Signed and normal areas

We already know that there are two types of areas, the area (which is the "normal" area we already know), and the signed area, which is the sum of bar areas, or the integral. As long as the graph of $f$ defining the border of the area is above the $x$ -axis, the area and the signed area are equal. However, as soon as the graph of $f$ is below the $x$ -axis, the bar area $\Delta x\cdot f(x)$ becomes negative, and the sum of the bar areas is a mixture of positive and negative numbers - which does not correspond to the "normal" area. The figure below illustrates this point:

Recipe 1

So if we want to have the area (the normal one, all positive), we have to divide the negative and positive parts, and determine their integral separately. For the picture above, we have to determine the area

A_{-}=\int_{-1}^2 f(x)\, dx

and

A_{+}=\int_2^3 f(x)\, dx

As $A_{-}$ is the sum of only negative bar areas, the integral will be negative, but its positive part will be the (normal) area of the region from $-1$ to $2$ . $A_{+}$ is the sum of only positive bar areas and the integral is already the (normal) area. Thus, the total (normal) area of the shaded region is the sum of the positive numbers:

|A_{-}|+A_{+}

Here, the vertical lines means that we have to take the absolute value of $A_{-}$ , which simply means to take the positive part of the number.

Of course we can generalise this to arbitrary many positive and negative subregions.

Example 1

The function equation of the graph $f$ shown above is

f(x)=\frac{1}{8}x^3-1

Find the (normal) area of the region enclosed by the graph of $f$ , the $x$ -axis, and the vertical lines at $x=-1$ and $x=3$ .

Solution

We first need to find the $x$ -intercept of $f$ . For this example it is obviously $x=2$ , which also follows from

\begin{array}{lll} \frac{1}{8}x^3-1 &=& 0\\ x^3&=& 8 \\ x&=& 2 \end{array}

The antiderivative of $f$ is

F(x)=\frac{1}{32}x^4-x

We then have

A_{-}=\int_{-1}^2 (\frac{1}{8}x^3-1)\, dx = F(2)-F(-1)=-2.53125

and

A_{+}=\int_{2}^3 (\frac{1}{8}x^3-1)\, dx = F(3)-F(2)=1.03125

Thus, the (normal) area of the region is

A=|A_{-}|+A_{+}=2.53125+1.03125=3.5625

Note that if add $A_{-}+A_{+}$ , we get $-1.5$ , which is simply the integral of $f$ from $-1$ to $3$ . Indeed:

\int_{-1}^3 (\frac{1}{8}x^3-1)\, dx = F(3)-F(-1)=-1.5

Exercise 1

Q1

Consider the function $f(x)=\frac{1}{2}x^2-1$ .

Determine $\int_{-2}^3 f(x)\, dx$
Determine the area of the region enclosed by the graph of $f$ , the $x$ -axis, and the vertical lines at $x=-2$ and $x=3$ .

Q2

The region $R$ is enclosed by the graph of $f(x)=-2(x-1)(x+1)$ , the $x$ -axis and the vertical lines at $x=-2$ and $x=3$ .

Determine the signed area $R$ .
Determine the area of $R$ .

Solution

A1

The antiderivative of $f$ is

F(x)=\frac{1}{6}x^3-x

$\int_{-2}^3 f(x)\, dx =F(3)-F(-2)=(\frac{1}{6}\cdot 3^3-3)-(\frac{1}{6}\cdot (-2)^3-(-2))=0.8\bar 3$
Draw graph (see below) to check if there are negative regions - indeed there are! We calculate the three regions separately. First we need to find the $x$ -intercepts of $f$ : Find $x$ with
$\begin{array}{lll} f(x) &= & 0\\ \frac{1}{2}x^2-1&= & 0\\ x^2&=& 2\\ x&=& \pm\sqrt{2} \end{array}$
Now we determine the areas with the integral:
$A_1=\int_{-2}^{-\sqrt{2}} f(x)\, dx = F(-\sqrt{2})-F(-2)=0.276142$ $A_2=\int_{-\sqrt{2}}^{\sqrt{2}} f(x)\, dx = F(\sqrt{2})-F(-\sqrt{2})=-1.88562$ $A_3=\int_{\sqrt{2}}^3 f(x)\, dx = F(3)-F(\sqrt{2})=2.44281$
Thus, the area is
$A=A_1+|A_2|+A_3=0.276142+1.88562+2.44281=4.604572$
Let's check if we get the result of (1), if we simply add the areas. Indeed, we have $A_1+A_2+A_3=0.276142-1.88562+2.44281=0.8\overline{3}$

A2

Draw the situation! Because of

f(x)=-2(x-1)(x+1)=-2x^2+2

the antiderivative is

F(x) = -\frac{2}{3}x^3+2x

$A=\int_{-2}^3 f(x)\, dx = F(3)-F(-2)=-\frac{40}{3}=-13.\overline{3}$ .
We have to calculate the negative and positive areas individually. First we need to find the $x$ -intercepts of $f$ : Find $x$ with
$f(x)=-2(x-1)(x+1)=0$
Clearly, this has to be at $x_1=-1$ and $x_2=1$ . Thus, we have the three areas
$A_1=\int_{-2}^{-1} f(x)\, dx = F(-1)-F(-2)=-2.\overline{6}$ $A_2=\int_{-1}^{1} f(x)\, dx = F(1)-F(-1)=\frac{8}{3}=2.\overline{6}$ $A_3=\int_{1}^{3} f(x)\, dx = F(3)-F(1)=-13.\overline{3}$
Thus, the area is $13.\overline{3}+2\cdot 2.\overline{6}=18.\overline{6}$