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I am stuck on this question:

Consider the following method for approximating $\int_{a}^{b}f(x)dx$. Partition the interval $[a,b]$ into $n$ equal subintervals. On each subinterval approximate the function $f$ by a quadratic polynomial that agrees with $f$ at both endpoints and at the midpoint of the subinterval.

(a) Explain why the integral of $f$ on the subinterval $[x_{i-1},x_{i}]$ is approximately equal to the expression:

$$\frac{x_{i}-x_{i-1}}{3}\left [ \frac{f(x_{i-1})}{2}+2f\left(\frac{x_{i-1}+x_{i}}{2}\right)+\frac{f(x_{i})}{2} \right ].$$

(b) Show that if we add up these approximations, we get Simpson's rule:

$$\int_{a}^{b}f(x)dx \approx \frac{2}{3}\text{MID}(n)+\frac{1}{3}\text{TRAP}(n)$$

Where MID and TRAP represent the mid-point and trapezoidal methods of integral approximation.

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up vote 1 down vote accepted

Write $h = (x_{i} - x_{i-1})/2$, consider a quadratic polynomial $p(x) = ax^2 + bx + c$. Since we only care about the integral, we may translate the problem to the interval $[-h,h]$. I would suggest that you first calculate the integral $$\int_{-h}^{h} p(x),$$ as this is going to simplify things later. Moving on to the conditions on $p(x)$, we see that \begin{align*} f(x_{i-1}) &= p(-h) = a(-h)^2 + b(-h) + c = ah^2 - bh + c \\ f\left( \frac{x_{i-1} + x_i}{2} \right) &= p(0) = a \cdot 0 + b \cdot 0 + c = c \\ f(x_i) &= p(h) = ah^2 + bh + c \end{align*} Now solve for whatever constants you need, and insert them into the expression you found for the integral of $p(x)$. This will give you the correct expression for the approximation to the integral of $f$. Problem (b) should be easy.

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Not too hard. I presume you know how to determine an interpolating polynomial?

The idea is to find the parabola $ax^2+bx+c$ passing through $x_{i-1}$, $\dfrac{x_{i-1}+x_i}{2}$, and $x_i$, and then integrate that over $[x_{i-1},x_i]$. You could, for instance, set up the (Vandermonde) system of linear equations


since you know the three points where the parabola to be integrated passes through. You should be getting something like

$\displaystyle p(x)=\frac{2\left(f\left(x_{i-1}\right)-2f\left(\frac{x_{i-1}+x_i}{2}\right)+f\left(x_i\right)\right)}{\left(x_i-x_{i-1}\right)^2}\left(x-\frac{x_{i-1}+x_i}{2}\right)^2+\frac{f\left(x_i\right)-f\left(x_{i-1}\right)}{x_i-x_{i-1}}\left(x-\frac{x_{i-1}+x_i}{2}\right)+f\left(\frac{x_{i-1}+x_i}{2}\right)$

and evaluating $\int_{x_{i-1}}^{x_i}p(u)\;\mathrm du$ in the usual way should yield Simpson's rule.

To show that a linear combination of the trapezoidal and midpoint rules give Simpson's rule, consider the trapezoidal and midpoint approximations

$$\begin{align*} \text{trap}&=\int_{x_{i-1}}^{x_i}\left(f\left(x_{i-1}\right)+\frac{f\left(x_i\right)-f\left(x_{i-1}\right)}{x_i-x_{i-1}}\left(u-x_{i-1}\right)\right)\;\mathrm du\\ &=\frac{x_i-x_{i-1}}{2}\left(f(x_{i-1})+f(x_i)\right)\\ \text{mid}&=\int_{x_{i-1}}^{x_i}f\left(\frac{x_{i-1}+x_i}{2}\right)\;\mathrm du\\ &=(x_i-x_{i-1})f\left(\frac{x_{i-1}+x_i}{2}\right) \end{align*}$$

and figure out appropriate values of $p$ and $q$ such that


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I'm afraid I am not familiar with interpolating polynomials. This is for a first year calculus course that assumes no knowledge of linear algebra. We have not been exposed to anything like the Vandermonde matrix. – J.Borges Feb 9 '12 at 11:15
But you do know how to solve $n$ linear equations in $n$ unknowns, don't you? That skill very much applies here. – J. M. Feb 9 '12 at 11:18
I think I left out something important. The question which preceded this one involved the following: Suppose that $a<b$. The purpose of this problem is to show that if $f$ is a quadratic polynomial, then we have $$\int_{a}^{b}f(x)dx=\frac{b-a}{3}\left [ \frac{f(a)}{2}+2f(\frac{a+b}{2})+\frac{f(b)}{2} \right ].$$ I was then asked to show that the equation holds for $f(x)=1$, $f(x)=x$, $f(x)=x^{2}$, and any general quadratic polynomial $f(x)=Ax^2 + Bx + C$ – J.Borges Feb 9 '12 at 11:23
That's odd, since that formula is precisely Simpson's rule! If you've proven that, it's a simple matter to let $a=x_{i-1}$ and $b=x_i$... – J. M. Feb 9 '12 at 11:36

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