An Intriguing Integral

Substitute $x=0$ and $y=0$ in the given functional equation to get $f(0)=0$ .

Since,

$\displaystyle f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim_{h\rightarrow 0} \frac{f(x)+f(h)+3xh-f(x)}{h}$

$\displaystyle \Rightarrow f'(x)=\lim_{h\rightarrow 0} \frac{f(h)+3xh}{h}$

Substituting $h=0$ gives 0/0 hence, apply l'Hôpital's rule.

$\displaystyle \Rightarrow f'(x)=\lim_{h\rightarrow 0} f'(h)+3x=3x+35$

Solving the differential equation and using the initial condition that $f(0)=0$ , we have

$\displaystyle f(x)=\frac{3x^2}{2}+35x \Rightarrow f(2)=\boxed{76}$

By differentiating both side w.r.t. $x$ , we obtain $f'(x+y)=f'(x)+3y$ . Continue differentiating both sides w.r.t. $y$ , we deduce that $f''(x+y)=3$ . Hence $f''(x)=3$ for all $x$ . Then $f'(x)=3x+c$ . Since $f'(0)=35$ , $c=35$ , and then $f(x)=\frac{3}{2}x^2+35x+d$ . By using the assumption $f(x+y)=f(x)+f(y)+3xy$ , we deduce that $d=0$ . Therefore, $f(x)=\frac{3}{2}x^2+35x$ and $f(2)=76$ .

First, we find $f(0)$ . $$f(0+0)=f(0)+f(0)+3(0)(0)$$ $$f(0)=2f(0)$$ $$f(0)=0$$ We know a value of the derivative of $f$ , so it would likely be helpful to take the derivative of $f$ . However, we need to get $f$ in terms of one variable, so we fix $y=c$ . So, we have $$f(x+c)=f(x)+f(c)+3xc.$$ Differentiating, we get $$f'(x+c)=f'(x)+3c$$ because $c$ is a constant. Differentiating again, we have $$f''(x+c)=f''(x).$$ Because this holds for any constant $c$ , we have that $f''(x)$ is constant. In other words, $$f''(x)=a,$$ for some constant $a$ . Integrating, we get $$f'(x)=ax+b,$$ for some constant $b$ . Because we know $f'(0)=35$ , we have $b=35$ . So, $$f'(x)=ax+35.$$ Integrating again, we get $$f(x)=\frac{a}{2}x^2+35x+k,$$ for some constant $k$ . We know $f(0)=0$ , so $k=0$ and $$f(x)=\frac{a}{2}x^2+35x.$$ To find out the value of $a$ , we observe $$f(0)=f(x+(-x))=f(x)+f(-x)-3x^2=0.$$ Thus, $$f(x)+f(-x)=3x^2.$$ Applying this fact to our equation $f(x)=\frac{a}{2}x^2+35x$ gives $$\left( \frac{a}{2}x^2 +35x \right) + \left( \frac{a}{2}x^2 -35x \right) = ax^2 = 3x^2.$$ Thus $a=3$ , and so $$f(x)=\frac{3}{2}x^2+35x$$ and $$f(2)=\frac{3}{2}(2)^2+35(2)=\fbox{76}.$$

Since $f'(0)$ is given to us, we aim to find an equation involving $f'(x)$ and the simplest way is to differentiate both sides of the given equation. Taking the partial derivative with respect to $x$ on both sides of the equation, we have $f'(x+y)=f'(x)+3y$ and substituting $x=0$ we get $f'(y)=f'(0)+3y=35+3y$ as given. Now, integrating both sides with respect to $y$ gives us $f(y)=\frac{3}{2}y^2+35y+C$ where $C$ is some constant.

