Let us see about calculus Riemann sums. In mathematics the Riemann calculus sums are way for roughing the whole area underneath a curve on a graph. This is also specified as integral. It is may be used to definite the integration operation.

The Riemann technique also used by geometry. It is one of the kinds of differential geometry. It is enormously extensive broad and theoretical generalization of the differential geometry of exterior. This geometry was originated by Bernhard Riemann in nineteenth century. These geometry agreements with geometries and it are varying from point to point.

Riemann Sum Formula

Let $f$ be any function defined on the interval [a,b] and let $N$ be a positive integer. The uniform partition of order $N$ of the interval [a, b] is the set of equally spaced points

$x_{i}$ = $a+i.$$\frac{b-a}{N} (0\leq i\leq N) that break the interval [a, b] into N equal-length sub intervals I_{1} = [x_{0},x_{1}], I_{2} = [x_{1},x_{2}],....I_{N} = [x_{N-1}, x_{N}] Here let \Delta x denote the common length \frac{b-a}{N} of these intervals. A choice of points associated with the uniform partition of the order N is a sequence S_{N} = (s_{1}, s_{2}...s_{n}) of points with s_{i} in I_{i} for each i=1, 2,....N. The expression R(f,S_{N}) = \sum _{i=1}^{N}f(s_{i}).\Delta x ....(1) is called a Riemann sum of f. The notation R(f, S_{N}) indicates that a Riemann sum depends on the function f and the choice of points S_{N}. Finding areas within shapes gets complicated fast outside of the standard geometric shapes like circles, squares triangles and trapezoids. Riemann added up the area under the complicated curves by using standard geometric shapes. Different methods of arranging rectangles under the curve results in more accurate, under-estimate or over-estimate of area. The smaller and more numerous the rectangle, the more accurate the estimates. A Riemann sum can be determined in three different ways as given below Left Riemann Sum The Riemann sum of a function f is defined in equation (1), From the definition we see that the Riemann sum depends on the function and the choice of points. The choice of pints in the given interval is arbitrary. If s_{i} = x_{i-1} for all i then R(f, S_{N}) is called a Left Riemann Sum. It can be written as R(f, S_{N}) = \sum _{i=1}^{N}f(x_{i-1}).\Delta x Lower Riemann Sum: Suppose f is continuous. Let l_{i} \epsilon I_{i} be the point at which f attains its minimum value on I_{i} The extreme value theorem says that the points l_{i} exist. We denotes L_{N}= (l_{1}, l_{2},......l(N)). The Riemann Sum R(f, L_{N}) = \sum _{i=1}^{N}f(l_{i}.\Delta x) is called as Lower Riemann Sum of order N. they are the least Riemann sum of order N. Right Riemann Sum If s_{i} = x_{i} for all i in equation (1) then R(f, S_{N}) is called a Right Riemann Sum. It can be written as R(f, S_{N}) = \sum _{i=1}^{N}f(x_{i}).\Delta x Upper Riemann Sum: Let u_{i} be the point in I_{i} at which f achieves its maximum value on I_{i}. The extreme value theorem says that the points u_{i} exist. We denotes U_{N}= (u_{1}, u_{2},......u(N)) The Riemann Sum R(f, U_{N}) = \sum _{i=1}^{N}f(u_{i}.\Delta x) is called as Upper Riemann Sum of order N. They are the greatest Riemann sum of order N. Midpoint Riemann Sum If s_{i} = \frac{x_{i-1}+x_{i}}{2} for all i in equation (1) then R(f, S_{N}) is called a Right Riemann Sum. It can be written as R(f, S_{N}) = \sum _{i=1}^{N}f($$\frac{x_{i-1}+x_{i}}{2}$$).\Delta x The midpoint Riemann sum is usually used to estimate the area under a curve within given boundaries because the midpoint Riemann sum gives a faster convergence to the limit than the upper and lower Riemann sums. Trapezoidal Riemann Sum The average of the left Riemann sum and right Riemann sum is the trapezoidal sum. R(f, S_{N}) = \sum _{i=1}^{N}$$\frac{f(x_{i-1})+f(x_{i})}{2}$.$\Delta x$
The trapezoidal methods uses trapezoid shapes rather than rectangle to compute area under the curve. The trapezoid shapes are connected point to point on the curve and can do a slightly better job of finding the area.

Riemann Sum Examples

The following problems are examples of Riemann sum.

Solved Examples

Question 1: If $f(x)$ = $x^{3}-2x$ find the Riemann sum for the function taking the sample point to be right end points and $a=0,b=3$ and $n=6$.
Solution:

Given $a=0,b=3$ and $n=6$

The interval width is $\Delta x$ = $\frac{b-a}{n}$ = $\frac{3-0}{6}$ = $\frac{1}{2}$

The right end points are $x_{1}$ = $0.5$, $x_{2}$ = $1$, $x_{3}$ = $1.5$, $x_{4}$ = $2$, $x_{5}$ = $2.5$ and $x_{6}$ = $3$.
Riemann sum is $R_{6}$ = $\sum _{i = 1}^{6}f(x_{i})\Delta x$

= $f(0.5)\Delta x + f(1)\Delta x + f(1.5)\Delta x+ f(2)\Delta x+ f(2.5)\Delta x+ f(3)\Delta x$

= $\frac{1}{2}$$(-0.875-1+0.375+4+10.625+21) = 34.125 Question 2: Estimate the area bounded by the graph of f(x) = x^{2}+ x + 2, in the x-axis and the vertical lines x = 0 and x = 2. 1) Using left Riemann sum with four intervals 2) Using right Riemann sum with four intervals 3) Using midpoint Riemann sum with four intervals. Solution: 1) Given a = 0,b = 2 and n=4 The interval width is \Delta x = \frac{b-a}{n} = \frac{2-0}{4} = \frac{1}{2} The left endpoints of each sub-intervals are 0, 0.5, 1, 1.5 so the left Riemann sum is R(f,S_{N}) = \sum _{i=1}^{4}f(s_{i}).\Delta x = (f(0)+f(0.5)+f(1)+f(1.5))$$\frac{1}{2}$

= $(2+2.75+4+5.75)$$\frac{1}{2} = 7.25 2) The Right endpoints of each sub-intervals are 0.5, 1, 1.5, 2 so the Right Riemann sum is R(f,S_{N}) = \sum _{i=1}^{4}f(s_{i}).\Delta x = f(0.5)+f(1)+f(1.5)+(f(2))$$\frac{1}{2}$

= $(2.75+4+5.75+8)$$\frac{1}{2} = 10.25 3)The midpoints of each sub-intervals are 0.25, 0.75, 1.25, 1.75 so the midpoint Riemann sum is R(f,S_{N}) = \sum _{i=1}^{4}f(s_{i}).\Delta x = f(0.25)+f(0.75)+f(1.25)+(f(1.75))$$\frac{1}{2}$

= $(2.3125+3.3125+4.8125+6.8125)$$\frac{1}{2}$

= $8.625$