Chapter 6.2, Problem 18E

To determine

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

Answer to Problem 18E

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

Explanation of Solution

Given:

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

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 semicircle is as follows:

  $A=\frac{1}{2}\cdot b\cdot h$

Here, the expression $b$ denotes the base of the triangle and h denotes the height of the triangle.

Calculation:

In the given integral, the integrand function is $f\left(x\right)=1-\left|x\right|$ that would be two straight lines as the function $f\left(x\right)=1-\left|x\right|$ is defined as

  $\begin{array}{l}1-\left|x\right|=\{\begin{array}{l}1-x,x\ge 0\\ 1-(-x),x<0\end{array}\\ 1-\left|x\right|=\{\begin{array}{l}1-x,x\ge 0\\ 1+x,x<0\end{array}\end{array}$

For values of x greater than or equal to 0 the value of function will be $1-x$ .

For values of x less than 0 the value of function will be $1+x$ .

  $f\left(x\right)=1-x$ is a straight line.

Put the value of x = 0, 1, 2 for coordinates of y , we get

For x = 0,

  $f\left(0\right)=1-0=1$

For x = 1,

  $f\left(1\right)=1-1=0$

For x = 2,

  $f\left(2\right)=1-2=-1$

Thus the coordinates are (0, 1), (1, 0), (2, -1).

  $f\left(x\right)=1+x$ is a straight line.

Put the value of x = 0, -1, -2 for coordinates of y , we get

For x = 0,

  $f\left(0\right)=1+0=1$

For x = -1,

  $f\left(-1\right)=1-1=0$

For x = -2,

  $f\left(-2\right)=1-2=-1$

Thus the coordinates are (0, 1), (-1, 0), (-2, -1).

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

  

It can be seen from Figure 1 that the area under a function is two triangles. The height of the left triangle is 1 unit and the base is 1 unit.

The height of the right triangle is on the left side is 1 unit and the length of the base is 1 unit.

The area of the region bounded by the $f\left(x\right)=1-\left|x\right|$ , $x=-1\text{ and }x=1$ is the sum of area of two triangles.

Use the formula of the triangle to obtain the area of the triangle.

  $\begin{array}{c}A=\frac{1}{2}\cdot 1\cdot 1+\frac{1}{2}\cdot 1\cdot 1\\ =\frac{1}{2}+\frac{1}{2}\\ =1\end{array}$

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

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

Conclusion:

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