Substituting this back to the original equation gives us $\frac{3}{2}(x+y)^2+35(x+y)+C=\frac{3}{2}x^2+35x+C+\frac{3}{2}y^2+35y+C+3xy$ and cancelling like terms on both sides gives $C=2C \Rightarrow C=0$ and hence $f(x)=\frac{3}{2}x^2+35x$ so $f(2)=\frac{3}{2}2^2+35(2)=76$

Let polynomial

$f(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \ldots + a_{2} x^{2} + a_{1} x + a_{0}$ ,

Then,

$f(y) = a_{n} y^{n} + a_{n-1} y^{n-1} + \ldots + a_{2} y^{2} + a_{1} y + a_{0}$ , and

$f(x + y) = a_{n} (x + y)^{n} + a_{n-1} (x + y)^{n-1} + \ldots + a_{2} (x + y)^{2} + a_{1} (x + y) + a_{0}$

---

Using the given equation, $f(x+ y) = f(x) + f(y) + 3xy$

We get $f(x+y) - (f(x) + f(y)) = 3xy$

$a_n((x + y)^{n} - (x^{n} + y^{n})) + \ldots + a_{2} ((x^{2} + y^{2} + 2xy) - (x^{2} + y^{2})) + a_{1}( (x + y) - (x + y)) + a_{0} - 2a_{0}= 3xy$

---

Since there is only term of $xy$ , and not $x^{2}y, xy^{2}$ , etc on the RHS, therefore the value of $n = 2$

From the above equation, we get $a_{0} - 2a_{0} = 0$ , therefore the constant term of $f(x), a_{0} = 0$

Also equating the $xy$ term gives $2a_{2}xy = 3xy$

$a_{2} = \frac{3}{2}$

$f(x)$ can be rewritten as $\frac{3}{2}x^{2} + a_1{x} + 0$

Here I will be replacing $a_{1}$ , with $a$ . So,

$f(x) = \frac{3}{2}x^{2} + ax$

Differentiating $f(x)$ will give

$f'(x) = \frac{3\times 2}{2}x + a$

$f'(x) = 3x + a$

Since $f'(0) = 35$ ,

$3 \times 0 + a = 35$

$a =35$

Now we know

$f(x) = \frac{3}{2}x^{2} + 35x$

Therefore,

$f(2) = \frac{3}{2} \times 2^{2} + 35 \times 2$

$f(2) = 6 + 70$

$f(2) = \boxed{76}$

putting x=y=0 we get,f(0)=0...now put x=x and y=h....f(x+h)=f(x)+f(h)+3xh now...(f(x+h)-f(x))/h=(f(h)-f(0))/h+ 3x...and h tends to 0....the equation becomes..f '(x)=f '(0)+3x=35+3x..after integrating..f(x)=35x+1.5x^2+c...c=0 as f(0)=0....f(2)=76!!

Firstly, check for x=y=0 in the given polynomial expression,

f(0)= f(0)+ f(0) ===>f(0)= \boxed{0}

Now, we know that,

*f'(x) = limit(h \rightarrow 0) \frac {f(x+h)- f(x)}{h} *

Simplifying above, we obtain,

f'(x)

    = limit(h\rightarrow0) \frac {[f(x) + f(h) + 3hx] -f(x) }{h}

   = limit(h\rightarrow0) \frac {f(h) + 3hx}{h}

   = limit(h\rightarrow0) \frac {3hx + f(h) - f(0)}{h}

   = limit(h\rightarrow0) \frac { 3x +  \frac {f(h)-f(0)}{h}}

Since, limit(h\rightarrow0) \frac {f(h)-f(0)}{h} = f'(0), we have,

f'(x) = 3x + f'(0) = 3x +35

\Rightarrow f(x) = 3/2 x^2 + 35x ... (since we get the constant of integration as 0 by using the condition f(0)=0)

Thus, \boxed{f(2)=76}

Differentiating f(x+y) = f(x) + f(y) + 3xy with respect to x we get,

    f'(x+y) = f'(x) + 3y

Now, put x=0. We get f'(y) = 3y + 35.

Integrating this equation we get, f(y) = 3/2 y^2 + 35y+ C, ............(i) where C is the constant of integration .

Putting x=y=0 in the original functional equation, we get that f(0)=0.

Putting y=0 in (i) we get f(y) = 3/2 y^2 + 35y

Putting y=2 in (i) we get, f(2) = 76.

f(2) = 76 is the required answer.

f'(x+y)=f'(x)+3y put x=0,y=x then

f'(x)=35+ 3x

y=35x+ 3x*x/2 +c

since f(0)=0

c=0

f(2)=76