Chapter 12, Problem 17CT

To determine

Find the area of the region bounded by the graph of the function.

Answer to Problem 17CT

The area of region is $\boxed{\frac{16}{3}}$ square unit.

Explanation of Solution

Given information:

Use the limit process to find the area of the region bounded by the graph of the function and the -axis over the specified interval.

  $f\left(x\right)=3-x^2$

  $\begin{array}{l}\mathrm{int}erval\\ \left[-1,1\right]\end{array}$

Calculation:

Consider the given function and corresponding interval ,

  $f\left(x\right)=3-x^2$

  $\begin{array}{l}\mathrm{int}erval\\ \left[-1,1\right]\end{array}$

From the interval,

  $\begin{array}{l}a=-1\\ b=1\end{array}$

Now calculate width and height by using standard formula,

  $\begin{array}{l}Width=\frac{b-a}{n}\\ Height=f\left(a+\frac{\left(b-a\right)i}{n}\right)\end{array}$

Substitute the given values,

  $Width=\frac{1-\left(-1\right)}{n}=\frac{2}{n}$

  $Height=f\left(-1+\frac{2i}{n}\right)$

Now height from the function,

  $f\left(x\right)=3-x^2$

  $\begin{array}{l}f\left(-1+\frac{2i}{n}\right)=3-{\left(-1+\frac{2i}{n}\right)}^2\\ =3-\left(1-\frac{4i}{n}+\frac{4i^2}{n^2}\right)\\ height=2+\frac{4i}{n}-\frac{4i^2}{n^2}\end{array}$

Now calculate the approximate area as sum of all areas of rectangles by using this formula.

  $\begin{array}{l}A={\displaystyle \sum _{i=1}^{n}f\left(a+\frac{\left(b-a\right)i}{n}\right)}\left(\frac{b-a}{n}\right)\\ ={\displaystyle \sum _{i=1}^n\left(\frac{4}{n}+\frac{8i}{n^2}-\frac{8i^2}{n^3}\right)}\end{array}$

Apply standard formula of summation,

  $\begin{array}{l}{\displaystyle \sum _{i=1}^{n}k=kn}\\ {\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{n}\\ {\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\end{array}$

Therefore,

  $\begin{array}{l}A=\frac{4}{n}{\displaystyle \sum _{i=1}^{n}1+}\frac{8}{n^2}{\displaystyle \sum _{i=1}^{n}i-}\frac{8}{n^3}{\displaystyle \sum _{i=1}^n{i}^2}\\ =\frac{4}{n}\left(n\right)+\frac{8}{n^2}\frac{n\left(n+1\right)}{2}-\frac{8}{n^3}\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ =4+\frac{\left(n+1\right)}{n}-\frac{4\left(2n^2+3n+1\right)}{3n^2}\\ A=\underset{n\to \infty }{\lim }4+\underset{n\to \infty }{\lim }\frac{\left(n+1\right)}{n}-\underset{n\to \infty }{\lim }\frac{4\left(2n^2+3n+1\right)}{3n^2}\end{array}$

Now evaluate the limit with conditions.

1. When degree of numerator is less than the degree of the denominator then limit exist and equal to zero.
2. When degree of numerator is equal to the degree of the denominator then limit exist and equal to the ratio of coefficients of the highest degree term.

  $\begin{array}{l}A=4+4-\frac{8}{3}\\ =8-\frac{8}{3}\\ =\frac{16}{3}\end{array}$

Hence the area of region is $\boxed{\frac{16}{3}}$ square unit.