Chapter 6.2, Problem 22E

To determine

To calculate: The integral ${\displaystyle {\int }_{\sqrt{2}}^{5\sqrt{2}}rdr}$ with the help of the graph of the integrand.

Answer to Problem 22E

The value of the integral ${\displaystyle {\int }_{\sqrt{2}}^{5\sqrt{2}}rdr}$ is $\text{24 units}$ .

Explanation of Solution

Given:

The integral is ${\displaystyle {\int }_{\sqrt{2}}^{5\sqrt{2}}rdr}$ .

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

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

Substitute $1$ for $r$ in $y=r$ to obtain the value of corresponding $y$ coordinate.

  $y=1$

Substitute $2$ for $r$ in $y=r$ to obtain the value of corresponding $y$ coordinate.

  $y=2$

Substitute $3$ for $x$ in $y=r$ to obtain the value of corresponding $y$ coordinate.

  $y=3$

The coordinates are $\left(1,1\right),\left(2,2\right)\text{ and }\left(3,3\right)$ .

Now draw the graph of $f\left(r\right)=r$ and make bounds of $r=\sqrt{2}\text{ and }r=5\sqrt{2}$ as shown in Figure 1.

  

It can be seen that the area under a curve from $r=\sqrt{2}$ to $r=5\sqrt{2}$ is trapezoid.

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

  $\begin{array}{c}b=5\sqrt{2}-\sqrt{2}\\ =4\sqrt{2}\end{array}$

It can be seen from Figure 1 that the parallel sides of the trapezoid are $h_1=\sqrt{2}$ and $h_2=5\sqrt{2}$ .

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

  $\begin{array}{c}A=\frac{1}{2}b\cdot \left(h_1+h_2\right)\\ =\frac{1}{2}\cdot 4\sqrt{2}\cdot \left(\sqrt{2}+5\sqrt{2}\right)\\ =\frac{1}{2}\cdot \left(4\sqrt{2}\right)\cdot \left(6\sqrt{2}\right)\\ =\frac{1}{2}\cdot \left(24\sqrt{2}\cdot \sqrt{2}\right)\\ =\frac{1}{2}\cdot \left(24\cdot 2\right)\\ =24\end{array}$

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

The value of the integral ${\displaystyle {\int }_{\sqrt{2}}^{5\sqrt{2}}rdr}$ is $\text{24 units}$ .

Conclusion:

Thus, the value of the integral ${\displaystyle {\int }_{\sqrt{2}}^{5\sqrt{2}}rdr}$ is $\text{24 units}$ .