Question

How do you calculate the left and right Riemann sum for the given function over the interval [2,6], for $f\left( x \right)=5{{x}^{2}}+3x+2$ ?

Answer 1

Since n is not given, we are going to assume that $\Delta x=1$. Then to find n, we use the formula of $\Delta x=\dfrac{b-a}{n}$. Then , to find the right Reimann sum, we use the following formula $sum=\Delta x\sum\limits_{r=1}^{n}{f\left( {{x}_{r}} \right)}$. Then , to find the left Reimann sum, we use the following formula $sum=\Delta x\sum\limits_{r=0}^{n-1}{f\left( {{x}_{r}} \right)}$. Finally, substitute all the values to get the final answer.

Complete step by step solution:
Since n is not given, we are going to assume that $\Delta x=1$. Then to find n, we use the formula of $\Delta x=\dfrac{b-a}{n}$.
$\Rightarrow \Delta x=\dfrac{b-a}{n}$
$\Rightarrow 1=\dfrac{6-2}{n}$
$\Rightarrow n=6-2=4$

Then , to find the right Reimann sum, we use the following formula $sum=\Delta x\sum\limits_{r=1}^{n}{f\left( {{x}_{r}} \right)}$.
$\Rightarrow sum=\Delta x\sum\limits_{r=1}^{n}{f\left( {{x}_{r}} \right)}$
$\Rightarrow sum=1\times \sum\limits_{r=1}^{4}{f\left( {{x}_{r}} \right)}$
$\Rightarrow sum=f\left( {{x}_{1}} \right)+f\left( {{x}_{2}} \right)+f\left( {{x}_{3}} \right)+f\left( {{x}_{4}} \right)$
Here, we have ${{x}_{0}}=2,{{x}_{1}}=3,{{x}_{2}}=4,{{x}_{3}}=5,{{x}_{4}}=6$.
Therefore, substituting these values, we get:
$\Rightarrow sum=f\left( 3 \right)+f\left( 4 \right)+f\left( 5 \right)+f\left( 6 \right)$.
But, $f\left( 2 \right)=28,f\left( 3 \right)=56,f\left( 4 \right)=94,f\left( 5 \right)=142,f\left( 6 \right)=200$. Therefore, substituting these values in the above equation, we get:
$\Rightarrow sum=56+94+142+200$.
$\Rightarrow sum=492$
Then , to find the left Reimann sum, we use the following formula $sum=\Delta x\sum\limits_{r=0}^{n-1}{f\left( {{x}_{r}} \right)}$.
$\Rightarrow sum=\Delta x\sum\limits_{r=0}^{n-1}{f\left( {{x}_{r}} \right)}$.
$\Rightarrow sum=1\times \sum\limits_{r=0}^{3}{f\left( {{x}_{r}} \right)}$.
$\Rightarrow sum=f\left( {{x}_{0}} \right)+f\left( {{x}_{1}} \right)+f\left( {{x}_{2}} \right)+f\left( {{x}_{3}} \right)$.
Here, we have ${{x}_{0}}=2,{{x}_{1}}=3,{{x}_{2}}=4,{{x}_{3}}=5,{{x}_{4}}=6$.
Therefore, substituting these values, we get:
$\Rightarrow sum=f\left( 2 \right)+f\left( 3 \right)+f\left( 4 \right)+f\left( 5 \right)$.
But, $f\left( 2 \right)=28,f\left( 3 \right)=56,f\left( 4 \right)=94,f\left( 5 \right)=142,f\left( 6 \right)=200$. Therefore, substituting these values in the above equation, we get:
$\Rightarrow sum=28+56+94+142$.
$\Rightarrow sum=320$
Therefore, we get the right Riemann sum as 492 and the right Riemann sum as 320.

Note: To do this question, you need to know the formulas for the left hand Riemann sum and the right hand Riemann sum. Also, you should be careful while doing all the substitutions. Riemann sum is the sum which is derived by the sum of all the areas of the divided regions.