# given $f(x)$ is continuous and $\int_0^{x^2}f(t)dt = x+sin\frac{\pi}{2}x$,solve $f(1)=?$

given $f(x)$ is continuous and $\int_0^{x^2}f(t)dt = x+\sin\frac{\pi}{2}x$,solve $f(1)=?$

we know $$[\int_0^{x^2}f(t)dt]^\prime= f(x^2) \cdot2x$$ and $$[x+\sin\frac{\pi}{2}x]^\prime = 1+ \frac{\pi}{2}\cos\frac{\pi}{2}x$$ therefore

$$f(x^2) \cdot2x = 1+ \frac{\pi}{2}\cos\frac{\pi}{2}x$$

substitute $x$ by $x=1$, we have: $$f(1) = \frac{1}{2}$$

however by substitute $x$ by $x = -1$ we have: $$f(1) = -\frac{1}{2}$$

more strangely , if I substitute $x$ by $x = 0$ we have: $$f(x^2)\cdot 2x = 0$$ while $$1 + \frac{\pi}{2} \cos\frac{\pi}{2}x \neq 0$$ what's wrong with it?

Thank you!

• Is it a mistake that you use $f(x)$ and $f(t)$? Jun 6, 2016 at 17:32
• @Colbi Seems like standard switching between the "default" variable and the "dummy" variable of integration to me.
– Ian
Jun 6, 2016 at 17:33
• My guess is that the stated relation is only true for $x>0$ not $x<0$. As an indication of this, note that if we were to substitute $y=x^2$ in the first line then to write the RHS in terms of $y$ we'd have to make some choice as to which sign of the square root is chosen. Jun 6, 2016 at 17:34

The issue is that the identity can only be valid for $x\geq 0$ or $x\leq 0$ but not both. The reason is simple: The LHS of the desired relation is even in $x$, but the RHS is odd! So the formula can't hold for all $x$, and we should limit ourselves to one particular sign.
Regarding the behavior at zero, the conclusion that $f(x^2)\cdot 2x\to 0$ as $x\to 0$ assumes that $f(0)$ exists. But if we rearrange the results shown in the OP to solve for $f(x^2)$, we have $$f(x^2)=\frac{1}{2x}\left(1+\frac{\pi}{2}\sin\frac{\pi}{2}x\right)$$ which diverges as $x\to 0$. More precisely, $f(x^2)\approx 1/(2x)$ for small $x$ and so $f(x^2)\cdot 2x\to 1$.