# B Average angle made by a curve with the $x-axis$

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1. Feb 24, 2017

### Kumar8434

The average angle made by a curve $f(x)$ between $x=a$ and $x=b$ is:
$$\alpha=\frac{\int_a^b\tan^{-1}{(f'(x))}}{b-a}$$
I don't think there should be any questions on that. Since $f'(x)$ is the value of $\tan{\theta}$ at every point, so $tan^{-1}{(f'(x))}$, should be the angle made by the curve at that point.

Now, I expected this to hold:
$$\tan^{-1}\left({\frac{f{(b})-f{(a)}}{b-a}}\right)=\alpha=\frac{\int_a^b\tan^{-1}{(f'(x))}}{b-a}$$
because, $\tan^{-1}\left({\frac{f{(b})-f{(a)}}{b-a}}\right)$ is also the 'average angle' made by the curve between $x=a$ and $x=b$.
It was true only approximately for $f(x)=\log{|\sec{x}|}$ when I checked for $a=0$ and $b=\frac{\pi}{4}$. It obviously holds for linear functions and I checked that it only approximately holds for quadratic functions. I don't know anything beyond high-school calculus, so couldn't check it for polynomials of degree greater than $2$.

I also tried root-mean-square instead of average angle but that expression too didn't hold accurately.

I took one step further and replaced $\tan^{-1}{x}$ with any function $g(x)$ and expected this to hold:
$$g\left({\frac{f{(b})-f{(a)}}{b-a}}\right)=k=\frac{\int_a^bg{(f'(x))}}{b-a}$$
But this one too only holds approximately for some $g(x)$ that I checked.
So, why don't these expressions hold as expected?

Last edited: Feb 24, 2017
2. Feb 24, 2017

### Staff: Mentor

arctan depends on x so you can't just pull it out of the integral over x like that.

In general, you can't pull any function dependent on x out of an integral over x.

If you changed the integral to a summation and ask yourself about this simpler example:

atan( f(x1) + f(x2) + f(x3) )

- vs -

atan( f(x1) ) + atan( f(x2) ) + atan( f(x3) )

3. Feb 24, 2017

### Staff: Mentor

Because your expectation was wrong.

4. Feb 24, 2017

### Staff: Mentor

Have you tried using the chain rule to evaluate the righthand side to see if it matches the lefthand side:

$$\tan^{-1}(f(b)-f(a)) = \int_a^b\tan^{-1}(f'(x)) dx$$

to see if you can derive your conclusion. It may also explain why the results are different from what you expected.

https://en.wikipedia.org/wiki/Chain_rule

In particular heck out the first proof which uses limits to prove the rule.

Last edited: Feb 24, 2017
5. Feb 24, 2017

### Staff: Mentor

Thread closed. As @Dale pointed out in post #3, your expectation is wrong. In future posts, please make more of an attempt on proving your conjectures instead of asking why something doesn't work.