Question

Use a labelled graph to demonstrate that the average radius $r_{k}ofthe{k}^{\text{th}}$frustum is given by if n frustums are used to approximate the area of the surface obtained by rotating the graph of a function $y=f\left(x\right)$around the $x$-axis on $[a,b]$

$r_k=\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}$

Short answer

The function $y=f\left(x\right)$on $[a,b]$by using the graph average radius of$kth$frustum is $r_k=\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}.$

Step-by-step solution

Given information

The function $y=f\left(x\right)$on $[a,b]$the average radiuslocalid="1661324495760" $r_k\mathrm{of}\mathrm{the}{k}^{\text{th}}$frustum is given by $r_k=\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}$.

Calculation

The graph below displays the solid's $kth$frustum, which was created by rotating the function's graph around the $x$-axis.

The radii $A,C$, and $B$in the aforementioned graph correspond to the frustum's radii at$x_{k-1}and{x}_k$, respectively. Therefore, the average of the lengths $A,C,$and $BD$equals the average radius of the $kth$frustum.

As a result, the Average Radius of the $kth$Frustum is given by,

$r_k=\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}.$