Chapter 12, Problem 101RE

To determine

To find: The area of the region under the given curve on $\left[-1,2\right]$ using limit summation formulas.

Answer to Problem 101RE

The required area is $15$ square units.

Explanation of Solution

Given:

The given curve is $f\left(x\right)=x^2+4$ .

Concept used:

Suppose $f$ is a continuous function and $f\left(x\right)\ge 0$ through out the interval $\left[a,b\right]$ . Then the area of region under the curve $y=f\left(x\right)$ over the interval is

  $A=\underset{n\to \infty }{\lim }\left[{\displaystyle \sum _{k=1}^{n}f\left(a+\frac{\left(b-a\right)k}{n}\right)\left(\frac{b-a}{n}\right)}\right]$

Formula used:

Summation formula for $k$ is given as follows:

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

Summation formula for $k^3$ is given as follows:

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

Summation formula for constant function is given as follows:

  ${\displaystyle \sum _{k=1}^{n}c}=cn$

Linear rule of summation is as follows:

  ${\displaystyle \sum _{k=1}^n\left(c{a}_k+c{b}_k\right)}=c{\displaystyle \sum _{k=1}^n{a}_k}+c{\displaystyle \sum _{k=1}^n{b}_k}$

Here, $c$ is any constant

Calculation:

Consider the given curve.

  $f\left(x\right)=x^2+4$

Here, the interval is $\left[-1,2\right]$ .

Therefore, $b=2,a=-1$

Therefore, width of each sub-interval will be as shown.

  $\begin{array}{c}\frac{b-a}{n}=\frac{2+1}{n}\\ =\frac{3}{n}\end{array}$

Now, the area of the region under the given curve can be calculated as shown.

Therefore, the exact area of the region under the given curve is $15$ square units.