Q. 50 - Calculus

Question

Use Definition 4.7 to prove that for any function $f$ and interval [a, b], the upper sum with $n$ rectangles is greater than or equal to the lower sum with $n$ rectangles.

Step-by-Step Solution

Answer

It is proved that upper sum with$n$ rectangles is greater than or equal to the lower sum with $n$ rectangles.

Step 1. Given Information

Upper and Lower Sums

Suppose f is a function that is continuous on the interval [a, b]. Given a positive integer n,

let $∆x=\frac{b-a}{n}$and $x_k=a+k∆x$.Then

(a) The $n$-rectangle upper sum for $f$on $\left[a,b\right]$ is where each$M_k$ is chosen so that is the maximum value of  $f$on $\left[x_{k-1},x_k\right]$.

(b) The n-rectangle lower sum for on  is where each $m_k$is chosen so that  is the minimum value of $f$on $\left[x_{k-1},x_k\right]$

Step 2. Proof of the Statement

Note that given any interval $\left[a,b\right]$and number $n$ of rectangles, we can write $∆x$ and $x_k$

in terms of a, b, and n. In practice, we will always need to use the explicit expressions$∆x=\frac{b-a}{n}$ and $x_k=a+k∆x$ (as well as using the definition of the function $f$ ) when

evaluating a Riemann sum. For example, the right sum expressed earlier is equal to

$\underset{k=1}{\overset{n}{\sum f\left(a+k\left(\frac{b-a}{n}\right)\right)\left(\frac{b-a}{n}\right)}}$

 The upper sum is always greater than or equal to the actual signed area.