Let $f:R \to R$ be a continuous function which satisfies $f\left( x \right) = \int_0^x {f\left( t \right)dt} $. Then the value of $f\left( {\ln 5} \right)$ is,

Hint: In this particular question differentiate the given equation w.r.t x according to the property that $\dfrac{d}{{dx}}\left( {\int_a^b {f\left( x \right)dx} } \right) = {\left( {f\left( x \right)} \right)_{x = b}} - {\left( {f\left( x \right)} \right)_{x = a}}$ so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given equation:
$f\left( x \right) = \int_0^x {f\left( t \right)dt} $............... (1)
Now we have to find out the value of $f\left( {\ln 5} \right)$.
So, differentiate equation (1) w.r.t x we have,
$\dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {\int_0^x {f\left( t \right)dt} } \right)$
Now as we know that $\dfrac{d}{{dx}}\left( {\int_a^b {f\left( x \right)dx} } \right) = {\left( {f\left( x \right)} \right)_{x = b}} - {\left( {f\left( x \right)} \right)_{x = a}}$ so use this property we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = {\left( {f\left( t \right)} \right)_{t = x}} - {\left( {f\left( t \right)} \right)_{t = 0}}$
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f\left( x \right) - f\left( 0 \right)$.............. (2)
Now in equation (1) substitute x = 0 we have,
$ \Rightarrow f\left( 0 \right) = \int_0^0 {f\left( t \right)dt} $
Now as we know that $\int_o^o {f\left( x \right)dx} = 0$ irrespective of the function.
$ \Rightarrow f\left( 0 \right) = 0$................ (3)
Now substitute this value in equation (2) we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f\left( x \right)$
Now let f (x) = y
$ \Rightarrow \dfrac{{dy}}{{dx}} = y$
$ \Rightarrow \dfrac{{dy}}{y} = dx$
Now integrate on both sides we have,
$ \Rightarrow \int {\dfrac{{dy}}{y}} = \int {dx} $
Now as we know that $\int {\dfrac{1}{x}dx} = \ln x + c,\int {1dx = x + c} $, where C is some arbitrary integration constant.
$ \Rightarrow \ln y + {c_1} = x + {c_2}$
$ \Rightarrow \ln y = x + {c_2} - {c_1}$
Let, ${c_2} - {c_1} = C$ new integration constant so we have,
$ \Rightarrow \ln y = x + C$
Now take antilog on both sides we have,
$ \Rightarrow y = {e^{x + C}}$
$ \Rightarrow y = {e^C}{e^x}$
Now let ${e^C} = A$ new constant.
$ \Rightarrow y = A{e^x}$
Now substitute the value of y we have,
$ \Rightarrow f\left( x \right) = A{e^x}$............... (4)
Now substitute x = 0 we have,
$ \Rightarrow f\left( 0 \right) = A{e^0} = A$
Now from equation (3) f (0) = 0 so we have,
$ \Rightarrow A = 0$
Now from equation (4), we have,
$ \Rightarrow f\left( x \right) = 0$
Now substitute in place of x, x = ln 5 so we have,
$ \Rightarrow f\left( {\ln 5} \right) = 0$
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic integration properties such as, $\int {\dfrac{1}{x}dx} = \ln x + c,\int {1dx = x + c} $, where C is some arbitrary integration constant, and always recall how to differentiate the definite integral which is stated above.