Question

The function $f\left(x\right)=-3x+9$ is defined on the interval
$[0,3]$
(a) Graph f.
In (b)–(e), approximate the area A under f from $0$to $3$
as follows:
(b) Partition $[0,3]$ into three sub intervals of equal length and choose u as the left endpoint of each sub interval.
(c) Partition $[0,3]$ into three sub intervals of equal length and choose u as the right endpoint of each sub interval.
(d) Partition $[0,3]$ into six sub intervals of equal length and choose u as the left endpoint of each sub interval.
(e) Partition $[0,3]$ into six sub intervals of equal length and choose u as the right endpoint of each sub interval.
(f) What is the actual area A?

Short answer

(a) The graph of the function $f\left(x\right)=-3x+9$defined on the $[0,3]$interval is:

(b) The area of the function by partitioning $[0,3]$into three sub intervals by choosing left end point is $18$

(c) The area of the function by partitioning $[0,3]$into three sub intervals by choosing right end point is $9$

(d) The area of the function by partitioning $[0,3]$into six sub intervals by choosing left end point is $15.75$

(e) The area of the function by partitioning $[0,3]$into six sub intervals by choosing right end point is $11.25$

(f) The actual area is$13.5$

Step-by-step solution

Part (a) Step 1. Given

The function:

$f\left(x\right)=-3x+9$

On the interval

Part (a) Step 2. Graph the function

Graph the given function on the interval

Part (b) Step 1. Choosing u as left end point

Partition $[0,3]$into three sub intervals, each of length $1$, and choose uas the left endpoint.

The sub intervals are $[0,1],[1,2],[2,3]$

Part (b) Step 2. Approximate the area.

Approximating the area of the function,

$$\begin{gathered} A=f\left(u_1\right)\frac{b-a}{3}+f\left(u_2\right)\frac{b-a}{3}+f\left(u_3\right)\frac{b-a}{3} \\ =f\left(0\right)\left(1\right)+f\left(1\right)\left(1\right)+f\left(2\right)1) \\ =9+6+3 \\ ={18}^{} \end{gathered}$$

Part (c) Step 1. Choosing u as right end point

Partition $[0,3]$into three sub intervals, each of length , and choose uas the right endpoint.

The sub intervals are$[0,1],[1,2],[2,3]$

Part (c) Step 2. Approximate the area.

Approximating the area of the function,

$$\begin{gathered} A=f\left(u_1\right)\frac{b-a}{3}+f\left(u_2\right)\frac{b-a}{3}+f\left(u_3\right)\frac{b-a}{3} \\ =f\left(1\right)1+f\left(2\right).1+f\left(3\right).1 \\ =6+3+0 \\ =9 \end{gathered}$$

Part (d) Step 1. Choosing u as left end point

Partition $[0,3]$ into six sub intervals, each of length 0.5, and choose uas the left endpoint.

The sub intervals are$[0,0.5],[0.5,1],[1,1.5],[1.5,2],[2,2.5],[2.5,3]$

Part (d) Step 2. Approximate the area.

Approximating the area of the function,

$A=f\left(u_1\right)\frac{b-a}{0.5}+f\left(u_2\right)\frac{b-a}{0.5}+f\left(u_3\right)\frac{b-a}{0.5}+f\left(u_4\right)\frac{b-a}{0.5}+f\left(u_5\right)\frac{b-a}{0.5}+f\left(u_6\right)\frac{b-a}{0.5}$

$$\begin{gathered} A=f\left(0\right)\frac{b-a}{6}+f(0.5)\frac{b-a}{6}+f\left(1\right)\frac{b-a}{6}+f(1.5)\frac{b-a}{6}+f\left(2\right)\frac{b-a}{6}+f(2.5)\frac{b-a}{6} \\ =9(0.5)+7.5(0.5)+6(0.5)+4.5(0.5)+3(0.5)+1.5(0.5) \\ =15.75 \end{gathered}$$

Part (e) Step 1. Choosing u as right end point

Partition $[0,3]$into six sub intervals, each of length $0.5$, and choose uas the right endpoint.

The sub intervals are$[0,0.5],[0.5,1],[1,1.5],[1.5,2],[2,2.5],[2.5,3]$

Part (e) Step 2. Approximate the area.

Approximating the area of the function,

$$\begin{gathered} A=f\left(u_1\right)\frac{b-a}{0.5}+f\left(u_2\right)\frac{b-a}{0.5}+f\left(u_3\right)\frac{b-a}{0.5}+f\left(u_4\right)\frac{b-a}{0.5}+f\left(u_5\right)\frac{b-a}{0.5}+f\left(u_6\right)\frac{b-a}{0.5} \\ =f(0.5)(0.5)+f\left(1\right)(0.5)+f(1.5)(0.5)+f\left(2\right)(0.5)+f(2.5)(0.5)+f\left(3\right)(0.5) \\ =11.25 \end{gathered}$$

Part (f) Step 1. Find the base and height

From the graph, the base is

The height of the triangle is

Part (f) Step 2. Find the area

Area of the triangle $=\frac{1}{2}bh$

$$\begin{gathered} =\frac{1}{2}\times 3\times 9 \\ =13.5 \end{gathered}$$