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="1647084671117" $f\left(x\right)=\ln x\left[3,7\right]$

Short answer

(a)

(b) The four subinterval are $\left[3,4\right],\left[4,5\right],\left[5,6\right],\left[6,7\right]$and the area is $5.886$

(c) The eight subinterval are $\left[3,\frac{7}{2}\right],\left[\frac{7}{2},4\right],\left[4,\frac{9}{2}\right],\left[\frac{9}{2},5\right],\left[5,\frac{11}{2}\right],\left[\frac{11}{2},6\right],\left[6,\frac{13}{2}\right],\left[\frac{13}{2},7\right]$and the area is $12.219$

(d)${\int }_a^{b}f\left(x\right)dx={\int }_3^7\ln xdx$

e) Using graphing utility the area found is role="math" localid="1647101992822" $6.326$

Step-by-step solution

Part (a) Step 1. Given

$f\left(x\right)=\ln x\left[3,7\right]$

Part (a) Step 2. Graph

Part (b) Step 1. Calculation

The area under the curve can be found using

$$\begin{gathered} A=\left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)\right) \\ where{u}_1,u_2,u_3,u_{4}are4equalinterval. \\ nowintervalwillbedecidedby \\ \left(\frac{b-a}{n}\right)=\frac{7-3}{4} \\ =1 \\ Therefore\left[3,4\right],\left[4,5\right],\left[5,6\right],\left[6,7\right]aretherespectiveintervals. \end{gathered}$$

$$\begin{gathered} Applyingtheformulaforareaweget \\ A=\left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)\right) \\ =\left(1\right)\left(\ln 3+\ln 4+\ln 5+\ln 6\right) \\ =5.886 \end{gathered}$$

Part (c) Step 1. Calculation

$$\begin{gathered} A=\left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)+f\left(u_5\right)+f\left(u_6\right)+f\left(u_7\right)+f\left(u_8\right)\right) \\ where{u}_1,u_2,u_3,u_4,u_5,u_6,u_7,u_{8}are8equalinterval. \\ nowintervalwillbedecidedby \\ \left(\frac{b-a}{n}\right)=\frac{6-2}{8} \\ =\frac{1}{2} \\ Therefore\left[3,\frac{7}{2}\right],\left[\frac{7}{2},4\right],\left[4,\frac{9}{2}\right],\left[\frac{9}{2},5\right],\left[5,\frac{11}{2}\right],\left[\frac{11}{2},6\right],\left[6,\frac{13}{2}\right],\left[\frac{13}{2},7\right]aretherespectiveintervals. \end{gathered}$$

$$\begin{gathered} A=\left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)+f\left(u_5\right)+f\left(u_6\right)+f\left(u_7\right)+f\left(u_8\right)\right) \\ =\left(\frac{1}{2}\right)\left(\ln 3+\ln \frac{7}{2}+\ln 4+\ln \frac{9}{2}+\ln 5+\ln \frac{11}{2}+\ln 6+\ln \frac{13}{2}\right) \\ =12.219 \end{gathered}$$

Part (d) Step 1. Area in integral form

$$\begin{gathered} f\left(x\right)=\ln x \\ \left[a,b\right]=\left[3,7\right] \\ {\int }_a^{b}f\left(x\right)dx={\int }_3^7\ln xdx \end{gathered}$$

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

The area comes out to be${\int }_3^7\ln xdx=6.326$