# 1.2 The definite integral  (Page 3/16)

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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

## 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 $f\left(x\right)=2-2{x}^{2}$ (shown in [link] ) on the interval $\left[0,2\right].$ Use $n=8$ and choose $\left\{{x}_{i}^{*}\text{}}$ 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

$\sum _{i=1}^{8}f\left({x}_{i}^{*}\right)\text{Δ}x=\left(\text{Area of rectangles above the}\phantom{\rule{0.2em}{0ex}}x\text{-axis}\right)-\left(\text{Area of rectangles below the}\phantom{\rule{0.2em}{0ex}}x\text{-axis}\right)$

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,

$\begin{array}{cc}{\int }_{0}^{2}f\left(x\right)dx\hfill & =\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}f\left({c}_{i}\right)\text{Δ}x\hfill \\ & ={A}_{1}-{A}_{2}.\hfill \end{array}$

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 $\left[-3,3\right].$

The function produces a straight line that forms two triangles: one from $x=-3$ to $x=0$ and the other from $x=0$ to $x=3$ ( [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

${\int }_{-3}^{3}2xdx={A}_{1}-{A}_{2}=9-9=0.$

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.

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

## 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.

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20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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