Chapter 12.5, Problem 47E

To determine

To find: The area of the boundaries of a parcel of land are two edges modelled by the coordinate axes using by the graphing utility to graph the equation.

Answer to Problem 47E

The area of region is $\text{63541}\text{.7}{\text{ft}}^{\text{2}}$

Explanation of Solution

Given information:

The given function is,

  $y=\left(-3.0\times {10}^{-6}\right)x^3+0.002x^2-1.05x+400$

Concept used:

The area of a region;

Let $f$ be continuous and nonnegative on the interval $\left[a,b\right]$ The area A of the region bounded by the graph of $f$ , the $\text{x-axis}$ ,and the vertical lines $x=a$ and $x=b$ is given by

  $A=\underset{x\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\underset{\text{height}}{\underbrace{\left(a+\frac{\left(b-a\right)i}{n}\right)}}}\underset{\text{width}}{\underbrace{\left(\frac{b-a}{n}\right)}}$

The given function is,

  $y=\left(-3.0\times {10}^{-6}\right)x^3+0.002x^2-1.05x+400$

The area of the region bounded by the graph of $y=\left(-3.0\times {10}^{-6}\right)x^3+0.002x^2-1.05x+400$ and the x-axis between $x=0$ and $x=500$

The dimensions of the rectangles;

$\begin{array}{c}width;\frac{b-a}{n}=\frac{500-0}{n}\\ =\frac{500}{n}\end{array}$

And,

  $\begin{array}{c}\text{Height}\text{;}f\left(a+\frac{\left(b-a\right)i}{n}\right)=f\left(0+\frac{\left(500-0\right)i}{n}\right)\\ =f\left(\frac{500i}{n}\right)\end{array}$

Given function; $y=\left(-3.0\times {10}^{-6}\right)x^3+0.002x^2-1.05x+400$

  $\begin{array}{c}f\left(x\right)=\left(-3.0\times {10}^{-6}\right)x^3+0.002x^2-1.05x+400\\ =\left(-3.0\times {10}^{-6}\right){\left(\frac{500i}{n}\right)}^3+0.002{\left(\frac{500i}{n}\right)}^2-1.05\left(\frac{500i}{n}\right)+400\\ =\left(-\frac{375i^3}{n^3}+\frac{500i^2}{n^2}-\frac{525i}{n}+400\right)\end{array}$

Approximate the areas as the sum of the areas of $n$ rectangles.

  $\begin{array}{c}\text{A=}\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{\text{i=1}}^{\text{n}}\text{f}\left(\text{a+}\frac{\left(\text{b-a}\right)\text{i}}{\text{n}}\right)\left(\frac{\text{b-a}}{\text{n}}\right)}\right)\\ =\underset{x\to \infty }{\lim }\left({\displaystyle \sum _{i=1}^n\left(-\frac{375i^3}{n^3}+\frac{500i^2}{n^2}-\frac{525i}{n}+400\right)\left(\frac{500}{n}\right)}\right)\end{array}$

Apply Summation formula and properties.

  ${\displaystyle \sum _{i=1}^{n}i=\frac{n\left(n+1\right)}{2}}$

  ${\displaystyle \sum _{i=1}^n{i}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

  ${\displaystyle \sum _{i=1}^n{i}^3}=\frac{n^2{\left(n+1\right)}^2}{4}$

Now,

  $\begin{array}{c}\text{A=}\underset{x\to \infty }{\lim }\left(\text{-}\left(\frac{\text{187500}}{{\text{n}}^{\text{4}}}{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{i}}^{\text{3}}}\right)\text{+}\left(\frac{\text{250000}}{{\text{n}}^{\text{3}}}{\displaystyle \sum _{\text{i=1}}^{\text{n}}{\text{i}}^{\text{2}}}\right)\text{-}\left(\frac{\text{262500}}{{\text{n}}^{\text{2}}}{\displaystyle \sum _{\text{i=1}}^{\text{n}}\text{i}}\right)\text{+200000}\right)\\ \text{=}\underset{x\to \infty }{\lim }\left(\text{-}\left(\frac{\text{187500}}{{\text{n}}^{\text{4}}}\text{×}\frac{{\text{n}}^{\text{2}}{\left(\text{n+1}\right)}^{\text{2}}}{\text{4}}\right)\text{+}\left(\frac{\text{250000}}{{\text{n}}^{\text{3}}}\text{×}\frac{\text{n}\left(\text{n+1}\right)\left(\text{2n+1}\right)}{\text{6}}\right)\text{-}\left(\frac{\text{262500}}{{\text{n}}^{\text{2}}}\text{×}\frac{\text{n}\left(\text{n+1}\right)}{\text{2}}\right)\text{+200000}\right)\\ \text{=}\underset{x\to \infty }{\lim }\left(\left(\text{-46875×}\frac{{\left(\text{n+1}\right)}^{\text{2}}}{{\text{n}}^{\text{2}}}\right)\text{+}\left(\frac{\text{125000}}{\text{3}}\text{×}\frac{{\text{2n}}^{\text{2}}\text{+3n+1}}{{\text{n}}^{\text{2}}}\right)\text{-}\left(\text{131250×}\frac{{\text{n}}^{\text{2}}\text{+n}}{{\text{n}}^{\text{2}}}\right)\text{+200000}\right)\end{array}$

Finally the exact area by taking the limit as $n$ approaches $\infty$

  $\begin{array}{c}A=\underset{x\to \infty }{\lim }\left[\left(-46875\times \frac{{\left(n+1\right)}^2}{n^2}\right)+\left(\frac{125000}{3}\times \frac{2n^2+3n+1}{n^2}\right)-\left(131250\times \frac{n^2+n}{n^2}\right)+200000\right]\\ =\left[-46875+\frac{125000}{3}-131250+200000\right]\\ =63541.7\end{array}$

Therefore, the exact area is $\text{63541}\text{.7}{\text{ft}}^{\text{2}}$