Question

The area bounded by $y=f\left(x\right)$, x-axis and the $y=1$, where $f\left(x\right)=1+\frac{1}{x}{\int }_1^{x}f\left(t\right)dt$

• 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 $\left(1-\frac{1}{e}\right)$

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

Multiply both sides by $x$, we get,

$xf\left(x\right)=x+{\int }_1^{x}f\left(x\right)dx$

Differentiating w.r.t.$x$., we get

$f\left(x\right)+x{f}^{\prime }\left(x\right)=1+f\left(x\right)$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x}$

Integrating both sides, we get

$\int f^{\prime }\left(x\right)dx=\int \frac{dx}{x}$

$\Rightarrow f\left(x\right)=ln\left(x\right)+C$ (2)

From $\left(1\right)$, we have

$f\left(1\right)=1$

Plugging in $\left(1\right)$, we have

$f\left(1\right)=ln\left(1\right)+C$

$\Rightarrow 1=C$

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

Thus, Area $A={\int }_{\frac{1}{e}}^1(lnx+1)dx$