Question

You are given the table below.

Answer 1

The integral value of $\int\limits_4^{20} {f\left( x \right)dx}$ is $\boxed{\int\limits_4^{20} {f\left( x \right)dx} = 284{\text{ uni}}{{\text{t}}^2}}.$

Further explanation:

The equation of the line can be expressed as follows,

$\boxed{\frac{{\left( {y - {y_1}} \right)}}{{{y_2} - {y_1}}} = \frac{{\left( {x - {x_1}} \right)}}{{{x_2} - {x_1}}}}$

Given:

The value of the function at different values of $x$ can be written as follows,

$\begin{aligned}x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,\,\,12\,\,\,\,\,\,\,\,\,\,16\,\,\,\,\,\,\,\,\,20\hfill\\f\left(x\right)\,\,\,\,\,\,11\,\,\,\,\,\,\,25\,\,\,\,\,\,\,\,\,\,\,16\,\,\,\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\,\,\,\,\,31 \hfill \\\end{aligned}$

Explanation:

The function is an increasing function if ${x_1} > {x_2}$, than $f\left( {{x_1}} \right) > f\left( {{x_2}} \right)$ and the function is decreasing if ${x_1} > {x_2}, {\text{than} f\left( {{x_1}} \right) < f\left( {{x_2}} \right).$

The $\int\limits_4^{20} {f\left( x \right)dx}$ can be obtained as follows,

$\int\limits_4^{20} {f\left( x \right)dx} = \int\limits_4^8 {f\left( x \right)dx} + \int\limits_8^{12} {f\left( x \right)dx} + \int\limits_{12}^{16} {f\left( x \right)dx} + \int\limits_{16}^{20} {f\left( x \right)dx}$

The equation of line between the points $\left( {4,11} \right) {\text{and} \left( {8,25} \right)$ can be obtained as follows,

$\begin{aligned}\frac{{y - 11}}{{25 - 11}} &= \frac{{x - 4}}{{8 - 4}}\\y- 11&=\frac{{14}}{4}\left( {x - 4} \right)\\y&= \frac{{7x}}{2}- 3\\\end{aligned}$

Similarly the equation between the points $\left( {8,25} \right) {\text{and} \left( {12,16} \right)$ can be expressed as follows,

$y = \dfrac{{ - 9x}}{4} + 43$

The equation between the points $\left( {12,16} \right) {\text{and} \left( {16,9} \right)$ can be expressed as follows,

$y = \dfrac{{ - 7x}}{4} + 37$

The equation between the points $\left( {16,9} \right) {\text{and} \left( {20,31} \right)$ can be expressed as follows,

$y = \dfrac{{11x}}{4} - 79$

The area under the line $y = \dfrac{{7x}}{2} - 3$ can be obtained as follows,

$\begin{aligned}\int\limits_4^8 {\left( {\frac{{7x}}{2} - 3} \right)} dx&= \left. {\frac{{7{x^2}}}{4}} \right|_{x = 4}^{x = 8} - \left. {3x} \right|_{x = 4}^{x = 8}\\&= \left( {112 - 24} \right) - \left( {28 - 12} \right)\\&= 72\\\end{aligned}$

Similarly the area under the line $y = \dfrac{{ - 9x}}{4} + 43$ can be obtained as follows,

$\begin{aligned}\int\limits_8^{12} {\left({- \frac{{9x}}{4} + 43} \right)} dx &= \left. { - \frac{{9{x^2}}}{8}} \right|_{x = 8}^{x = 12} + 4\left. {3x} \right|_{x = 8}^{x = 12}\\&= \left( { - 162 + 516} \right) - \left( {72 - 344} \right)\\&= 82\\\end{aligned}$

The area under the line $y = \dfrac{{ - 7x}}{4} + 37$ can be obtained as follows,

$\begin{aligned}\int\limits_{12}^{16} {\left({ - \frac{{7x}}{4} + 37} \right)} dx &= \left.{ - \frac{{7{x^2}}}{8}} \right|_{x = 12}^{x = 16} - \left. {3x} \right|_{x = 12}^{x = 16}\\&= \left( { - 224 + 592} \right) - \left( {126 - 444} \right)\\&= 50\\\end{aligned}$

The area under the line $y ={ \dfrac{{11x}}{4} - 79$ can be obtained as follows,

$\begin{aligned}\int\limits_{16}^{20} {\left( {\dfrac{{11x}}{2} - 79} \right)} dx &= \left. {\dfrac{{11{x^2}}}{4}} \right|_{x = 16}^{x = 20} - \left. {79x} \right|_{x = 16}^{x = 20}\\&= \left( {1100 - 1580} \right) - \left( {704 - 1264} \right)\\&= 80\\\end{aligned}$

Total area can be obtained as follows,

$\begin{aligned}\int\limits_4^{20} {f\left( x \right)dx}&= 72 + 82 + 50 + 80\\&= 284{\text{ unit}}{{\text{s}}^2}\\\end{aligned}$

The integral value of $\int\limits_4^{20} {f\left( x \right)dx}$ is $\boxed{\int\limits_4^{20} {f\left( x \right)dx} = 284{\text{ uni}}{{\text{t}}^2}}.$

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: College

Subject: Mathematics

Chapter: Integrals

Keywords: function, decreasing, increasing, monotonic, integral, monotone, data is not monotone, table, sketch, possible, graph, integration, area under the curve.

Answer 2

$\int\limits^{20}_4 {f(x)} \, dx = area from4 to 20$

n = 4 means that you are going to divide the area in 4 equal parts, that means that the size of the intervals is [20-4] / 4 = 16 / 4 = 4.

Now you can estimate the area from 4 to 20, by adding the area from 4 to 8 + the area from 8 to 12 +  the area from 12 to 16 + the area from 16 to 20

And estimate each small area by a rectangle or a trapezoid.

As a trapezoid, area of each region is [f(x) +  f(x+4)] /2 [x +4 - x] = [f(x) +  f(x+4)] /2 [4] =

=[f(x) +  f(x+4)] / 8

Area from 4 to 8: [11 +25] / 8 = 4.50
Area from 8 to 12: [25+16]/8 = 5.125
Area from 12 to 16 = [16+9]/8 = 3.125
Area from 16 to 20 = [9+31]/8 =   5.00

Total area = 17.75

Answer: 17.75