Chapter 6.2, Problem 20E

To determine

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

Answer to Problem 20E

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

Explanation of Solution

Given:

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

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:

  $A=\frac{1}{2}\cdot \pi r^2$

The are of the rectangle is as follows:

  $A=l\cdot b$

Here, l is the length and bis the breadth.

Calculation:

In the given integral, the integrand function is $f\left(x\right)=1+\sqrt{1-x^2}$ .

The general equation of the semicircle is $y=\sqrt{r^2-x^2}$ .

It can be observed that the function $f\left(x\right)=1+\sqrt{1-x^2}$ is the function of the semicircle with the addition of 1 unit in y.

Draw the graph of the function $y=1+\sqrt{1-x^2}$ with the help of the method of semicircle.

Compare the equation $y=\sqrt{r^2-x^2}$ with the equation $y=1+\sqrt{1-x^2}$ .

The value of $r$ is 1.

Now draw the graph of $f\left(x\right)=1+\sqrt{1-x^2}$ by moving upward 1 unit and then draw the semicircle with the help of compass and ruler of radius 1 unit.

Now 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 $x=1$ is the addition of semicircle and rectangle.

It can be seen from Figure 1 that the radius of the semicircle is 1.

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

  $\begin{array}{c}A=\frac{1}{2}\pi r^2\\ =\frac{1}{2}\pi \cdot 1^2\\ =\frac{\pi }{2}\end{array}$

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

  $\begin{array}{c}A=l\cdot b\\ =2\cdot 1\\ =2\end{array}$

Therefore, by the concept the value of the integral is equal to the area of the semicircle and rectangle.

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

Conclusion:

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