# Riemann sums, finding the lower sum?

just doing some review and I am a bit confused on how I pick the upper vs lower sum of a Riemann sum. I get for the upper sum I choose the maximum value of $f$ on $[x_{k-1},x_k]$ and the lower sum is the minimum of $[x_{k-1},x_k]$.

Could someone give me a worked out example of an upper or lower sum using Riemann's definition.

Do I plug in values of the interval $[a,b]$ and see where it is increasing and decreasing? But which formula do I plug this in, the left sum or right sum?

• What do you mean by 'left sum' and 'right sum' in this context? – DominikS Jul 27 '17 at 20:24
• I think what you are referring to in the last paragraph is that when a function is monotone, the maximum value occurs at an endpoint of a subinterval (and also the minimum value occurs at the other such endpoint). – hardmath Jul 27 '17 at 20:38

Suppose that $f\colon[-1,1]\longrightarrow\mathbb R$ is defined by $f(x)=x^2$. Now, take $x_0=-1$, $x_1=-\frac13$, $x_2=\frac13$, and $x_3=1$. Then:

1. in $[x_0,x_1]$, the minimum of $f$ is $\frac19$ and the maximum is $1$;
2. in $[x_1,x_2]$, the minimum of $f$ is $0$ and the maximum is $\frac19$;
3. in $[x_2,x_3]$, the minimum of $f$ is $\frac19$ and the maximum is $1$.

Therefore, if $P=\{x_0,x_1,x_2\}$, then the lower sum of $f$ with respect to $P$ is$$\frac23\times\frac19+\frac23\times0+\frac23\times\frac{1}{9}=\frac4{27}$$and the upper sum is$$\frac23\times1+\frac23\times\frac19+\frac23\times\frac19=\frac{38}{27}.$$

Finding "Upper sum, Lower sum" you don't "plug into" **either the "Left sum" or "Right sum". Those are completely different things.

In all four cases, you divide the total interval into "n" sub-intervals.

For the "upper sum" you take f(x) as the largest value of f in the sub-interval and multiply by the length of the sub-interval, then sum.

For the "lower sum" you take f(x) as the smallest value of f in the sub-interval and multiply by the length of the sub-interval, then sum.

For the "left sum" you take f(x) for x the end of the sub-interval and multiply by length of the sub-interval, then sum.

For the "right sum" you take f(x) for x at the right end of the interval and multiply by the length of the sub-interval, then sum.

For example, take the interval to be from x= 0 to x= 1 and divide into 4 equal intervals, f(x)= x^2.

The left end of the four intervals are x= 0, x= 1/4, x= 1/2, x= 3/4. The squares are 0, 1/16, 1/4, 9/16. Multiply each by 1/4, the length of each sub-interval and we have 0, 1/64, 1/16, 9/64. Adding, 0+ 1/64+ 4/64+ 9/64= 14/64.

Because this is an increasing function, y= x^2 is always least at the left so that left sum is the same as the "lower sum".

The right end of the four intervals are x= 1/4, x= 1/2, x= 3/4, and x= 1. The squares are 1/16, 1/4, 9/16, and 1. Multiply each by 1/4, the length of each sub-interval and we have 1/64, 1/16, 9/64, 1/4. Adding, 1/16+ 1/4+ 9/64+ 1/4= 30/64.

Because this is an increasing function, y= x^2 is always highest at the right so that the right sum is th The right end o

An example of a lower sum with intervals of length $1$ for the monotonically increasing function $f(x) = x^2$ (on $x\in[0,k]$):

$$\int_0^{k}x^2dx \leq\sum_{0}^{k-1} n^2$$

It is easier to verify that this is a lower sum because for an interval where the function is monotonic the smallest function value will allways be at one of the end points. In our case with a monotonically increasing function, the lowest end: $$f(x_k)\leq f(\xi) \,\,\forall\,\, \xi \in[x_{k},x_{k+1}]$$

For more advanced problems we may need to search for an estimate of the maximum or minimum function value. That can sometimes be a rather difficult problem!