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Question: How do you calculate the left and right Riemann Sum for the given function over the interval \([1,5]...

How do you calculate the left and right Riemann Sum for the given function over the interval [1,5][1,5],using n=4n = 4 for f(x)=3xf(x) = 3x?

Explanation

Solution

This sum is a bit difficult if the student is not aware about the concept. Before starting the sum the student should know about the concept of Riemann Sum. A Riemann sum is a certain kind of approximation of an integral by a finite sum. It has two types of sum, Left Riemann Sum(LRS) and Right Riemann sum(RRS). A Right Riemann Sum uses rectangles whose top-right vertices are on the curve.A left Riemann Sum(LRS) uses rectangles whose top-left vertices are on the curve.

Complete Step by Step Solution:
In this sum we will calculate LRS & RRS separately.
It is given f(x)=3xf(x) = 3x
Also , we have n=4n = 4, we want to calculate over the interval [1,5][1,5]with 44strips, thus from the given data
x=514=1\vartriangle x = \dfrac{{5 - 1}}{4} = 1
In this particular sum we have a fixed Interval, but a Riemann sum can have a varying size partition width.
The values of the function are as follows:

xx1122334455
f(x)f(x)33669912121515

Left Riemann Sum
LRS = r=14f(x)×x\sum\limits_{r = 1}^4 {f(x) \times \vartriangle x}

x×f(1)+f(2)+f(3)+f(4)\Rightarrow \vartriangle x \times \\{ f(1) + f(2) + f(3) + f(4)\\}
1×3+6+9+12\Rightarrow 1 \times \\{ 3 + 6 + 9 + 12\\}
30\Rightarrow 30

Right Riemann Sum
RRS =r=25f(x)×x\sum\limits_{r = 2}^5 {f(x) \times \vartriangle x}

x×f(2)+f(3)+f(4)+f(5)\Rightarrow \vartriangle x \times \\{ f(2) + f(3) + f(4) + f(5)\\}
1×6+9+12+15\Rightarrow 1 \times \\{ 6 + 9 + 12 + 15\\}
42\Rightarrow 42
We will have to find out the actual value for comparing the accuracy
Area = 153x.dx\int\limits_1^5 {3x.dx}
=3×[x22]15= 3 \times {[\dfrac{{{x^2}}}{2}]_1}^5
=32×(5212)= \dfrac{3}{2} \times ({5^2} - {1^2})
=36= 36

Thus the Answer for Riemann Sums are
LRS =3030
RRS= 4242

Note:
The students should know that the Riemann sum is used for approximation. It is used for approximating the area of functions or lines on a graph, but also used for approximations of lengths and other dimensions. The Riemann sum gives an approximate answer and the student should not rely completely on it. If a sum on Riemann sum is asked, the student should always verify the answer by performing integration. Average of LRS & RRS should be close to the value obtained by integration.