1.2 The definite integral  (Page 3/16)

Area and the definite integral

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

Net signed area

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

Taking the limit as $n\to \infty ,$ 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 [link] . Then,

The quantity $A_1-A_2$ 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.

Finding the net signed area

Find the net signed area between the curve of the function $f\left(x\right)=2x$ and the x -axis over the interval $\left[\mathrm{-3},3\right].$

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

$A_1=\frac{1}{2}3\left(6\right)=9,$

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

$A_2=\frac{1}{2}\left(3\right)\left(6\right)=9,$

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

${\displaystyle {\int }_{\mathrm{-3}}^{3}2xdx}=A_1-A_2=9-9=0.$

Analysis

If A ₁ is the area above the x -axis and A ₂ 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.

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

6

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Total area

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