Question

Let f(x) is continuous function as shown in figure. If the area bounded by the curve $y=f\left(x\right)$, $y=x\sqrt{x}$ and line segment AB is equal to area bounded by $y=f\left(x\right)$, y-axis and line segment AC and ${\int }_0^{1}f\left(x\right)dx=\frac{1}{3}$ and $f\left(\frac{1}{4}\right)=\frac{a}{b}$ (where a and b have no common factors) then value of $\lfloor a/b\rfloor$ is (where $\lfloor \rfloor$ denotes greatest integer function)

Answer 1

${\int }_0^{\alpha }f\left(x\right)dx-{\int }_0^{\alpha }x\sqrt{x}dv=\alpha f\left(\alpha \right)-{\int }_0^{\alpha }f\left(x\right)dx$
$f\left(\alpha \right)-\alpha \sqrt{\alpha }=\alpha f^{{}^{\prime }}\left(\alpha \right)+f\left(\alpha \right)-f\left(\alpha \right)$
$\frac{dy}{dx}-\frac{y}{x}=-\sqrt{x}$ $(\alpha \to x)$
If $e^{-\int \frac{dx}{x}}=\frac{1}{x}$
$y.\frac{1}{x}=\int -\sqrt{x}\times \frac{1}{x}dx+c$
$\frac{y}{x}=-2\sqrt{x}+c\Rightarrow y=-2x\sqrt{x}+cx$
${\int }_0^1(-2x\sqrt{x}+cx)dx=\frac{1}{3}$
$\Rightarrow c=\frac{34}{15}$
So $f\left(\frac{1}{4}\right)=-2\times \frac{1}{4}\times \frac{1}{2}+\frac{34}{15}\times \frac{1}{4}=-\frac{1}{4}+\frac{17}{30}=\frac{19}{60}$