Question

Let f: $[1,\infty )\to [2,\infty )$ be a differentiable function such that $f(1)=2$. If $6\underset{1}{\overset{x}{\int }}f(t)dt=3xf(x)-x^3$ for all $x\ge$ 1, then the value of $f(2)$ is____

Answer 1

Answer: 6

$6\underset{1}{\overset{x}{\int }}f(t)dt=3xf(x)-x^3$
$\Rightarrow 6f(x)=3f(x)+3x{f}^{\prime }(x)-3x^2$
$\Rightarrow 3f(x)=3x{f}^{\prime }(x)-3x^2$
$\Rightarrow x{f}^{\prime }(x)-f(x)=x^2$
$\Rightarrow x\frac{dy}{dx}-y=x^2$
$\Rightarrow \frac{dy}{dx}-\frac{1}{x}y=x$ ...(i)
$I.F=e^{\int \frac{1}{x}dx}=e^{-{\log }_{e}x}$
Multiplying (i) both sides by $\frac{1}{x}$
$\frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=1$
$\Rightarrow \frac{d}{dx}\left(y\cdot \frac{1}{x}\right)=1$
integrating
$\frac{y}{x}=x+c$
Put $x=1,y=2$
$\Rightarrow 2=1+c$
$\Rightarrow c=1$
$\Rightarrow y=x^2+x$
$\Rightarrow f(x)=x^2+x$
$\Rightarrow f(2)=6$
Note: If we put $x=1$ in the given equation we get $f(1)=1/3$.