3
$\begingroup$

How can I calculate this integral?

$$\int_0^{2014} \frac{f(x)}{f(x) + f(2014 - x)}dx$$

$\endgroup$
4
  • 1
    $\begingroup$ If you know that every function will give same value(maybe by options), then put $f(x)=1$. $$\int_0^{2014}\frac 1 2 dx=1007$$ $\endgroup$
    – evil999man
    May 20, 2014 at 13:09
  • $\begingroup$ Thanks. That's help me. But if I don't know? $\endgroup$
    – MathDav
    May 20, 2014 at 13:19
  • $\begingroup$ trying it another time for f(x)=x will satisfy you that the value is constant. If you just have to submit the answer, then it is an excellent approach(if you can't solve it formally) $\endgroup$
    – evil999man
    May 20, 2014 at 13:33
  • 2
    $\begingroup$ Let $$ \mathcal{I}_1=\int_0^{2014}\frac{f(x)}{f(x)+f(2014-x)}dx. $$ Then, using property $$ \int_b^af(x)\ dx=\int_b^af(a+b-x)\ dx, $$ the integral turns out to be $$ \mathcal{I}_2=\int_0^{2014}\frac{f(2014-x)}{f(2014-x)+f(x)}dx. $$ Since $\mathcal{I}_1=\mathcal{I}_2$, adding $\mathcal{I}_1$ and $\mathcal{I}_2$ yields $$ 2\mathcal{I}=\int_0^{2014}\ dx\quad\Rightarrow\quad \mathcal{I}=1007. $$ $\endgroup$
    – Tunk-Fey
    May 20, 2014 at 17:24

2 Answers 2

8
$\begingroup$

Use substitution $ x' = 2014 - x $, add the two integrals, you get $ 2I = \displaystyle \int_0^{2014} \mathrm{d}x = 2014 \Rightarrow I = 1007 $

$\endgroup$
1
  • $\begingroup$ Why that get similar integral? $\endgroup$
    – MathDav
    May 20, 2014 at 13:03
5
$\begingroup$

Note that for $b=2014$, the object you want is

$$W:=\int_0^b \frac{f(x)}{f(x)+f(b-x)} dx = \int_0^b \frac{f(x)+f(b-x)}{f(x)+f(b-x)} dx - \int_0^b \frac{f(b-x)}{f(x)+f(b-x)} dx\\ = b - \int_0^b \frac{f(b-x)}{f(x)+f(b-x)} dx.$$

Now, by substitution of $t=b-x$,

$$ \int_0^b \frac{f(b-x)}{f(x)+f(b-x)} dx = -\int_b^0 \frac{f(t)}{f(b-t)+f(t)} dt=\int_0^b \frac{f(t)}{f(t)+f(b-t)}dt. $$

So plugging that back into the first equation we get $W=b-W$, such that $W=b/2$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .