Question

Consider the function $f\left(x\right)$ such that $f\left(x\right)=x^2+{\int }_0^1(x+t)f\left(t\right)dt$, then the area bounded by the $x-$axis, the tangent and normal at $x=0$ on the curve $y=f\left(x\right)$ is,

• A. $\frac{3757}{60}$
• B. $\frac{3757}{90}$
• C. $\frac{3757}{180}$
• D. $\frac{3757}{120}$

Answer 1

Answer: (C)

$\frac{3757}{180}$

$y\left(x\right)=x^2+ax+b,$

where $a={\int }_0^{1}y\left(t\right)dt,b={\int }_0^{1}t.y\left(t\right)dt$

$\therefore a={\int }_0^1(t^2+at+b)dt,b={\int }_0^{1}t.(t^2+at+b)dt={\int }_0^1(t^3+a{t}^2+bt)dt$

$\Rightarrow a={[\frac{t^3}{3}+\frac{a{t}^2}{2}+bt]}_0^1,b={[\frac{t^4}{4}+\frac{a{t}^3}{3}+\frac{b{t}^2}{2}]}_0^1$

$a=\frac{1}{3}+\frac{a}{2}+b$ and $b=\frac{1}{4}+\frac{a}{3}+\frac{b}{2}$

$\Rightarrow \frac{a}{2}-b=\frac{1}{3}$ and $\frac{b}{2}-\frac{a}{3}=\frac{1}{4}$

By solving we get

substituting $b=\frac{a}{2}-\frac{1}{3}$ in $\frac{b}{2}-\frac{a}{3}=\frac{1}{4}$ we get

$\frac{b}{2}-\frac{a}{3}=\frac{1}{4}$

$\Rightarrow \frac{(\frac{a}{2}-\frac{1}{3})}{2}-\frac{a}{3}=\frac{1}{4}$

$\Rightarrow \frac{a}{4}-\frac{1}{6}-\frac{a}{3}=\frac{1}{4}$

$\Rightarrow \frac{-a}{12}=\frac{10}{24}$

$\Rightarrow a=-5$

$\therefore b=\frac{a}{2}-\frac{1}{3}=\frac{-5}{2}-\frac{1}{3}=\frac{-17}{6}$

$\therefore f\left(x\right)=x^2-5x-\frac{17}{6}$

$f\left(x\right)=x^2-5x-\frac{17}{6}$

$f^{\prime }\left(x\right)=2x-5$

Tangent at $x=0$ is $y+\frac{17}{6}=-5x$ which meets the $x-$axis at $(\frac{-17}{30},0)$

The normal $x=0$ is $y+\frac{17}{6}=\frac{x}{5}$ which meets the $x-$axis at $(\frac{85}{6},0)$

Area$=\frac{1}{2}\left(\frac{17}{6}\right)(\frac{85}{6}+\frac{17}{30})=\frac{3757}{180}$sq.units