Chapter 6.2, Problem 19E

To determine

To calculate: The integral ${\displaystyle {\int }_{-1}^1\left(2-\left|x\right|\right)}dx$ with the help of the graph of the integrand.

Answer to Problem 19E

Thevalue of the integral ${\displaystyle {\int }_{-1}^1\left(2-\left|x\right|\right)}dx$ is $\text{3 units}$ .

Explanation of Solution

Given:

The integral is ${\displaystyle {\int }_{-1}^1\left(2-\left|x\right|\right)}dx$ .

Concept used:

If a nonnegative function $y=f\left(x\right)$ is integrable over the interval $\left[a,b\right]$ , then the area $A$ lies below the curve $y=f\left(x\right)$ that is bounded from $a\text{ to }b$ is equal to the integral of $f$ from a to b that can be mathematically expressed as $A={\displaystyle {\int }_a^{b}f\left(x\right)dx}$ .

Formula used:

The area of the trapezium is as follows:

  $A=\frac{1}{2}\cdot \left(\text{sum of parallel sides}\right)\cdot \text{height}$

Calculation:

In the given integral, the integrand function $f\left(x\right)=2-\left|x\right|$ is the modulus function that can be expressed by the definition of modulus function as,

  $f\left(x\right)=2-\left|x\right|=\{\begin{array}{l}2-(x),x\ge 0\\ 2-(-x),x<0\end{array}$

Draw the graph of the function $f\left(x\right)=2-\left|x\right|$ with the help of coordinates.

Substitute 0 for $x$ in $y=2-\left|x\right|$ to obtain the value of corresponding $y$ coordinate.

  $\begin{array}{l}y=2-\left|0\right|\\ y=2\end{array}$

Substitute 1 for $x$ in $y=2-\left|x\right|$ to obtain the value of corresponding $y$ coordinate.

  $\begin{array}{l}y=2-\left|1\right|\\ y=2-1\\ y=1\end{array}$

Substitute -1 for $x$ in $y=2-\left|x\right|$ to obtain the value of corresponding $y$ coordinate.

  $\begin{array}{l}y=2-\left|-1\right|\\ y=2-1\\ y=1\end{array}$

Substitute 2 for $x$ in $y=2-\left|x\right|$ to obtain the value of corresponding $y$ coordinate.

  $\begin{array}{l}y=2-\left|2\right|\\ y=2-2\\ y=0\end{array}$

Substitute $-2$ for $x$ in $y=2-\left|x\right|$ to obtain the value of corresponding $y$ coordinate.

  $\begin{array}{l}y=2-\left|-2\right|\\ y=2-2\\ y=0\end{array}$

Substitute 3 for $x$ in $y=2-\left|x\right|$ to obtain the value of corresponding $y$ coordinate.

  $\begin{array}{l}y=3-\left|2\right|\\ y=3-2\\ y=1\end{array}$

Substitute $-3$ for $x$ in $y=2-\left|x\right|$ to obtain the value of corresponding $y$ coordinate.

  $\begin{array}{l}y=-3-\left|-2\right|\\ y=3-2\\ y=1\end{array}$

It can be seen that there are two lines intersects at point $\left(0,2\right)$ that is the vertex and the highest point of the function.

Now draw the graph of $f\left(x\right)=2-\left|x\right|$ with the help of coordinates and make bounds of $x=-1\text{ and }x=1$ as shown in Figure 1.

  

It can be seen that the area under a curve from $x=-1$ to $x=1$ are two trapezoids.

The base or height of the left trapezoid can be calculated as,

  $\begin{array}{c}b=0-\left(-1\right)\\ =1\end{array}$

It can be seen from Figure 1 that the parallel sides of the left trapezoid are $h_1=1$ and $h_2=2$ .

Use the formula of the trapezium to obtain the area of the left trapezoid $A_1$ .

  $\begin{array}{c}{A}_1=\frac{1}{2}b\cdot \left(h_1+h_2\right)\\ =\frac{1}{2}\cdot 1\cdot \left(1+2\right)\\ =\frac{1}{2}\cdot \left(1\right)\cdot \left(3\right)\\ =\frac{3}{2}\end{array}$

The base or height of the right trapezoid can be calculated as,

  $\begin{array}{c}b=1-0\\ =1\end{array}$

It can be seen from Figure 1 that the parallel sides of the right trapezoid are $h_1=1$ and $h_2=2$ .

Use the formula of the trapezium to obtain the area of the right trapezoid $A_2$ .

  $\begin{array}{c}{A}_2=\frac{1}{2}b\cdot \left(h_1+h_2\right)\\ =\frac{1}{2}\cdot 1\cdot \left(1+2\right)\\ =\frac{1}{2}\cdot \left(1\right)\cdot \left(3\right)\\ =\frac{3}{2}\end{array}$

Find the sum of $A_1\text{ and }{A}_2$ to find the area under a curve.

Therefore, by the concept the value of the integral is equal to the sum of the area of two trapezoids.

  $\begin{array}{l}A=A_1+A_2\\ =\frac{3}{2}+\frac{3}{2}\\ =3\end{array}$

The value of the integral ${\displaystyle {\int }_{-1}^1\left(2-\left|x\right|\right)}dx$ is $\text{3 units}$ .

Conclusion:

Thus, thevalue of the integral ${\displaystyle {\int }_{-1}^1\left(2-\left|x\right|\right)}dx$ is $\text{3 units}$ .