed Area

Let us return to the Riemann sum.

Consider, for example, the function [latex]f(x)=2-2x^2[/latex] (shown in Figure 2) on the interval [latex][0,2][/latex]. Use [latex]n=8[/latex] and choose [latex]x_i^*[/latex] as the left endpoint of each interval. Construct a rectangle on each subinterval of height [latex]f(x_i^*)[/latex] and width [latex]\Delta x[/latex].

When [latex]f(x_i^*)[/latex] is positive, the product [latex]f(x_i^*) \Delta x[/latex] represents the area of the rectangle, as before. When [latex]f(x_i^*)[/latex] is negative, however, the product [latex]f(x_i^*) \Delta x[/latex] represents the negative of the area of the rectangle.

The Riemann sum then becomes

Taking the limit as [latex]n\to \infty[/latex], the Riemann sum approaches the area between the curve above the [latex]x[/latex]-axis and the [latex]x[/latex]-axis, minus the area between the curve below the [latex]x[/latex]-axis and the [latex]x[/latex]-axis, as shown in Figure 3.

Then,

The quantity [latex]A_1-A_2[/latex] is called the net signed area.

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

Show Solution

The function produces a straight line that forms two triangles: one from [latex]x=-3[/latex] to [latex]x=0[/latex] and the other from [latex]x=0[/latex] to [latex]x=3[/latex] (Figure 4). Using the geometric formula for the area of a triangle, [latex]A=\frac{1}{2}bh[/latex], the area of triangle [latex]A_1[/latex], above the axis, is

[latex]A_1=\frac{1}{2}3(6)=9[/latex],

where [latex]3[/latex] is the base and [latex]2(3)=6[/latex] is the height. The area of triangle [latex]A_2[/latex], below the axis, is

[latex]A_2=\frac{1}{2}(3)(6)=9[/latex],

where [latex]3[/latex] is the base and [latex]6[/latex] is the height. Thus, the net area is

[latex]\displaystyle\int_{-3}^3 2x dx=A_1-A_2=9-9=0[/latex].

Analysis

If [latex]A_1[/latex] is the area above the [latex]x[/latex]-axis and [latex]A_2[/latex] is the area below the [latex]x[/latex]-axis, then the net area is [latex]A_1-A_2[/latex]. Since the areas of the two triangles are equal, the net area is zero.

Starting at timestamp 8:40, watch the following video to see the worked solution to this example.

 

You can view the transcript for this segmented clip of “5.2 The Definite Integral” here (opens in new window).

Total Area

One application of the definite integral is finding displacement when given a velocity function. If [latex]v(t)[/latex] 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 [latex]70[/latex] mph for [latex]2[/latex] hours, then it is [latex]140[/latex] mi away from its original position (Figure 6). Using integral notation, we have

[latex]\displaystyle\int_0^2 70 dt=140[/latex]

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 [latex]60[/latex] mph for [latex]2[/latex] hours, it is [latex]120[/latex] mi north of its starting position. If the car then turns around and travels south at a speed of [latex]40[/latex] mph for [latex]3[/latex] hours, it will be back at it starting position (Figure 7).

Again, using integral notation, we have

[latex]\begin{array}{ll} \displaystyle\int_0^2 60 dt + \displaystyle\int_2^5 -40 dt & =120-120 \\ & =0 \end{array}[/latex]

In this case the displacement 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 [latex]x[/latex]-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

[latex]\begin{array}{ll} \displaystyle\int_0^2 |60| dt + \displaystyle\int_2^5 |-40| dt & = \displaystyle\int_0^2 60 dt + \displaystyle\int_2^5 40 dt \\ & =120+120 \\ & =240 \end{array}[/latex]

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

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

Show Solution

Calculate the [latex]x[/latex]-intercept as [latex](2,0)[/latex] (set [latex]y=0[/latex], solve for [latex]x[/latex]). To find the total area, take the area below the [latex]x[/latex]-axis over the subinterval [latex][0,2][/latex] and add it to the area above the [latex]x[/latex]-axis on the subinterval [latex][2,6][/latex] (Figure 8).

We have

[latex]\displaystyle\int_0^6 |(x-2)| dx = A_2+A_1[/latex]

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

[latex]A_2=\frac{1}{2}bh=\frac{1}{2} \cdot 2 \cdot 2=2[/latex]

[latex]A_1=\frac{1}{2}bh=\frac{1}{2} \cdot 4 \cdot 4=8[/latex]

The total area, then, is

[latex]A_1+A_2=8+2=10[/latex].

 

Starting at timestamp 11:34, watch the following video to see the worked solution to this example.

 

You can view the transcript for this segmented clip of “5.2 The Definite Integral” here (opens in new window).