Chapter 6.2, Problem 21E

To determine

To calculate: The integral ${\displaystyle {\int }_{\pi }^{2\pi }\theta d\theta }$ with the help of the graph of the integrand.

Answer to Problem 21E

The value of the integral ${\displaystyle {\int }_{\pi }^{2\pi }\theta d\theta }$ is $\frac{3{\pi }^2}{2}\text{ units}$ .

Explanation of Solution

Given:

The integral is ${\displaystyle {\int }_{\pi }^{2\pi }\theta d\theta }$ .

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 $f\left(\theta \right)=\theta$ that would be straight line, since it is a linear function.

Draw the graph of the function $f\left(\theta \right)=\theta$ with the help of coordinates.

Substitute $\frac{\pi }{2}$ for $x$ in $y=\theta$ to obtain the value of corresponding $y$ coordinate.

  $f\left(\theta \right)=\frac{\pi }{2}$

Substitute $\frac{3\pi }{2}$ for $x$ in $y=\theta$ to obtain the value of corresponding $y$ coordinate.

  $f\left(\theta \right)=\frac{3\pi }{2}$

Substitute $\frac{5\pi }{2}$ for $x$ in $y=\theta$ to obtain the value of corresponding $y$ coordinate.

  $f\left(\theta \right)=\frac{5\pi }{2}$

The coordinates are $\left(\frac{\pi }{2},\frac{\pi }{2}\right),\left(\frac{3\pi }{2},\frac{3\pi }{2}\right)\text{ and }\left(\frac{5\pi }{2},\frac{5\pi }{2}\right)$ .

Now draw the graph of $f\left(\theta \right)=\theta$ and make bounds of $x=\pi \text{ and }x=2\pi$ as shown in Figure 1.

  

It can be seen that the area under a curve from $x=\pi$ to $x=2\pi$ is trapezoid.

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

  $\begin{array}{c}b=2\pi -\pi \\ =\pi \end{array}$

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

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

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

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

The value of the integral ${\displaystyle {\int }_{\pi }^{2\pi }\theta d\theta }$ is $\frac{3{\pi }^2}{2}\text{ units}$ .

Conclusion:

Thus, the value of the integral ${\displaystyle {\int }_{\pi }^{2\pi }\theta d\theta }$ is $\frac{3{\pi }^2}{2}\text{ units}$ .