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 curve $y=f\left(x\right)$ and the $x-$axis is,

• A. $\frac{341}{3}\sqrt{109}$
• B. $\frac{109}{18}\sqrt{\frac{109}{3}}$
• C. $341\sqrt{\frac{109}{3}}$
• D. $\frac{109}{9}\sqrt{\frac{109}{3}}$

Answer 1

The correct option is B $\frac{109}{18}\sqrt{\frac{109}{3}}$

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

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

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

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

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

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

$\Rightarrow a=-5$

$\therefore b=\frac{a}{2}-\frac{1}{3}=\frac{-5}{2}-\frac{1}{3}=\frac{-17}{6}$
$\therefore f\left(x\right)=x^2-5x-\frac{17}{6}$

$x^2-5x-\frac{17}{6}=0$ gives the roots

$\alpha =\frac{5}{2}-\frac{1}{2}\sqrt{\frac{109}{3}}$

and $\beta =\frac{5}{2}+\frac{1}{2}\sqrt{\frac{109}{3}}$

$A=$Area$=-{\int }_{\alpha }^{\beta }f\left(x\right)dx$

$=-{\int }_{\alpha }^{\beta }(x^2-5x-\frac{17}{6})dx$

$=-{[\frac{x^3}{3}-\frac{5x^2}{2}-\frac{17x}{6}]}_{\alpha }^{\beta }$

$=\frac{-(\beta -\alpha )}{6}[2({\alpha }^2+{\beta }^2+\alpha \beta )-15(\alpha +\beta )-17]$

$=\frac{-(\beta -\alpha )}{6}[{(\alpha +\beta )}^2-2\alpha \beta -15(\alpha +\beta )-17]$

But $\alpha +\beta =5,\alpha \beta =\frac{-17}{6}$

$\therefore$ Area$=\frac{-1}{6}\sqrt{\frac{109}{3}}[50+\frac{17}{3}-75-17]=\frac{109}{18}\sqrt{\frac{109}{3}}$