Question

Consider the function $f\left(x\right)=\sqrt{1-x^2}$whose domain is the interval $[-1,1]$

(a) Graph f.

(b) Approximate the area under the graph of f from $-1$to $1$by dividing $[-1,1]$into five subintervals, each of equal length.

(c) Approximate the area under the graph of f from $-1$to $1$by dividing $[-1,1]$into ten subintervals, each of equal length.

(d) Express the area as an integral.

(e) Evaluate the integral using a graphing utility.

(f) What is the actual area?

Short answer

Part (a) The graph is

Part (b) Approx area for five intervals is$0.8$

Part (c) Approx area for ten intervals is$0.6$

Part (d) The integral is${\int }_a^{b}f\left(x\right)dx={\int }_{-1}^1\sqrt{1-x^2}dx$

Part (d) By using graphical utility, area is$1.5708$

Part (f) Actual area is$\frac{\mathrm{\pi }}{2}$

Step-by-step solution

Step 1. Given information

A function is given as$f\left(x\right)=\sqrt{1-x^2}$

Part (a) Step 1. Plotting graph

The graph is

Part (b) Step 2. Approx area

Area under curve can be determined by

$$\begin{gathered} A\approx \sum _{i=1}^{n}f\left(u_i\right)\left(\frac{b-a}{n}\right) \\ \approx \left(\frac{b-a}{2}\right)\left(f\left(u_1\right)+f\left(u_n\right)\right) \end{gathered}$$

where $[a,b]$ is he interval, nis the number of intervals, u_i is the end points of the first sub-interval, f(u_i)is the height of that endpoint u_n is the nth endpoint.

Now substitute the values as,

$$\begin{gathered} \frac{b-a}{n}=\frac{1-(-1)}{5} \\ =\frac{2}{5} \end{gathered}$$

The sub-intervals of equal length $\frac{2}{5}:\left[-1,-\frac{3}{5}\right],\left[-\frac{3}{5},-\frac{1}{5}\right],\left[-\frac{1}{5},\frac{1}{5}\right],\left[\frac{1}{5},\frac{3}{5}\right]\text{and}\left[\frac{3}{5},1\right]$

Put $u_1=-1\text{and}{u}_n=u_5=\frac{3}{5}$

$$\begin{gathered} f\left(x\right)=\sqrt{1-x^2} \\ f(-1)=\sqrt{1-(-1{)}^2} \\ =\sqrt{1-1} \\ =0 \end{gathered}$$

Now put $u_n=\frac{3}{5}$

$$\begin{gathered} f\left(\frac{3}{5}\right)=\sqrt{1-{\left(\frac{3}{5}\right)}^2} \\ =\sqrt{1-\frac{9}{25}} \\ =\frac{4}{5} \end{gathered}$$

Now substitute $f\left(u_i\right)=0\&f\left(u_n\right)=\frac{4}{5}$

$$\begin{gathered} A\approx \left(\frac{b-a}{2}\right)\left(f\left(u_1\right)+f\left(u_n\right)\right) \\ \approx \left(\frac{1-(-1)}{2}\right)\left(0+\frac{4}{5}\right) \\ =\frac{4}{5} \end{gathered}$$

The approx area is$0.8$

Part (c) Step 1. Approx area

Here $n=10$and $\frac{b-a}{n}=\frac{1-(-1)}{10}=\frac{2}{10}$.

Now substitute the values $f\left(0\right)=f\left(u_1\right)=0\&u_n=\frac{4}{5}$,

$$\begin{gathered} f\left(x\right)=\sqrt{1-x^2} \\ f\left(\frac{4}{5}\right)=\sqrt{1-{\left(\frac{4}{5}\right)}^2} \\ =\sqrt{1-\frac{16}{25}} \\ =\frac{3}{5} \end{gathered}$$

Now substitute $f\left(u_n\right)=\frac{3}{5}\&f\left(u_1\right)=0$

$$\begin{gathered} A\approx \left(\frac{b-a}{2}\right)\left(f\left(u_1\right)+f\left(u_n\right)\right) \\ \approx 1\left(0+\frac{3}{5}\right) \\ \approx 1\left(\frac{3}{5}\right) \end{gathered}$$

The approx area in ten intervals is$0.6$

Part (d) Step 1. Area as intergral

The integral of area for given conditions is${\int }_a^{b}f\left(x\right)dx={\int }_{-1}^1\sqrt{1-x^2}dx$

Part (e) Step 1. Evaluate integral

By using graphical utility, value of integral is${\int }_{-1}^1\sqrt{1-x^2}dx=1.5708$

Part (f) Step 1. Actual area

The area of function $f\left(x\right)=\sqrt{1-x^2}$is half of the area of circle of radius one.

$$\begin{gathered} A=\frac{\pi r^2}{2} \\ =\frac{\mathrm{\pi }{\left(1\right)}^2}{2} \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$