Chapter 14.5, Problem 9AYU

To determine

To solve: The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$,

a. Graph $f$.

In (b)–(e), approximate the area $A$ under $f$ from 0 to 6 as follows:

Answer to Problem 9AYU

Solution:

a.

Explanation of Solution

Given:

The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$.

Calculation:

$f\left(0\right)=3\left(0\right)=0$; $f\left(1\right)=3\left(1\right)=3$; $f\left(2\right)=3\left(2\right)=6$; $f\left(3\right)=3\left(3\right)=9$; $f\left(4\right)=3\left(4\right)=12$; $f\left(5\right)=3\left(5\right)=15$; $f\left(6\right)=3\left(6\right)=18$

a. Graph $f\left(x\right)=3x$.

To determine

To solve: The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$,

b. Partition $\left[0,6\right]$ into three subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 9AYU

Solution:

b. 36

Explanation of Solution

Given:

The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$.

Calculation:

$f\left(0\right)=3\left(0\right)=0$; $f\left(1\right)=3\left(1\right)=3$; $f\left(2\right)=3\left(2\right)=6$; $f\left(3\right)=3\left(3\right)=9$; $f\left(4\right)=3\left(4\right)=12$; $f\left(5\right)=3\left(5\right)=15$; $f\left(6\right)=3\left(6\right)=18$

b. Partition $\left[0,6\right]$ into three subintervals of equal length 2 and choose $u$ as the left endpoint of each subinterval.

The area $A$ is approximated as

$A=f\left(0\right)2+f\left(2\right)2+f\left(4\right)2$

$=0\left(2\right)+6\left(2\right)+12\left(2\right)$

$=0+12+24$

$=36$

To determine

To solve: The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$,

c. Partition $\left[0,6\right]$ into three subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

Answer to Problem 9AYU

Solution:

c. 72

Explanation of Solution

Given:

The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$.

Calculation:

$f\left(0\right)=3\left(0\right)=0$; $f\left(1\right)=3\left(1\right)=3$; $f\left(2\right)=3\left(2\right)=6$; $f\left(3\right)=3\left(3\right)=9$; $f\left(4\right)=3\left(4\right)=12$; $f\left(5\right)=3\left(5\right)=15$; $f\left(6\right)=3\left(6\right)=18$

c. Partition $\left[0,6\right]$ into three subintervals of equal length 2 and choose $u$ as the right endpoint of each subinterval.

The area $A$ is approximated as

$A=f\left(2\right)2+f\left(4\right)2+f\left(6\right)2$

$=6\left(2\right)+12\left(2\right)+18\left(2\right)$

$=12+24+36$

$=72$

To determine

To solve: The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$,

d. Partition $\left[0,6\right]$ into six subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 9AYU

Solution:

d. 45

Explanation of Solution

Given:

The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$.

Calculation:

$f\left(0\right)=3\left(0\right)=0$; $f\left(1\right)=3\left(1\right)=3$; $f\left(2\right)=3\left(2\right)=6$; $f\left(3\right)=3\left(3\right)=9$; $f\left(4\right)=3\left(4\right)=12$; $f\left(5\right)=3\left(5\right)=15$; $f\left(6\right)=3\left(6\right)=18$

d. Partition $\left[0,6\right]$ into six subintervals of equal length 1 and choose $u$ as the left endpoint of each subinterval.

The area $A$ is approximated as

$A=f\left(0\right)1+f\left(1\right)1+f\left(2\right)1+f\left(3\right)1+f\left(4\right)1+f\left(5\right)1$

$=0\left(1\right)+3\left(1\right)+6\left(1\right)+9\left(1\right)+12\left(1\right)+15\left(1\right)$

$=0+3+6+9+12+15$

$=45$

To determine

To solve: The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$,

e. Partition $\left[0,6\right]$ into six subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

Answer to Problem 9AYU

Solution:

e. 63

Explanation of Solution

Given:

The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$.

Calculation:

$f\left(0\right)=3\left(0\right)=0$; $f\left(1\right)=3\left(1\right)=3$; $f\left(2\right)=3\left(2\right)=6$; $f\left(3\right)=3\left(3\right)=9$; $f\left(4\right)=3\left(4\right)=12$; $f\left(5\right)=3\left(5\right)=15$; $f\left(6\right)=3\left(6\right)=18$

e. Partition $\left[0,6\right]$ into six subintervals of equal length 1 and choose $u$ as the right endpoint of each subinterval.

The area $A$ is approximated as

$A=f\left(1\right)1+f\left(2\right)1+f\left(3\right)1+f\left(4\right)1+f\left(5\right)1+f\left(6\right)1$

$=3\left(1\right)+6\left(1\right)+9\left(1\right)+12\left(1\right)+15\left(1\right)+18\left(1\right)$

$=3+6+9+12+15+18$

$=63$

To determine

To solve: The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$,

f. What is the actual area $A$?

Answer to Problem 9AYU

Solution:

f. 54

Explanation of Solution

Given:

The function $f\left(x\right)=3x$ is defined on the interval $\left[0,6\right]$.

Calculation:

$f\left(0\right)=3\left(0\right)=0$; $f\left(1\right)=3\left(1\right)=3$; $f\left(2\right)=3\left(2\right)=6$; $f\left(3\right)=3\left(3\right)=9$; $f\left(4\right)=3\left(4\right)=12$; $f\left(5\right)=3\left(5\right)=15$; $f\left(6\right)=3\left(6\right)=18$

f. The actual area under the graph of $f\left(x\right)=3x$ from 0 to 6 is the area of a right triangle whose base is of length 6 and whose height is 18. The actual area $A$ is

Therefore

$A= \frac{1}{2}\text{base }\times \text{ height}$

$A= \frac{1}{2}\left(6\right)\left(18\right)=54$