3
$\begingroup$

I need a simple proof for the identity

$$\int_a^{b} f(x)dx=(b-a)\frac{f(a)+f(b)}{2} + (b-a)^2\frac{f'(a)-f'(b)}{12}+O((b-a)⁴)$$

I do not want to use Bernoulli's numbers, as this is not the general formula. Taylor would be great, for example.

Thank you!

$\endgroup$
  • $\begingroup$ Have you tried integration by parts? $\endgroup$ – Mhenni Benghorbal May 2 '14 at 5:56
  • $\begingroup$ Yes, I've tried, it didn't work... maybe I'm missing something $\endgroup$ – Lessa121 May 2 '14 at 6:08
  • $\begingroup$ You may need to do some manipulations to reach the desired form. $\endgroup$ – Mhenni Benghorbal May 2 '14 at 6:11
2
$\begingroup$

Firstly, as you may shift and translate by doing a linear change of variables in $x$, let's assume our bounds are from $0$ to $1$. We will integrate by parts three times (though I missed the third time, Robert Israel did not).

Typically when we integrate $\displaystyle \int u \mathrm d v$ by parts, although we recognize that we can take any antiderivative of $v$, we typically choose the constant to be $0$. But now we'll choose our constants so that when we perform the integration $\displaystyle \int_0^1 v \mathrm{d}x$, we get $0$.

We start with $\displaystyle \int_0^1 f(x)\mathrm{d}x$. I'll let $u = f(x)$ and $\mathrm{d}v = \mathrm{d}x$, so that $\mathrm{d}u = f'(x)$ and $v = x - \frac{1}{2}$. We get

$$ \begin{align} \int_0^1 f(x) \mathrm{d}x &= f(x)\left(x-\frac 12\right) \biggr\rvert_0^1 - \int_0^1 f'(x)\left( x - \frac 12 \right) \mathrm{d}x \\ &= (1-0)\frac{f(1) + f(0)}{2} - \int_0^1 f'(x) \left( x - \frac 12 \right) \mathrm{d}x. \end{align}$$

We do this once more, but choosing the constant from integrating $x - \frac 12$ to be $\frac{1}{12}$, for the same reason as before. Then we see that

$$\begin{align} \int_0^1 f'(x)\left(x - \frac 12\right) \mathrm{d}x &= f''(x)\left( \frac{x^2}{2} - \frac{x}{2} + \frac{1}{12}\right)\biggr\rvert_0^1 - \int_0^1 f''(x) \left( \frac{x^2}{2} - \frac{x}{2} + \frac{1}{12}\right)\mathrm{d}x \\ &= (1-0)^2\frac{f'(1) - f'(0)}{12} - \int_0^1 f''(x) \left( \frac{x^2}{2} - \frac{x}{2} + \frac{1}{12}\right)\mathrm{d}x. \end{align}$$

It just so happens to be that when we integrate by parts a third time, the next term vanishes. Because this has become routine, I omit this calculation. Putting these together, we get the expansion of they type that you want.

$$ \int_0^1 f(x) \mathrm{d}x = (1-0)\frac{f(1) + f(0)}{2} + (1 - 0)^2 \frac{f'(0) - f'(1)}{12} - \int_0^1 f'''(x)\left( \frac{x^3}{6} - \frac{x^2}{4} + \frac{x}{12}\right)\mathrm{d}x.$$

Now $f$ is smooth, and so $f'''(x)$ is bounded on $[0,1]$. So the error is of size $\displaystyle O \left[ \int_0^1 \left(\frac{x^3}{6} - \frac{x^2}{4} + \frac{x}{12}\right)\mathrm{d}x\right] = O \left[\int_0^1 x^3 \mathrm{d}x \right] = O[(1-0)^4],$ where here and throughout I've left the bounds to the correct powers they would appear if we hadn't shifted our bounds to $[0,1]$. As a final note, this is in fact with Bernouilli polynomials, but without ever using their theory or words.

$\endgroup$
  • $\begingroup$ Probably better to do it from $-r$ to $r$ rather than $0$ to $1$. Symmetry is your friend. Also you'll want to do one more integration by parts. $\endgroup$ – Robert Israel May 2 '14 at 6:34
  • $\begingroup$ I'm solving it with -r and r, but I do not understand why one more integration by parts would help... $\endgroup$ – Lessa121 May 2 '14 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.