Mathematics - Definite Integration Question with Solution | TestHub

Given, $f(1)=\frac{1}{3}$ and $6{\int }_1^{x}f(t)dt$ $=3xf(x)-x^3$, for all $x\ge 1$ Using (Newton-Leibnitz formula), On Differentiating both sides, $$\begin{array}{rl}& 6f(x)\cdot 1-0-3f(x)+3x{f}^{\prime }(x)-3x^2\\ \Rightarrow & 3x{f}^{\prime }(x)-3f(x)=3x^2\\ \Rightarrow & f^{\prime }(x)-\frac{1}{x}f(x)=x\end{array}$$ $$\Rightarrow \frac{x{f}^{\prime }(x)-f(x)}{x^2}=1\Rightarrow \frac{d}{dx}\left\{\frac{f(x)}{x}\right\}=1$$ On integrating both sides, $$\begin{array}{ll}& \frac{f(x)}{x}=x+C\left[\because f(1)=\frac{1}{3}\right]\\ \Rightarrow & \frac{1}{3}=1+C\Rightarrow C=-\frac{2}{3}\\ & \text{ Now, }f(x)=x^2-\frac{2}{3}x\\ \Rightarrow & f(2)=4-\frac{4}{3}=\frac{8}{3}\end{array}$$ Note Here, $f(1)=2$ does not satisfy given function. $$\therefore f(1)=\frac{1}{3}$$ For that, $f(x)=x^2-\frac{2}{3}x$ and $3f\left(2\right)=3\left(4-\frac{4}{3}\right)=3\left(\frac{8}{3}\right)$=8