Question

How do you use the limit process to find the area of the region between the graph $y=16-{{x}^{2}}$ and the x-axis over the interval $\left[ 1,3 \right]$?

Answer 1

In order to find the solution of the given question that is to use the limit process to find the area of the region between the graph $y=16-{{x}^{2}}$ and the x-axis over the interval apply the limit definition of the definite integral which states that $\int_{a}^{b}{f\left( x \right)dx=\displaystyle \lim_{n \to \infty }}\sum\limits_{i=1}^{n}{f\left( {{x}_{i}} \right)}\Delta x$ where, for each positive integer n, we let $\Delta x=\dfrac{b-a}{n}$
And for $i=1,2,3,...,n,$ we let ${{x}_{i}}=a+i\Delta x$ (These ${{x}_{i}}$ are the right endpoints of the subintervals.)

Complete step by step answer:
According to the question, we have to find the following using limit process:
$\int_{1}^{3}{\left( 16-{{x}^{2}} \right)dx}$
To find the value of above expression using limit process we will use the formula that is $\int_{a}^{b}{f\left( x \right)dx=\displaystyle \lim_{n \to \infty }}\sum\limits_{i=1}^{n}{f\left( {{x}_{i}} \right)}\Delta x$ where, for each positive integer n, we let $\Delta x=\dfrac{b-a}{n}$
And for $i=1,2,3,...,n,$ we let ${{x}_{i}}=a+i\Delta x$.
So first we have to find $\Delta x$
For each n, we get
$\Delta x=\dfrac{b-a}{n}=\dfrac{3-1}{n}=\dfrac{2}{n}$
Then find the value of ${{x}_{i}}$, we will have:
${{x}_{i}}=a+i\Delta x=1+i\dfrac{2}{n}=1+\dfrac{2i}{n}$
Now we will find the value of $f\left( {{x}_{i}} \right)$, we will get:
$f\left( {{x}_{i}} \right)=16-{{\left( {{x}_{i}} \right)}^{2}}=16-{{\left( 1+\dfrac{2i}{n} \right)}^{2}}$
$\Rightarrow f\left( {{x}_{i}} \right)=16-\left( 1+\dfrac{4i}{n}+\dfrac{4{{i}^{2}}}{{{n}^{2}}} \right)=15-\dfrac{4i}{n}-\dfrac{4{{i}^{2}}}{{{n}^{2}}}$
After this evaluate $\sum\limits_{i=1}^{n}{f\left( {{x}_{i}} \right)}\Delta x$ in order to evaluate the sums, we will get:
$\sum\limits_{i=1}^{n}{f\left( {{x}_{i}} \right)}\Delta x=\sum\limits_{i=1}^{n}{\left( 15-\dfrac{4i}{n}-\dfrac{4{{i}^{2}}}{{{n}^{2}}} \right)}\dfrac{2}{n}$
$\Rightarrow \sum\limits_{i=1}^{n}{f\left( {{x}_{i}} \right)}\Delta x=\sum\limits_{i=1}^{n}{\left( \dfrac{30}{n}-\dfrac{8i}{{{n}^{2}}}-\dfrac{8{{i}^{2}}}{{{n}^{3}}} \right)}$
$\Rightarrow \sum\limits_{i=1}^{n}{f\left( {{x}_{i}} \right)}\Delta x=\dfrac{30}{n}\sum\limits_{i=1}^{n}{\left( 1 \right)-}\dfrac{8}{{{n}^{2}}}\sum\limits_{i=1}^{n}{\left( i \right)}-\dfrac{8}{{{n}^{3}}}\sum\limits_{i=1}^{n}{\left( {{i}^{2}} \right)}$
Applying the summation formulas in the above expression we will get:
$\Rightarrow \sum\limits_{i=1}^{n}{f\left( {{x}_{i}} \right)}\Delta x=\dfrac{30}{n}\left( n \right)-\dfrac{8}{{{n}^{2}}}\left( \dfrac{n\left( n+1 \right)}{2} \right)-\dfrac{8}{{{n}^{3}}}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)$
Now putting the above values in the formula $\int_{a}^{b}{f\left( x \right)dx=\displaystyle \lim_{n \to \infty }}\sum\limits_{i=1}^{n}{f\left( {{x}_{i}} \right)}\Delta x$, we will get:
$\Rightarrow \int_{1}^{3}{\left( 16-{{x}^{2}} \right)dx=}\displaystyle \lim_{n \to \infty }\left( 30-4\left( \dfrac{n\left( n+1 \right)}{{{n}^{2}}} \right)-\dfrac{4}{3}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{{{n}^{3}}} \right) \right)$
Now we know that $\displaystyle \lim_{n \to \infty }\left( \dfrac{n\left( n+1 \right)}{{{n}^{2}}} \right)=1$ and $\displaystyle \lim_{n \to \infty }\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{{{n}^{3}}} \right)=2$, substituting these values in the above expression we will have:
$\Rightarrow \int_{1}^{3}{\left( 16-{{x}^{2}} \right)dx=}30-4\left( 1 \right)-\dfrac{4}{3}\left( 2 \right)$
Simplifying the above expression by opening the brackets and taking the LCM we will have:
$\Rightarrow \int_{1}^{3}{\left( 16-{{x}^{2}} \right)dx=}\dfrac{90}{3}-\dfrac{12}{3}-\dfrac{8}{3}$
After solving it further we will get:
$\Rightarrow \int_{1}^{3}{\left( 16-{{x}^{2}} \right)dx=}\dfrac{70}{3}$
Therefore, the area of the region between the graph $y=16-{{x}^{2}}$ and the x-axis over the interval $\left[ 1,3 \right]$ that is $\int_{1}^{3}{\left( 16-{{x}^{2}} \right)dx}$equal to $\dfrac{70}{3}$.

Note:
Students can go wrong by applying the wrong formula of limit definition of definite integral that is they apply $\int_{a}^{b}{f\left( x \right)dx=\displaystyle \lim_{n \to \infty }}\sum\limits_{i=1}^{n}{f\left( \Delta x \right)}{{x}_{i}}$which is completely wrong and leads to the wrong answer. It’s important to remember that the formula of limit definition of definite integral is $\int_{a}^{b}{f\left( x \right)dx=\displaystyle \lim_{n \to \infty }}\sum\limits_{i=1}^{n}{f\left( {{x}_{i}} \right)}\Delta x$ where, for each positive integer n, $\Delta x=\dfrac{b-a}{n}$ for $i=1,2,3,...,n,$ and ${{x}_{i}}=a+i\Delta x$.