10-seconds challenge-8

Calculus Level 4

Let $f: \mathbb R \rightarrow \mathbb R$ be a continuous function $:$

$f(x)=\int_{0}^x f(t) \, dt.$

Find $f(\ln5)$ .

This is a part of 10-seconds challenge .

The answer is 0.

2 solutions

Rishabh Jain
Mar 8, 2016

Differentiating both sides:- $f'(x)=f(x)$ Multiplying by $e^{-x}$ on both sides:- $e^{-x}f'(x)-e^{-x}f(x)=0$ $\dfrac{d}{dx}(e^{-x}f(x))=0$ $\implies e^{-x}f(x)=C$ For some real constant $C$ . $\implies f(x)=Ce^x\cdots (1)$ Putting $x=0$ in the given condition (i.e $f(x)=\int_{0}^x f(t) \, dt$ ), we get $f(0)=0\cdots (2)$ . From $1$ and $2$ :- $f(x)=0~\forall ~x\in\mathfrak R$ Hence, $f(\ln 5)=\boxed 0$

Yes, exactly! You got it! (+1)

I would omit the passage "Now since we know (1) represents an exponential equation which is always greater than 0"; it's not correct and not needed for your argument: $f(x)=Ce^x$ together with $f(0)=0$ does the job.

Otto Bretscher - 5 years, 3 months ago

Thanks for helping.. Yet again ;-)

Rishabh Jain - 5 years, 3 months ago

Now it's perfect!

Otto Bretscher - 5 years, 3 months ago

This solution is incorrect. From $f'(x)=f(x)$ you cannot infer that $f(x)=e^{x+C}$ .

We could write $f(x)=Ce^x$ ... but do we need to justify why that works?

Otto Bretscher - 5 years, 3 months ago

No... $\dfrac{f'(x)}{f(x)}=1$ Integrating both sides $\ln(f(x))=x+C$ $\implies f(x)=e^{x+C}$ Isn't it correct?? Edit:- Its the same thing you have let $e^C$ as another constant!!

Rishabh Jain - 5 years, 3 months ago

There is a flaw in each step: You are dividing by 0 and then you are taking the $\ln$ of 0.

The proper way to do this kind of problem is by Integrating Factors

Otto Bretscher - 5 years, 3 months ago

@Otto Bretscher – Thanks.. I'll delete my solution..:-)

Rishabh Jain - 5 years, 3 months ago

@Rishabh Jain – Write a better one! You can do it!

Otto Bretscher - 5 years, 3 months ago

@Otto Bretscher – Now is it correct?

Rishabh Jain - 5 years, 3 months ago

A Former Brilliant Member
Jan 2, 2019

Put $x = 0$ , clearly we get $f(0) = 0$

Now differentiate on both sides, we get

$f'(x) = f(x)$

Now, $\dfrac {f'(x) }{f(x)} = 1$

Now, integrate $\displaystyle \int \dfrac {f'(x) }{f(x)} dx = \displaystyle \int 1 dx$

$\Rightarrow \ln (f(x)) = x + C$

Where $C$ is the constant of integration

We get, $f(x) = e^{x+C} \Rightarrow f(0) = e^{0+C}\Rightarrow e^C = 0$

Therefore, $f(x) = e^x e^C = e^x(0) = 0$

Hence, $f(\ln 5) = \boxed{0}$

10-seconds challenge-8

The answer is 0.

2 solutions

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