# Integration formula for cubic polynomial $\int_a^bq(x)dx=\frac{b-a}{2}(q(b)+q(a))-\frac{(b-a)^2}{12}(q'(b)-q'(a))$

Show that $$\forall a,b\in \mathbb{R}$$, with $$a, we have$$\int \limits _a^bq(x)\,dx=\frac{b-a}{2}(q(b)+q(a))-\frac{(b-a)^2}{12}(q'(b)-q'(a)),$$ where $$q\in \mathcal{P}_3$$ is a cubic polynomial.

I've tried doing integration by parts with $$1$$ and $$q(x)$$. But that just gives $$b-a$$ instead of $$\dfrac{b-a}{2}$$.

• huh I think I saw this.. Jul 5 at 21:44
• By linearity it suffices to check this for $q(x)=1,x,x^2,x^3$. Jul 5 at 21:52
• That answers my question, thanks. Jul 5 at 22:01
• Note that the RHS is the Trapezoid Rule approximation of the integral plus an error term.
– Dan
Jul 5 at 22:28

For the cubic polynomial $$q(x)$$, we have $$q'''(a)= q'''(b)= q’’’$$ and \begin{align} &q'(a)-\ q'(b)=q''(b)(a-b)+\frac12 q'''(a-b)^2\\ &q'(b)-\ q'(a)=q''(a)(b-a)+\frac12 q'''(b-a)^2\\ \end{align} which result in $$q''(b)+q''(a)=\frac{2(q'(b)-q'(a))}{b-a}$$. Integrate the following integral by parts twice \begin{align} &\int_a^b\left(x-\frac{a+b}{2}\right)q'(x)dx \\ =& \int_a^b \frac{q'(x)}2d[(x-\frac{a+b}{2})^2] = \frac{(b-a)^2}8(q'(b)-q'(a)) - \int_a^b \frac{q''(x)}6d[(x-\frac{a+b}{2})^3]\\ = &\ \frac{(b-a)^2}8(q'(b)-q'(a)) - \frac{(b-a)^3}{48}(q''(b)+q''(a))=\frac{(b-a)^2}{12}(q'(b)-q'(a))\tag1 \\ \end{align}

On the other hand, an alternative integration by parts yields the following

$$\int_a^b\left(x-\frac{a+b}{2}\right)q'(x)dx =\frac{b-a}2(q(b)+q(a))-\int_a^b q(x)dx\tag2$$ Combine (1) and (2) to obtain $$\int_a^b q(x)dx = \frac{b-a}{2}(q(b)+q(a)) - \frac{(b-a)^2}{12}(q'(b) - q'(a))$$

HINT:

Let $$\phi(t) = q (a + t(b-a))$$. Show that

$$\int_0^1 \phi(t) \, dt = \frac{1}{2}( \phi(0) + \phi(1)) - \frac{1}{12}(\phi'(1) - \phi'(0))$$

for $$\phi$$ ( $$q$$) polynomial of degree $$\le 3$$. This is the Euler-Maclaurin formula of order $$3$$.

$$\bf{Added:}$$ Proof of the above formula.

Consider $$\psi(t)$$ any function $$C^4$$ we have

$$\left(\psi^{(3)}(t) \phi(t) - \psi^{(2)}(t) \phi'(t) + \psi'(t) \phi^{(2)}(t) - \psi(t)\phi^{(3)}(t)\right)' = \\=\psi^{(4)}(t) \phi(t) - \psi(t) \phi^{(4)}(t)$$

Therefore

$$\int_0^1 \psi^{(4)}(t) \phi(t)\, dt = \int_0^1 \psi(t) \phi^{(4)}(t)\, dt + \\ +\left(\psi^{(3)}(t) \phi(t) - \psi^{(2)}(t) \phi'(t) + \psi'(t) \phi^{(2)}(t) - \psi(t)\phi^{(3)}(t)\right)\mid_0^1$$

Take now in the formula above $$\psi(t) = \frac{1}{4!} (B_4(t)- B_4(0)) = \frac{1}{24}( t^4 - 2 t^3 + t^2)$$ where $$B_n(t)$$ is the Bernoulli polynomial of index $$n$$.

We have $$\psi^{(4)}(t)\equiv 1\\ \psi^{(3)}(1) = -\psi^{(3)}(0) =\frac{1}{2}\\ \psi^{(2)}(1) = \psi^{(2)}(0) = \frac{1}{12}\\ \psi'(1) = \psi'(0) = 0 \\ \psi(1) = \psi(0) = 0$$

We conclude

$$\int_0^1\phi(t)\, dt = \frac{1}{2}( \phi(0) + \phi(1)) - \frac{1}{12} (\phi'(1)- \phi'(0)) +\\ + \frac{1}{24}\int_0^1 \phi^{(4)}(t)\, t^2(1-t)^2 \,dt$$

$$\bf{Added:}$$ We can show the formula above considering $$\phi(t)= t^n$$. Assuming $$n>0$$ we get LHS $$= \frac{1}{n+1}$$ while RHS $$=$$

$$\frac{1}{2} -\frac{n}{12} + \frac{(n-3)(n-2)}{12 (n+1)}= \frac{6 (n+1) -n(n+1) +(n-3)(n-2)}{12(n+1)}=\frac{1}{n+1}$$