Question

a function f is defined over an interval $\left[a,b\right]$

(a) Graph f, indicating the area A under f from a to b.

(b) Approximate the area A by partitioning $\left[a,b\right]$

into four subintervals of equal length and choosing u as the left

endpoint of each subinterval.

(c) Approximate the area A by partitioning$\left[a,b\right]$

into eight

subintervals of equal length and choosing u as the left

endpoint of each subinterval.

(d) Express the area A as an integral.

(e) Use a graphing utility to approximate the integral.

localid="1647079709839" $f\left(x\right)=\sqrt{x}\left[0,4\right]$

Short answer

(a)

(b) The four subinterval are $\left[0,1\right],\left[1,2\right],\left[2,3\right],\left[3,4\right]$and the area is $4.416$

(c) The eight subinterval are $\left[0,\frac{1}{2}\right],\left[\frac{1}{2},1\right],\left[1,\frac{3}{2}\right],\left[\frac{3}{2},2\right],\left[2,\frac{5}{2}\right],\left[\frac{5}{2},3\right],\left[3,\frac{7}{2}\right],\left[\frac{7}{2},4\right]$and the area is $4.765$

(d) Area under the curve as an integral is given by${\int }_a^{b}f\left(x\right)dx={\int }_0^4\sqrt{x}dx$

e) Using graphing utility the area found is$\frac{16}{3}$

Step-by-step solution

Part (a) Step 1. Given

$f\left(x\right)=\sqrt{x}\left[0,4\right]$

Part (a) Step 2. Graph

Part (b) Step 1. Calculation

The area under the curve can be found using

$$\begin{gathered} A=\left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)\right) \\ where{u}_1,u_2,u_3,u_{4}are4equalinterval. \\ nowintervalwillbedecidedby \\ \left(\frac{b-a}{n}\right)=\frac{4-0}{4} \\ =1 \\ Therefore\left[0,1\right],\left[1,2\right],\left[2,3\right],\left[3,4\right]aretherespectiveintervals. \end{gathered}$$

$$\begin{gathered} Applyingtheformulaforareaweget \\ A=\left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)\right) \\ =\left(1\right)\left(1+\sqrt{2}+\sqrt{3}\right) \\ =4.416 \end{gathered}$$

Part (c) Step 1. Calculation

$$\begin{gathered} A=\left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)+f\left(u_5\right)+f\left(u_6\right)+f\left(u_7\right)+f\left(u_8\right)\right) \\ where{u}_1,u_2,u_3,u_4,u_5,u_6,u_7,u_{8}are8equalinterval. \\ nowintervalwillbedecidedby \\ \left(\frac{b-a}{n}\right)=\frac{4-0}{8} \\ =\frac{1}{2} \\ Therefore\left[0,\frac{1}{2}\right],\left[\frac{1}{2},1\right],\left[1,\frac{3}{2}\right],\left[\frac{3}{2},2\right],\left[2,\frac{5}{2}\right],\left[\frac{5}{2},3\right],\left[3,\frac{7}{2}\right],\left[\frac{7}{2},4\right]aretherespectiveintervals. \end{gathered}$$

$$\begin{gathered} A=\left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)+f\left(u_5\right)+f\left(u_6\right)+f\left(u_7\right)+f\left(u_8\right)\right) \\ =\left(\frac{1}{2}\right)\left(1+\sqrt{2}+\sqrt{3}+\frac{\sqrt{2}+\sqrt{6}+\sqrt{10}+\sqrt{14}}{2}\right) \\ =4.765 \end{gathered}$$

Part (d) Step 1. Area in integral form

$$\begin{gathered} f\left(x\right)=\sqrt{x} \\ \left[a,b\right]=\left[0,4\right] \\ {\int }_a^{b}f\left(x\right)dx={\int }_0^4\sqrt{x}dx \end{gathered}$$

Part (e) Step 1. Area using a graphing utility

The area comes out to be${\int }_0^4\sqrt{x}dx=\frac{16}{3}$