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The average angle made by a curve $f(x)$ between $x=a$ and $x=b$ is: $$\alpha=\frac{\int_a^b\tan^{-1}{(f'(x))}dx}{b-a}$$ 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: $$\tan^{-1}\left({\frac{f{(b})-f{(a)}}{b-a}}\right)=\alpha=\frac{\int_a^b\tan^{-1}{(f'(x))dx}}{b-a}$$ 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 held 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: $$\beta=\sqrt{\frac{\int_a^b(\tan^{-1}{{(f'(x))}})^2dx}{b-a}}$$ and expected the same to hold but that expression to didn't hold accurately.

I took one step further and replaced $\tan^{-1}{x}$ with any function $g(x)$ and expected this to hold: $$g\left({\frac{f{(b})-f{(a)}}{b-a}}\right)=k=\frac{\int_a^bg{(f'(x))}dx}{b-a}$$ But this one too only holds approximately for some $g(x)$ that I checked. I had worked with $f(x)=x^2$ to obtain the approximate formula for $g(a+b)$ in this previous post of mine:When do these expansions of $log(a+b)$, $tan^{-1}(a+b)$, $sin^{-1}(a+b)$, etc work? because for $f(x)=x^2$, the LHS of these expressions reduces to $g(a+b)$.

So, why don't these expressions hold as expected?

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Here's an extremal example for you.

Take a piece-wise linear curve that is horizontal for $x$ in $[0,1]$ and almost vertical (say of height 1000) for $x \in [1,2]$. The average angle is approximately $\frac{0 \deg}{2} + \frac{90 \deg}{2}$, which is approximately $45 \deg$, that is very different from $90 \deg$.

This doesn't work with the average slope because $\frac{0}{2} + \frac{ 1000}{2} = 500$ as expected.

This example can be smoothed by a bump function.

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    How is the average angle 45 deg? Average angle=$\frac{0 deg}{(1-0)}+\frac{90 deg}{(2-1)}=90 deg$. And, average slope=$\frac{0}{(1-0)}+\frac{1000}{(2-1)}=1000$. And arctan(average slope)=arctan(1000)$\approx$90 deg. – 2017-02-24
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    @Dove b=2, a=0, so you have to divide by two. – 2017-02-25
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    @Dove if you want to calculate the averages for the two pieces separately, you can not then just add them to obtain the total average. You have to add them with weights corresponding to how long horizontally those pieces are. Think of average speed. If you drive 60km/h for 1 minute and 40 km/h for 1 minute your average speed is not 100 km/h, but 50 km/h – 2017-02-25
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    But why do these equations hold approximately if it's completely wrong to do that? – 2017-02-26
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    @Dove If you divided by $\frac12(b-a)$ instead of $b-a$ in your equations, the way you (in effect) did in the comment above, they wouldn't have come out anywhere near what you expected. – 2017-02-26