# Riemann left,right and midpoint sums

Find left, right and midpoint Riemann sums for

$\displaystyle\int_{1}^{2} \frac{1}{x} dx$

$P = 2, \frac{5}{2},3,4$

Using: $f(x_i)\Delta x$

Please check my work:

$(\frac{1}{2})(\frac{1}{2})+(\frac{2}{5})(\frac{1}{2})+(\frac{1}{3})(1)$ for the left Riemann sum.

$(\frac{2}{5})(\frac{1}{2})+(\frac{1}{3})(\frac{1}{2})+(\frac{1}{4})(1)$ for the right Riemann sum.

And using the midpoint formula: $\frac{b+a}{n}$

$(\frac{4}{9})(\frac{1}{2})+(\frac{4}{11})(\frac{1}{2})+(\frac{2}{7})(1)$ for the midpoint Riemann sum.

Is this correct? One thing I don't quite understand and dismissed was the original integral from [1,2] was that needed? I thought that if an integral is given in a problem like this that it will follow accordingly to the partitions but this does not so I am unsure if I have ruined the solutions because of not understanding some key piece of information.

Thank you

• I'm sorry, but I can make no sense of either the problem as stated or your work. To do a Riemann sum, you need a partition of the interval $[1,2]$ to be summed. One might suppose $P$ should give this partition, but it doesn't. None of the numbers listed is even in $[1,2]$. What is $P$ supposed to be? A Riemann sum is of the form $$\sum f(\xi_i)(x_{i+1} - x_i)$$ In particular, one factor in each term is the width of the partition interval, which should add up in total to the width of the integration interval $(2 -1) = 1$ This is not the case for your sums. Commented Nov 27, 2015 at 20:58

To further illustrate the points in my comment. One partition of $[1,2]$ has partition points $\frac 5 4,\frac 3 2, \frac 7 4$. Here is how to answer the question for that partition:

The partition divides $[1,2]$ into four subintervals: $\left[1, \frac 5 4\right], \left[\frac 5 4, \frac 3 2\right], \left[\frac 3 2,\frac 7 4\right], \left[\frac 7 4,2\right]$

• $\left[1, \frac 5 4\right]$: $\Delta x = \frac 5 4 - 1 = \frac 1 4$. Left pt is $1$, midpt is $\frac 9 8$, right pt is $\frac 5 4$.
• $\left[\frac 5 4,\frac 3 2\right]$: $\Delta x = \frac 3 2 - \frac 5 4 = \frac 1 4$. Left pt is $\frac 5 4$, midpt is $\frac{11} 8$, right pt is $\frac 3 2$.
• $\left[\frac 3 2, \frac 7 4\right]$: $\Delta x = \frac 7 4 - \frac 3 2 = \frac 1 4$. Left pt is $\frac 3 2$, midpt is $\frac{13} 8$, right pt is $\frac 7 4$.
• $\left[\frac 7 4, 2\right]$: $\Delta x = 2 - \frac 7 4 = \frac 1 4$. Left pt is $\frac 7 4$, midpt is $\frac{15} 8$, right pt is $2$.

(Note that the sum of the widths of the 4 partition intervals $= \left(\frac 1 4\right) + \left(\frac 1 4\right) +\left(\frac 1 4\right) +\left(\frac 1 4\right) = 1$, the width of the complete interval $[1,2]$.)

So the Left Riemann sum is: $$\left(\frac 1 {1}\right)\left(\frac 1 4\right) + \left(\frac 1 {\frac 5 4}\right)\left(\frac 1 4\right) + \left(\frac 1 {\frac 3 2}\right)\left(\frac 1 4\right) + \left(\frac 1 {\frac 7 4}\right)\left(\frac 1 4\right)\\=\left(1 + \frac 4 5 + \frac 2 3 + \frac 4 7\right)\left(\frac 1 4\right) = \frac{319}{420}\approx 0.7595$$ The Mid Riemann sum is: $$\left(\frac 1 {\frac 9 8}\right)\left(\frac 1 4\right) + \left(\frac 1 {\frac{11} 8}\right)\left(\frac 1 4\right) + \left(\frac 1 {\frac{13} 8}\right)\left(\frac 1 4\right) + \left(\frac 1 {\frac{15} 8}\right)\left(\frac 1 4\right)\\=\left(\frac 8 9 + \frac 8 {11} + \frac 8 {13} + \frac 8 {15}\right)\left(\frac 1 4\right) = \frac{4448}{6435}\approx 0.6912$$ The Right Riemann sum is:

$$\left(\frac 1 {\frac 5 4}\right)\left(\frac 1 4\right) + \left(\frac 1 {\frac{3} 2}\right)\left(\frac 1 4\right) + \left(\frac 1 {\frac{7} 4}\right)\left(\frac 1 4\right) + \left(\frac 1 {2}\right)\left(\frac 1 4\right)\\=\left(\frac 4 5 + \frac 2 3 + \frac 4 7 +\frac 1 2\right)\left(\frac 1 4\right) = \frac{533}{840}\approx 0.6345$$

For comparison, the actual value of the integral is $\frac 3 4 = 0.75$.

Without knowing what the partition you are supposed to use is, I cannot show you exactly how to answer. But it should be something similar to this.