91Ó°ÊÓ

We are given a function$f\left(x\right)=\sqrt{16-x^2}$in the interval$[0,4].$

We need to plot the graph.

Now, the graph of the function$f\left(x\right)=\sqrt{16-x^2}$is

but the interval is$[0,4]$So, the required graph is the shaded graph,

We know, the formula to calculate the area is

$\text{Area}A=f\left(u_1\right)\frac{b-a}{n}+⋯+f\left(u_n\right)\frac{b-a}{n}$

$$\begin{gathered} f\left(u_1=0\right)=4 \\ f\left(u_2=0.5\right)=3.969 \\ f\left(u_3=1\right)=3.873 \\ f\left(u_4=1.5\right)=3.708 \\ f\left(u_5=2\right)=3.464 \\ f\left(u_6=2.5\right)=3.122 \\ f\left(u_7=3\right)=2.646 \\ f\left(u_8=3.5\right)=1.936 \end{gathered}$$

$$\begin{gathered} \text{Area}A=4(0.5)+3.969(0.5)+3.873(0.5)+3.708(0.5)+3.464(0.5)+3.122(0.5)+2.646(0.5)+1.936(0.5) \\ =0.5(4+3.969+3.873+3.708+3.464+3.122+2.646+1.936) \\ =13.359 \end{gathered}$$

The formula to calculate the actual area is:

$$\begin{gathered} {∫}_a^{b}f\left(x\right)dx \\ ={∫}_0^4\sqrt{16-x^2}dx \\ \text{let}x=4\sin u \\ \text{for}x=4,4\sin \left(\frac{\mathrm{π}}{2}\right)=4 \\ \text{for}x=0,4\sin \left(0\right)=0 \\ \text{and}dx=4\cos udu \\ Therefore, \\ {∫}_0^4\sqrt{16-x^2}dx=={∫}_0^{\frac{\mathrm{π}}{2}}\sqrt{16-(4\sin u{)}^2}4\cos udu \\ ={∫}_0^{\frac{\mathrm{π}}{2}}\sqrt{16-16{\sin }^{2}u}4\cos udu \\ ={∫}_0^{\frac{\mathrm{π}}{2}}\sqrt{16\left(1-{\sin }^{2}u\right)}4\cos udu \\ ={∫}_0^{\frac{1}{2}}\sqrt{16\left({\cos }^{2}u\right)}4\cos udu \\ ={∫}_0^{\frac{\mathrm{π}}{2}}\left(4\cos u\right)4\cos udu \\ ={∫}_0^{\frac{\mathrm{π}}{2}}16{\cos }^{2}udu \\ =16{∫}_0^{\frac{\mathrm{π}}{2}}{\cos }^{2}udu \\ =16{∫}_0^{\frac{\mathrm{π}}{2}}\frac{1+\cos 2u}{2}du \\ =16·\frac{1}{2}{∫}_0^{\frac{\mathrm{π}}{2}}(1+\cos 2u)du \\ =8\left({∫}_0^{\frac{\mathrm{π}}{2}}1du+{∫}_0^{\frac{\mathrm{π}}{2}}\cos \left(2u\right)du\right) \\ 8\left({\left.u\right|}_0^{\frac{\mathrm{π}}{2}}+{∫}_0^{\frac{\mathrm{π}}{2}}\cos \left(2u\right)du\right) \\ =8\left(\frac{\mathrm{π}}{2}-0+{∫}_0^{\frac{\mathrm{π}}{2}}\cos \left(2u\right)du\right) \\ =8\left(\frac{\mathrm{π}}{2}+{∫}_0^{\frac{\mathrm{π}}{2}}\cos \left(2u\right)du\right) \end{gathered}$$

Solving

$$\begin{gathered} {∫}_0^{\frac{\mathrm{π}}{2}}\cos \left(2u\right)du \\ \text{let}t=2u \\ dt=2du⇒\frac{1}{2}dt=du \\ \mathbf{=}{\mathbf{∫}}_{\mathbf{0}}^{\frac{\mathbf{π}}{\mathbf{2}}}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{cos}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathit{d}\mathit{t} \\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{∫}}_{\mathbf{0}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{cos}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathit{d}\mathit{t} \\ ={\left.\frac{1}{2}\sin t\right|}_0^{\frac{\mathrm{π}}{2}} \\ ={\left.\frac{1}{2}\sin \left(2u\right)\right|}_0^{\frac{\mathrm{π}}{2}} \\ =\frac{1}{2}\left[\sin \left(\frac{2\mathrm{π}}{2}\right)-\sin \left(2\right(0\left)\right)\right] \\ =\frac{1}{2}\left[\sin \right(\mathrm{π})-0]=0 \end{gathered}$$

Therefore,

$$\begin{gathered} {∫}_0^4\sqrt{16-x^2}dx==8\left(\frac{\mathrm{π}}{2}+{∫}_{-\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{2}}\cos \left(2u\right)du\right) \\ =8\left(\frac{\mathrm{π}}{2}+0\right)=4\mathrm{π} \\ =12.566 \end{gathered}$$

The value obtained from approximation is greater than the value obtained of the actual area.