Originality!!

It is given that:

$f'(1) = \tan {\frac {\pi}{6}} = \frac {1}{\sqrt{3}} \quad \Rightarrow f'(2) = \tan {\frac {\pi}{3}} = \sqrt{3} \quad \Rightarrow f'(3) = \tan {\frac {\pi}{4}} = 1$

and we need to solve the integral below:

$\displaystyle \int _2 ^3 {f'(x)f '' (x)\space dx} + \int _1 ^3 {f''(x)\space dx}$

Now consider:

$\frac {d}{dx} \left( f'(x)f'(x) \right) = f'(x)f''(x) + f''(x)f'(x) = 2 f'(x)f''(x)$

$\Rightarrow \displaystyle \int {f'(x)f '' (x)\space dx} = \frac {1}{2} f'(x)f'(x) + C$

It is obvious that: $\displaystyle \int {f''(x)\space dx} = f'(x) + C$

Therefore,

$\displaystyle \int _2 ^3 {f'(x)f '' (x)\space dx} + \int _1 ^3 {f''(x)\space dx} = \left[ \frac {1}{2} f'(x)f'(x) \right] _{x=2} ^{x=3} + \left[ f'(x) \right] _{x=1} ^{x=3}$

$= \dfrac {1}{2} (1-3) + 1 - \dfrac {1}{\sqrt{3}} = \dfrac {-1}{\sqrt{3}} \quad \Rightarrow a = \boxed {\sqrt{3}}$

@Parth Lohomi I follow you already . ;)

Well not a level 5 I suppose!!!!

If we look at the first integrand, we see an $f^{ ' }\left( x \right)$ and an $f^{ '' }\left( x \right)$ . This should remind us of chain rule, and we should think about how we would differentiate a function to get $f^{ ' }\left( x \right)$ and $f^{ '' }\left( x \right)$ . We should realize that we need to differentiate $f^{ ' }\left( x \right)$ to some power in order to obtain the $f^{ '' }\left( x \right)$ , and then we realize that power has to be 2.

The indefinite integral of $\int { f'\left( x \right) f''\left( x \right) = } \frac { 1 }{ 2 } { f'\left( x \right) }^{ 2 }$

Just by simple integration, we see that the indefinite integral of the other integrand is $f^{ ' }\left( x \right)$ .

We have bounds given to us, so we now need to find the definite values of these indefinite integrals. If the tangent at (1, f(1)) makes an angle of pi/6 with the +x axis, we know the slope, and therefore the quantity $f^{ ' }(1)$ , is arctan(pi/6)= $\frac { 1 }{ \sqrt { 3 } }$ . The same principle can be applied to the other two points, and we find $f^{ ' }(2)\quad =\arctan { \frac { \pi }{ 3 } }=\quad \sqrt { 3 }$ and $f^{ ' }(3)\quad =\arctan { \frac { \pi }{ 4 } } =\quad 1$ .

Then we go to our integrals and plug in our values:

$\int _{ 2 }^{ 3 }{ f'\left( x \right) f''\left( x \right) } =\frac { 1 }{ 2 } { \left( f'\left( 3 \right) \right) }^{ 2 }-\frac { 1 }{ 2 } { \left( f'\left( 2 \right) \right) }^{ 2 }$

and

$\int _{ 1 }^{ 3 }{ f''\left( x \right) } =f'\left( 3 \right) \quad -\quad f'\left( 1 \right)$

Plugging in our values and adding the two integrals we get it to come out to $\frac { -1 }{ \sqrt { 3 } }$ , so our answer is a = $\sqrt { 3 } \approx \quad 1.732$ .