Question

Use Definitions 4.6 and 4.8 to prove that for any function f and interval $[a,b]$, the trapezoid sum with n trapezoids is always the average of the left sum and the right sum with n rectangles.

Short answer

It is provedthat for any function f and interval $[a,b]$, the trapezoid sum with n trapezoids is always the average of the left sum and the right sum with n rectangles.

Step-by-step solution

Step 1. Given Information 

We are given thatfor any function f and interval $[a,b]$, the trapezoid sum with n trapezoids is always the average of the left sum and the right sum with n rectangles.

Step 2. Proving the statement 

The left-sum defined for n rectangles on $[a,b]$is $\sum _{k=1}^{n}f\left(x_{k-1}\right)\Delta x$.

Where, $\Delta x=\frac{b-a}{n},x_k=a+k\Delta x$.

The right sum defined for n rectangles on $[a,b]$is $\sum _{k=1}^{n}f\left(x_k\right)\Delta x$.

The average of left sum and right sum is,

$\frac{{\sum }_{k=1}^{n}f\left(x_{k-1}\right)\Delta x+{\sum }_{k=1}^{n}f\left(x_k\right)\Delta x}{2}=\sum _{k=1}^n\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}\Delta x$

The trapezoid sum for n rectangles on $[a,b]$is $\sum _{k=1}^n\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}\Delta x$.

Since f is increasing,

$x_{k-1}\le x_k$

Hence Proved.