Let [tex] f \ : \ \mathbb{R}\to \mathbb{R} [/tex] be a continuous function which satisfies ,

[tex]\implies \sf f(x) = \displaystyle \int_0^x f(t) dt . [/tex]

Then the value of [tex] f(ln \ 5)[/tex] is ? ​

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Explanation:

According to the Fundamental Theorem of Calculus, we can say

[tex]\displaystyle f(x) = \int_{0}^{x} f(t)dt\\\\\\\displaystyle f'(x) = \frac{d}{dx}\int_{0}^{x} f(t)dt\\\\\\\displaystyle f'(x) = f(x)\\\\[/tex]

Then think about a function in which its derivative is itself. The first thing that should come to mind is the exponential function [tex]f(x) = e^x[/tex]. There aren't any other functions that fit the description. The proof of this is lengthy and involves ODE (ordinary differential equations), so I'm not sure how far along in your calculus class you are.

Anyways,

[tex]f(x) = e^x\\\\\\f(\ln(5)) = e^{\ln(5)}\\\\\\f(\ln(5)) = 5\\\\\\[/tex]

The exponential base 'e' cancels out with the natural log.

We are given with a function and a condition of the respective function too, consider the given what we have ;

[tex]{:\implies \quad \displaystyle \sf f(x)=\int_{0}^{x}f(t)\: dt}[/tex]

By using the variable changing property we can change the Integrand and it will yield ;

[tex]{:\implies \quad \displaystyle \sf f(x)=\int_{0}^{x}f(x)\: dx}[/tex]

Now , let assume that

[tex]{:\implies \quad \displaystyle \sf \int f(x)dx=F(x)}[/tex]

Now, differentiating both sides wr.t.x (using Newton - Leibnitz's rule for Differentiating a function under the integral sign)

[tex]{:\implies \quad \displaystyle \sf \dfrac{df}{dx}=\dfrac{d}{dx}\bigg\{F(x)\bigg|_{0}^{x}\bigg\}}[/tex]

[tex]{:\implies \quad \displaystyle \sf \dfrac{df}{dx}=\dfrac{d}{dx}\{F(x)\}-\dfrac{d}{dx}\{F(0)\}}[/tex]

[tex]{:\implies \quad \displaystyle \sf \dfrac{df}{dx}=\dfrac{d}{dx}\{F(x)\}}[/tex]

[tex]{:\implies \quad \displaystyle \sf \dfrac{df}{dx}=f(x)}[/tex]

So now, we have a function whose derivative is equal to the respective function, now if you differentiate both sides again you will get [tex]{\bf{f^{\prime \prime}(x)=f^{\prime}(x)}}[/tex] and similarly you can get [tex]{\bf{f^{n}(x)=f^{n-1}(x)}}[/tex] . And their is only one function which satisfies this condition for f(x) defined from [tex]{\mathbb{R}}[/tex] to [tex]{\mathbb{R}}[/tex] i.e the exponential function, you just can prove this fact easily by using the Maclaurin's series, so we now arrived on the conclusion that [tex]{\bf{f(x)=e^x}}[/tex] , so now [tex]{\bf{f\{ln(5)\}}}[/tex] will be [tex]{\bf{e^{ln(5)}}}[/tex] i.e hence on simplifying [tex]{\bf{f\{ln(5)\}=\boxed{\bf 5}}}[/tex]

This is the required answer

Note :- If you want to prove the fact that for [tex]{\bf f\: : \: \mathbb{R}\to \mathbb{R}}[/tex], such that [tex]{\bf{f^{n}(x)=f^{n-1}(x)}}[/tex] is [tex]{\bf{f(x)=e^x}}[/tex], you can prove it by using the Maclaurin's series, and if you want it's proof too, I will prefer you seeing once here ;

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