Question

How do you calculate the left and right Riemann sum for the given function over the interval $\left[ 0,\ln 2 \right]$, using $n=40$ for ${{e}^{x}}$?

Answer 1

Hint: We find the points for left and right Riemann sum in case of interval $\left[ 0,\ln 2 \right]$, using $n=40$ for ${{e}^{x}}$. We find the width of the rectangles. We use them in the formula of \[\sum\limits_{i=0}^{n-1}{\Delta xf\left( a+i\Delta x \right)}\] and \[\sum\limits_{i=1}^{n}{\Delta xf\left( a+i\Delta x \right)}\] to find the solution.

Complete step by step solution:
A Riemann sum can be visualized as a division of (approximately) the area under the curve $f\left( x \right)$ on $\left[ a,b \right]$ into $n$ adjacent rectangles spanning the interval, where the ${{k}^{th}}$ rectangle has width $\Delta x=\dfrac{b-a}{n}$.
This Riemann sum can be expressed with respect to both the starting points or the ending points of the rectangles. They are divided into two parts where they are called left and right Riemann sum.
The formula for left Riemann sum is \[\sum\limits_{i=0}^{n-1}{\Delta xf\left( a+i\Delta x \right)}\].
The formula for left Riemann sum is \[\sum\limits_{i=1}^{n}{\Delta xf\left( a+i\Delta x \right)}\].
We need to calculate the left and right Riemann sum for the given function over the interval $\left[ 0,\ln 2 \right]$, using $n=40$ for ${{e}^{x}}$. Here $f\left( x \right)={{e}^{x}}$. The interval is $\left[ a,b \right]\equiv \left[ 0,\ln 2 \right]$.
We first find the width of the rectangles where $\Delta x=\dfrac{\ln 2-0}{40}=\dfrac{\ln 2}{40}$.
We have to find the functional values of \[f\left( a+i\Delta x \right),i=0(1)40\].
Therefore, for $f\left( x \right)={{e}^{x}}$, we have
$f\left( 0 \right)={{e}^{0}}=1$
$f\left( 0+\dfrac{\ln 2}{40} \right)={{e}^{\dfrac{\ln 2}{40}}}={{2}^{\dfrac{1}{40}}}$
$f\left( 0+\dfrac{2\ln 2}{40} \right)={{e}^{\dfrac{2\ln 2}{40}}}={{2}^{\dfrac{2}{40}}}$
……………………
…………………….
$f\left( 0+\dfrac{40\ln 2}{40} \right)={{e}^{\ln 2}}=2$
Now we find the left Riemann sum value where
\[\begin{align}
  & \sum\limits_{i=0}^{n-1}{\Delta xf\left( a+i\Delta x \right)} \\
 & =\dfrac{\ln 2}{40}\left( 1+{{2}^{\dfrac{1}{40}}}+{{2}^{\dfrac{2}{40}}}+...+{{2}^{\dfrac{39}{40}}} \right) \\
 & =\dfrac{\ln 2}{40}\times \dfrac{2-1}{{{2}^{\dfrac{1}{40}}}-1} \\
 & =\dfrac{\ln 2}{40\left( {{2}^{\dfrac{1}{40}}}-1 \right)} \\
 & =0.991 \\
\end{align}\]
Now we find the right Riemann sum value where
\[\begin{align}
  & \sum\limits_{i=1}^{n}{\Delta xf\left( a+i\Delta x \right)} \\
 & =\dfrac{\ln 2}{40}\left( {{2}^{\dfrac{1}{40}}}+{{2}^{\dfrac{2}{40}}}+...+2 \right) \\
 & =\dfrac{\ln 2}{40}\times \dfrac{{{2}^{\dfrac{1}{40}}}\left( 2-1 \right)}{{{2}^{\dfrac{1}{40}}}-1} \\
 & =\dfrac{{{2}^{\dfrac{1}{40}}}\ln 2}{40\left( {{2}^{\dfrac{1}{40}}}-1 \right)} \\
 & =1.008 \\
\end{align}\]
Therefore, left and right Riemann sums for the given function over the interval $\left[ 0,\ln 2 \right]$, using $n=40$ for ${{e}^{x}}$ are $0.991,1.008$ respectively.

Note: A Riemann sum is an approximation of a region's area, obtained by adding up the areas of multiple simplified slices of the region. It is applied in calculus to formalize the method of exhaustion, used to determine the area of a region. This process yields the integral, which computes the value of the area exactly.