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="1647079335733" role="math" $f\left(x\right)=\frac{1}{x}\left[1,5\right]$

Short answer

(a)

(b) The four subinterval are $\left[1,2\right],\left[2,3\right],\left[3,4\right],\left[4,5\right]$and the area is $\frac{25}{12}$.

(c) The eight subinterval are $\left[1,\frac{3}{2}\right],\left[\frac{3}{2},2\right],\left[2,\frac{5}{2}\right],\left[\frac{5}{2},3\right],\left[3,\frac{7}{2}\right],\left[\frac{7}{2},4\right],\left[4,\frac{9}{2}\right],\left[\frac{9}{2},5\right]$and the area is $\frac{4609}{2520}$

(d) (Area under the curve as an integral is given by${\int }_a^{b}f\left(x\right)dx={\int }_1^5\frac{1}{x}dx$

e) Using graphing utility the area found is$1.609$.

Step-by-step solution

Part (a) Step 1. Given

$f\left(x\right)=\frac{1}{x}\left[1,5\right]$

Part (a) Step 2. Graph

Part (b) Step 1. Calculation

The area under the curve can be found using

$$\begin{gathered} A=\left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)\right) \\ where{u}_1,u_2,u_3,u_{4}are4equalinterval. \\ nowintervalwillbedecidedby \\ \left(\frac{b-a}{n}\right)=\frac{5-1}{4} \\ =1 \\ Therefore\left[1,2\right],\left[2,3\right],\left[3,4\right],\left[4,5\right]aretherespectiveintervals. \end{gathered}$$

$$\begin{gathered} Applyingtheformulaforareaweget \\ A=\left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)\right) \\ =\left(1\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) \\ =\frac{25}{12} \end{gathered}$$

Part (c) Step 1. Calculation

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

$$\begin{gathered} A=\left(\frac{b-a}{n}\right)\left(f\left(u_1\right)+f\left(u_2\right)+f\left(u_3\right)+f\left(u_4\right)+f\left(u_5\right)+f\left(u_6\right)+f\left(u_7\right)+f\left(u_8\right)\right) \\ =\left(\frac{1}{2}\right)\left(1+\frac{2}{3}+\frac{1}{2}+\frac{2}{5}+\frac{1}{3}+\frac{2}{7}+\frac{1}{4}+\frac{2}{9}\right) \\ =\left(\frac{1}{2}\right)\left(\frac{4609}{1260}\right) \\ =\frac{4609}{2520} \end{gathered}$$

Part (d) Step 1. Area in integral form

$$\begin{gathered} f\left(x\right)=x^3 \\ \left[a,b\right]=\left[1,5\right] \\ {\int }_a^{b}f\left(x\right)dx={\int }_1^5\frac{1}{x}dx \end{gathered}$$

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

The area comes out to be${\int }_1^5\frac{1}{x}dx=1.609$