Question

How do you calculate the right hand and left hand Riemann sum using \[4\] sub intervals of \[f\left( x \right) = 3x\] on the interval \[\left[ {1,5} \right]\]?

Answer 1

Hint: Here we will first find all the subintervals of the given function on the given interval. For this we divide the length of \[\left[ {1,5} \right]\] by \[4\] and from this value we will get the partition interval of the given interval. Putting the elements of the partition interval into the given function, we will get the \[4\] subintervals of the function. For the left-hand Riemann sum, we will multiply the first element of each of the \[4\] subintervals with their lengths and find their sum. And for the right-hand Riemann sum, we will multiply the second element of each of the intervals with their lengths and find the sum.

Complete step-by-step solution:
According to the question, the given interval is \[\left[ {1,5} \right]\]. From this we have
\[\begin{array}{l}a = 1\\b = 5\end{array}\]
Now if we divide this intervals into \[4\] equal subintervals, then the length of each of the sub intervals is given by
\[h = \dfrac{{b - a}}{n}\]
\[ \Rightarrow h = \dfrac{{5 - 1}}{4}\]
Subtracting the terms, we get
\[ \Rightarrow h = 1\]
So the partition interval of the given interval becomes
\[P = \left[ {1,1 + h,1 + 2h,1 + 3h,1 + 4h} \right]\]
Putting \[h = 1\] we get
\[P = \left[ {1,2,3,4,5} \right]\]
Putting each element in the given function, we get the sub intervals of \[f\left( x \right)\] as \[\left[ {f\left( 1 \right),f\left( 2 \right)} \right];\left[ {f\left( 2 \right),f\left( 3 \right)} \right];\left[ {f\left( 3 \right),f\left( 4 \right)} \right];\left[ {f\left( 4 \right),f\left( 5 \right)} \right]\]
According to the question, \[f\left( x \right) = 3x\]. So the sub-intervals becomes
\[\begin{array}{l}\left[ {3\left( 1 \right),3\left( 2 \right)} \right];\left[ {3\left( 2 \right),3\left( 3 \right)} \right];\left[ {3\left( 3 \right),3\left( 4 \right)} \right];\left[ {3\left( 4 \right),3\left( 5 \right)} \right]\\ \Rightarrow \left[ {3,6} \right];\left[ {6,9} \right];\left[ {9,12} \right];\left[ {12,15} \right]\end{array}\]
From these sub-intervals, we get the greatest lower bounds (glb) as
\[{m_1} = 3,{m_2} = 6,{m_3} = 9,{m_4} = 12\]
Also, the least upper bounds (lub) are
\[{M_1} = 6,{M_2} = 9,{M_3} = 12,{M_4} = 15\]
Now, we know that the left hand Riemann sum is given by
\[L = {m_1}{h_1} + {m_2}{h_2} + {m_3}{h_3} + {m_4}{h_4}\]
Since all the intervals are of equal length of \[h = 1\], so we have
\[\begin{array}{l} \Rightarrow L = 1\left( {{m_1} + {m_2} + {m_3} + {m_4}} \right)\\ \Rightarrow L = {m_1} + {m_2} + {m_3} + {m_4}\end{array}\]
Putting the values of greatest lower bounds, we get
\[\begin{array}{l} \Rightarrow L = 3 + 6 + 9 + 12\\ \Rightarrow L = 30\end{array}\]
Similarly, we get the right hand Riemann sum as
\[\begin{array}{l}R = {M_1} + {M_2} + {M_3} + {M_4}\\ \Rightarrow R = 6 + 9 + 12 + 15\\ \Rightarrow R = 42\end{array}\]

Hence, the left hand and the right-hand Riemann sums are \[30\] and \[42\] respectively.

Note:
A subinterval is an interval that is present inside another interval. It is not necessary to divide the given interval into equal subintervals. The given interval can be divided in any manner. We divided it into equal subintervals for easing out the calculations. Riemann sum is used to find the area under the curve on a particular graph. We do this by finding the endpoints of the right and left of the subintervals.