Finding the area under curve

Question

Given,

$$ f(x)=1+\frac{1}{x}\int_1^xf(t)dt$$

If $y=f(x)$ , find area enclosed under $y=1$ and co-ordinate axes.

I am not able to evaluate $f(x)$.

Answer 1

Using Newton Leibniz,

$$f'(x)=-\frac{1}{x^2}\int_1^xf(t) dt + \frac{f(x)}{x}$$

Putting $\int_1^xf(x)$ from parent equation,

$$f'(x)=\frac{1}{x}$$

$$f(x)=\log{x}+c$$

Note that on putting $x=1$, integral collapses and $f(x)$ becomes $1$, so $f(1)=1$

Using this, $c=1$.

$$f(x)=\log{x}+1$$

Rest you can proceed.

Answer 2

As an alternative to above solution multiply by x. And then differentiate using Leibnitz rule. We have $f (x)+xf'(x)=1+f (x) $ from here we have $f'(x)=\frac {1}{x} $ and from here hope you can continue.