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Use the formula for the area of a trapezoid to evaluate ${\int}_{2}^{4}\left(2x+3\right)dx}.$
18 square units
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?
Let us return to the Riemann sum. Consider, for example, the function $f\left(x\right)=2-2{x}^{2}$ (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 $f\left({x}_{i}^{*}\right)\text{\Delta}x$ represents the area of the rectangle, as before. When $f\left({x}_{i}^{*}\right)$ is negative, however, the product $f\left({x}_{i}^{*}\right)\text{\Delta}x$ 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.
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
where 3 is the base and $2\left(3\right)=6$ is the height. The area of triangle A _{2} , below the axis, is
where 3 is the base and 6 is the height. Thus, the net area is
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.
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
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.
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