Question

Show that if \(f\) is a polynomial of degree 3 or lower, then Simpson's Rule gives the exact value of\(\int_a^b f (x)dx\).

Short answer

A polynomial of degree 3 or lower gives the same value as with the Simpson’s rule.

Step-by-step solution

Definition

Simpson's Rule\(\int_a^b f (x)dx \approx {S_n} = \frac{{\Delta x}}{3}\left( {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 4f\left( {{x_3}} \right) + L + 2f\left( {{x_{n - 2}}} \right) + 4f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right)\)

Where\(x\)is even and\(\Delta x = \frac{{b - a}}{n}\).

Assumptions

Let the\(x\)values be \({x_{i - 1}},{x_i}\)and\({x_{i + 1}}\). The two subinterval are\({x_{i - h}},{x_i}\)and\({x_{i + h}}\)where\(\Delta x = h = \frac{{b - a}}{h}\).

Now, consider a polynomial of degree 3 \(f(x) = A{x^3} + B{x^2} + Cx + D\)and integrate it over a subinterval (say) \(\left( {{x_{i - h}},{x_{i + h}}} \right)\).

Integrate

Therefore\(\int_{{x_{j - 6}}}^{{x_{i + h}}} {(A} {x^3} + B{x^2} + Cx + D)dx\)

Let\(u = x - {x_i}\)\( \Rightarrow x = u + {x_i}\) and\(du = dx\).

When \(x = {x_i} + h\)then \(u = {x_i} + h - {x_i} = h\)

When\(x = {x_i} - h\)then \(u = {x_i} - h - {x_i} = - h\)

Therefore;

\(\begin{aligned}{c}\int_{{x_{i - h}}}^{{x_{i + h}}} A {x^3} + B{x^2} + Cx + Ddx = \int_{ - h}^h A {\left( {u + {x_i}} \right)^3} + B{\left( {u + {x_i}} \right)^2} + C\left( {u + {x_i}} \right) + Ddu\\ = \int_{ - h}^h a {u^3} + b{u^2} + du + ddu\end{aligned}\)

Where\(a = A,b = 3A{x_i} + B,c = 3Ax_i^2 + 2B{x_i} + C,d = Ax_i^3 + Bx_i^2 + C{x_i} + D\)

Simplify

Now let\(g(x) = a{x^3} + b{x^2} + cx + d\), this will be used later in the proof.

\(\int_{ - h}^h {(a} {u^3} + b{u^2} + du + d)du\)can be split into odd and even integrals:

\(\int_{ - h}^h {(a} {u^3} + b{u^2} + du + d)du = \int_{ - h}^h {(a} {u^3} + cu)du + \int_{ - h}^h {(b} {u^2} + d)du\)

The first integral is odd over\(( - h,h)\)so it evaluates to zero, whereas the second is even, Using the properties\(\int_{ - a}^a f (x)dx = 2\int_0^a f (x)dx\)if\(f\)is an even function & \(\int\limits_{ - a}^a f (x)dx = 0\), if\(f\) is an odd function we have:

\(\int_{ - h}^h b {u^2} + ddu = 2\int_{ - h}^h {(b} {u^2} + d)du\)

Integrating

Using the integral formula\(\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C\)we have:

\(\begin{aligned}{c}2\int_{ - h}^h b {u^2} + ddu = 2\left( {\frac{{b{u^3}}}{3} + du} \right)_0^h\\ = 2\left( {\frac{{b{h^3}}}{3} + dh} \right)\\ = \frac{h}{3}\left( {2b{h^2} + bh} \right)\\ = \frac{h}{3}(g( - h) + 4g(0) + g(h))\end{aligned}\)

This is exactly the same as\(\frac{h}{3}\left( {f\left( {{x_{i - 1}}} \right) + 4f\left( {{x_i}} \right) + f\left( {{x_{i + 1}}} \right)} \right)\)which is Simpson's rule with two subintervals.

Therefore, if a polynomial is of degree 3 or lower, then Simpson's rule gives the exact value of\(\int_a^b f (x)dx\).