Chapter 3.4, Problem 20AYU

Calculus: Simpson's Rule The figure shows the graph of $y=a{x}^2+bx+c$. Suppose that the points $\left(-h,y_0\right)\text{, }\left(0,y_1\right),$ and $(h,y_2)$ are on the graph. It can be shown that the area enclosed by the parabola, the $x\text{-axis}$, and the lines $x=-h$ and $x=h$ is

$\text{Area }=\frac{h}{3}(\text{2}a{h}^2+\text{ 6}c)$

Show that this area may also be given by

$\text{Area = }\frac{h}{3}(y0+4y1+y2)$

To determine

To prove: That the area enclosed by the parabola, the $x\text{-axis}$, and the lines $x=h$ and $x=-h$ is $Area=\frac{h}{3}\left(y_0+4y_1+y_2\right)$.

Answer to Problem 20AYU

Here, in the given question, we can see that the graph of the function $y=a{x}^2+bx+c$ is subdivided into the 2 equally spaced segments of width $h$.

Therefore, using Simpson’s rule, the area enclosed by the given function, the $x\text{-axis}$ and the lines $x=h$ and $x=-h$ is

$A=\frac{h}{3}\left(y_0+4y_1+y_2\right)$

Explanation of Solution

Given:

The graph of the function $y=a{x}^2+bx+c$ is given as

Formula used:

Here, in order to prove the given equation, we have to use the Simpson’s rule.

Consider a function $y=f(x)$ which is subdivided into $n$ equal segments of width $h$ Then using Simpson’s rule, we can approximate the area of the function $f$ as

$A=\frac{h}{3}\left(y_0+4(y_1+y_3+y_5+\mathrm{...})+2(y_2+y_4+y_6+\mathrm{....})+y_n\right)$

Proof:

Here, in the given question, we can see that the graph of the function $y=a{x}^2+bx+c$ is subdivided into the 2 equally spaced segments of width $h$.

Therefore, using Simpson’s rule, the area enclosed by the given function, the $x\text{-axis}$ and the lines $x=h$ and $x=-h$ is

$A=\frac{h}{3}\left(y_0+4y_1+y_2\right)$