Find $\alpha, \beta, \gamma$ such that accuracy degree of $Q(f) = \alpha f(0) + \beta f(\frac{1}{2}) + \gamma f(1)$ is as high as possible

Question

Given the Quadrature Formula

$$Q(f) = \alpha f(0) + \beta f(\frac{1}{2}) + \gamma f(1)$$ for the approximation of $$I(f) = \int_{0}^{1} f(x)dx$$

We're asked to

1. Determine the $\alpha, \beta, \gamma$ such that the degree of accuracy of this rule is as high as possible.
2. Transform the Quadrature Rule in an arbitrary Interval [a,b].
3. Using above results, derive Simpson's Rule

Answer 1

1. We have three unknowns, thus we need at least three equations. Thus the degree of accuracy is at least three.

$$Q[x^0=1] = \alpha + \beta + \gamma \overset{!}{=} I(x^0=1) = 1$$

$$Q[x] = \frac{\beta}{2} + \gamma \overset{!}{=} I[x] = \frac{1}{2}$$

$$Q[x^2] = \frac{\beta}{4} + \gamma \overset{!}{=} I[x^2] = \frac{1}{3}$$

From the second and third equation, we get $$\beta = \frac{4}{6} \rightarrow \gamma = \frac{1}{6} \rightarrow \alpha = \frac{1}{6}$$

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2. If we let $$x = a + (b-a)\xi$$ for $\xi \in [0, 1]$ we get $\frac{dx}{d \xi} = (b-a)$ and so $$ \int_{0}^{1}f(a + (b-a)\xi)(b-a) d\xi$$

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3. From 1., we have $$\frac{1}{6}(f(0) + 4f(\frac{1}{2}) + f(1))$$ Using our result from 2., we get $$\frac{b-a}{6}(f(a) + 4f(\frac{a+b}{2}) + f(b))$$ which is Simpson's Rule.