Question

What is midpoint Riemann sum?

Answer 1

Hint: Before giving the answer to the above question, first I will like to tell you what the Riemann sum. In mathematics, a Riemann sum is a certain kind of approximation of an integral by a finite sum. The very common application of Riemann sum is that it is the approximating the area of functions or lines on a graph, but also the length of curves and other approximations.

Complete step by step solution:
Let $f:[a,b]\to \mathbb{R}$ be a function defined on a closed interval $[a,b]$ of the real numbers,$\mathbb{R}$, and $P=\{[x_0,x_1],[x_1,x_2]..........[x_{n-1},x_n]\}$ be the partition of I, where $a=x_0<x_1<x_2<...........<x_n=b$ .
A Riemann sum S of $f$ over I with partition P is defined as $S=\sum _{i=1}^{n}f\left({x^{\ast }}_i\right)\mathrm{\Delta }{x}_i$ where $\mathrm{\Delta }{x}_i=x_i-x_{i-1}$ and ${x^{\ast }}_i\in [x_{i-1},x_i]$.
One might produce different Riemann sums depending on which ${x^{\ast }}_i$`s chosen. In the end this will not matter, if the function is Riemann integrable, when the difference or width of the summands $\mathrm{\Delta }{x}_i$approaches zero.
Now we will discuss the midpoint Riemann sum. As we discussed above the Riemann sum is an approximation of the area under a curve by dividing it into multiple simple shapes. And in the midpoint Riemann sum, the height of each rectangle is equal to the value of the function at the midpoint of its base.
Let's take an example to simplify it. Let for the interval $[1,3]$ and for $n=4$.
Now first as we do in Riemann sum, we will find $\mathrm{\Delta }x$.
$\begin{array}{rl}& \Rightarrow \mathrm{\Delta }x={\displaystyle \frac{b-a}{n}}\\ & \Rightarrow \mathrm{\Delta }x={\displaystyle \frac{3-1}{4}}={\displaystyle \frac{1}{2}}\end{array}$
Now the end points of the subintervals are:
$1,{\displaystyle \frac{3}{2}},2,{\displaystyle \frac{5}{2}},3$
Now the left Riemann sum uses the left endpoints to find the height of the rectangle. The midpoint sum uses the midpoints of the subintervals:
$[1,{\displaystyle \frac{3}{2}}][{\displaystyle \frac{3}{2}},2][2,{\displaystyle \frac{5}{2}}][{\displaystyle \frac{5}{2}},3]$
The midpoint of an interval is the average of the endpoints, that is:
$\Rightarrow m_1={\displaystyle \frac{1}{2}}(1+{\displaystyle \frac{3}{2}})={\displaystyle \frac{5}{4}}$
Similarly for others endpoints are:
$\begin{array}{rl}& \Rightarrow m_2={\displaystyle \frac{1}{2}}({\displaystyle \frac{3}{2}}+2)={\displaystyle \frac{7}{4}}\\ & \Rightarrow m_3={\displaystyle \frac{1}{2}}(2+{\displaystyle \frac{5}{2}})={\displaystyle \frac{9}{4}}\\ & \Rightarrow m_4={\displaystyle \frac{1}{2}}({\displaystyle \frac{5}{2}}+3)={\displaystyle \frac{11}{4}}\end{array}$
Now for any function$f$, we get the Riemann sum:
$\Rightarrow sum=\mathrm{\Delta }x(f\left(m_1\right)+f\left(m_2\right)+f\left(m_3\right)+f\left(m_4\right))$
Now putting values, we get
$\Rightarrow sum={\displaystyle \frac{1}{2}}(f\left({\displaystyle \frac{5}{4}}\right)+f\left({\displaystyle \frac{7}{4}}\right)+f\left({\displaystyle \frac{9}{4}}\right)+f\left({\displaystyle \frac{11}{4}}\right))$

Note: The point to be noted is that the midpoint Riemann sum is one for which we evaluate the function we are integrating at the midpoint of each interval, and use those values to determine the height of the rectangles. And if the curve is decreasing then the right sum are underestimates and the left sums are overestimates.