# Find $f(1)$ from equation $\int_{0}^{x}f(t)dt = x+\int_{x}^{1}tf(t)dt$

My brother brought me this same question, also on this website.

$$\int_{0}^{x}f(t)dt = x+\int_{x}^{1}tf(t)dt \tag{1}$$

When I try to solve it, I also did same thing. Differentiation leads to

$$f(x) = 1-xf(x)\\ f(x) = \frac{1}{1+x} \\ \int_{0}^{1}f(x) dx = \ln 2$$

But try putting $x = 1$ in original equation $(1)$,

$$\int_{0}^{1} f(t) dt = 1$$

so what has happened? We get two contradictory results. Is it still true that $f(1) = 1/2$? Where is the mistake?

• This reminds me of an old question from India's prestigious JEE for admission to top government engineering colleges. See math.stackexchange.com/q/1495336/72031 Commented Mar 12, 2018 at 6:31
• +1 to you for checking that $f(x) =1/(1+x)$ leads to a contradiction. Most students stop at the answer $f(x) =1/(1+x)$. Commented Mar 12, 2018 at 6:34
• Wow again we have 3 close votes. Where are my popcorns? Commented Mar 15, 2018 at 16:23

You are correct. The problem is with the person who made the question.

The functional equation is incorrect.

The correct functional equation for this function must be :

$$\int_{0}^{x}f(t)dt = x+\int_{x}^{1}tf(t)dt + \ln 2 -1$$

There are always issues, if you are differentiating an expression, and there are constants.

Since taking derivative destroys existence of any constant, often there are mistakes in questions, if not formatted correctly.

• Does that imply that original functional equation has no solutions? Commented Feb 14, 2018 at 0:02
• @wilkersmon Yes. Definitely. Commented Feb 14, 2018 at 5:09
• @KingTut Check this out Commented Mar 10, 2018 at 14:03
• +1 for the starting line. Settings questions for an exam/book is not as easy as it may appear. Commented Mar 12, 2018 at 6:38

If two functions have the same derivative, all you can say about them is that they differ by a constant. So the $f$ you found is the solution to an equality of the form $$\int_{0}^{x}f(t)dt = x+\int_{x}^{1}tf(t)dt + c,$$ where $c$ is not necessarily $0$. In particular, by differentiating you lost the chance of identifying $f(1)$.

• Well... $c = \ln 2 -1$, atleast for this case. Commented Feb 13, 2018 at 14:45
• Indeed. $\ \ \ \$ Commented Feb 13, 2018 at 14:46
• I'm not sure what you are aiming for. If you differentiate, you lose constants. If you square, you lose minus signs. They are things one needs to be aware of when processing an equation. Commented Feb 13, 2018 at 14:56
• Yes, now i understand. Commented Feb 13, 2018 at 15:48