Chapter 6.2, Problem 14E

To determine

To calculate: The integral ${\displaystyle {\int }_{\frac{1}{2}}^{\frac{3}{2}}\left(-2x+4\right)}dx$ with the help of the graph of the integrand.

Answer to Problem 14E

The value of the integral ${\displaystyle {\int }_{\frac{1}{2}}^{\frac{3}{2}}\left(-2x+4\right)}dx$ is $\text{2}\text{unit}$ .

Explanation of Solution

Given:

The integral is ${\displaystyle {\int }_{\frac{1}{2}}^{\frac{3}{2}}\left(-2x+4\right)}dx$ .

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 is $f\left(x\right)=-2x+4$ that would be straight line, since it is a linear function.

Draw the graph of the function $f\left(x\right)=-2x+4$ with the help of coordinates.

Substitute 0 for $x$ in $y=-2x+4$ to obtain the value of corresponding $y$ coordinate.

  $\begin{array}{l}y=-2\left(0\right)+4\\ y=4\end{array}$

Substitute 1 for $x$ in $y=-2x+4$ to obtain the value of corresponding $y$ coordinate.

  $\begin{array}{l}y=-2\left(1\right)+4\\ y=-2+4\\ y=2\end{array}$

Substitute 2 for $x$ in $y=-2x+4$ to obtain the value of corresponding $y$ coordinate.

  $\begin{array}{l}y=-2\left(2\right)+4\\ y=-4+4\\ y=0\end{array}$

The coordinates are $\left(0,4\right),\left(1,2\right)\text{ and }\left(2,0\right)$ .

Now draw the graph of $f\left(x\right)=-2x+4$ and make bounds of $x=\frac{1}{2}\text{ and }x=\frac{3}{2}$ as shown in Figure 1.

  

It can be seen that the area under a line from $x=\frac{1}{2}$ to $x=\frac{3}{2}$ is a trapezoid.

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

  $\begin{array}{c}b=\frac{3}{2}-\left(\frac{1}{2}\right)\\ =1\end{array}$

It can be seen from Figure 1 that the parallel sides of the trapezoid are $h_1=1$ and $h_2=3$ .

Use the formula of the trapezium to obtain the area of the trapezoid.

  $\begin{array}{c}A=\frac{1}{2}b\cdot \left(h_1+h_2\right)\\ =\frac{1}{2}\cdot 1\cdot \left(1+3\right)\\ =\frac{1}{2}\cdot 1\cdot 4\\ =2\end{array}$

Therefore, by the concept the value of the integral is equal to the area of the trapezoid.

The value of the integral ${\displaystyle {\int }_{\frac{1}{2}}^{\frac{3}{2}}\left(-2x+4\right)}dx$ is $\text{2}\text{unit}$ .

Conclusion:

Thus, the value of the integral ${\displaystyle {\int }_{\frac{1}{2}}^{\frac{3}{2}}\left(-2x+4\right)}dx$ is $\text{2}\text{unit}$ .