Chapter 6.2, Problem 16E

To determine

To calculate: The integral ${\displaystyle {\int }_{-4}^0\sqrt{16-x^2}}dx$ with the help of the graph of the integrand.

Answer to Problem 16E

The value of the integral ${\displaystyle {\int }_{-4}^0\sqrt{16-x^2}}dx$ is $4\pi \text{units}$ .

Explanation of Solution

Given:

The integral is ${\displaystyle {\int }_{-4}^0\sqrt{16-x^2}}dx$ .

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:

  $A=\frac{1}{2}\cdot \pi r^2$

Here, the expression $r$ denotes the radius.

Calculation:

In the given integral, the integrand function is $f\left(x\right)=\sqrt{16-x^2}$ that would be semicircle, where the values of y are positive.

The equation of circle for the semicircle $\sqrt{16-x^2}$ is $y^2=16-x^2\Rightarrow x^2+y^2=16$ ,

Comparing this with the standard equation of circle $x^2+y^2=r^2$ , we get r = 4.

Now draw the graph of $f\left(x\right)=\sqrt{16-x^2}$ and make bounds of $x=-4\text{ and }x=0$ as shown in Figure 1.

  

It can be seen that the area under a curve from $x=-4$ to $x=0$ is a semicircle.

It can be seen from Figure 1 that the radius of the semicircle is 4. The area of the region is also half of the semicircle.

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

  $\begin{array}{c}A=\frac{1}{2}\pi r^2\\ =\frac{1}{2}\pi \cdot 4^2\\ =8\pi \end{array}$

The area of the region is half of the area of the semicircle, thus the area of the region under the curve is

  $\begin{array}{c}A=\frac{1}{2}(8\pi )\\ =4\pi \end{array}$

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

The value of the integral ${\displaystyle {\int }_{-4}^0\sqrt{16-x^2}}dx$ is $4\pi \text{units}$ .

Conclusion:

Thus, the value of the integral ${\displaystyle {\int }_{-4}^0\sqrt{16-x^2}}dx$ is $4\pi \text{units}$ .