Question

The area bounded by $y=f\left(x\right),x-axis$ and the line $y=1$, where $f\left(x\right)=1+\frac{1}{x}\underset{1}{\overset{x}{\int }}f\left(t\right)dt$ is

• A. $2\left(e+1\right)$
• B. $\left(1-\frac{1}{e}\right)$
• C. $2\left(e-1\right)$
• D. None of these

Answer 1

The correct option is B $\left(1-\frac{1}{e}\right)$

$f\left(x\right)=1+\frac{1}{x}{\int }_I^{x}f\left(t\right)dt$

Here, $f\left(1\right)=1+{\int }_1^{1}f\left(t\right)dt=1$

$f^{\prime }\left(x\right)=\frac{d\left(f\right(x)}{dx}=\frac{d}{dx}(1+\frac{1}{x}{\int }_I^{x}f(t\left)dt\right)$

$=0+\frac{d}{dx}\left(\frac{1}{x}{\int }_I^{x}f\right(t\left)dt\right)$

$=\frac{1}{x}\frac{d}{dx}\left({\int }_I^{x}f\right(t\left)dt\right)+{\int }_I^{x}f\left)t\right)dt\frac{d}{dx}\left(\frac{1}{x}\right)$

Using Newton's Lebinitz Rule

$\frac{df\left(x\right)}{dx}=\frac{1}{x}\frac{d\left(x\right)}{dx}.f\left(x\right)-\frac{1}{x}\frac{d\left(1\right)}{dx}.f\left(1\right)-\frac{1}{x}\left[f\right(x)-1]x$

$=\frac{f\left(x\right)}{x}+\frac{1-f\left(x\right)}{x}=\frac{1}{x}$

$\therefore df\left(x\right)=\int \frac{1}{x}dx$

$f\left(x\right)=4x+4c$

$\therefore f\left(1\right)=1=4\left(1\right)+4c=4c$

$\therefore c=e$

Hence, $f\left(x\right)=4x+1$

$\therefore f\left(x\right)=0\Rightarrow 4x=-1\Rightarrow x=\frac{1}{e}$

$\therefore$ Area bounded=$(1-0)(1-0)-{\int }_{\frac{1}{e}}^{1}f\left(x\right)dx$

$=1-(x4x{)}_{\frac{1}{e}}^1=1-0+\frac{1}{e}4(\frac{1}{e})$

$=1-\frac{1}{e}$