# Calculate the integral without knowing the integrand [closed]

How can I calculate this integral?

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

• If you know that every function will give same value(maybe by options), then put $f(x)=1$. $$\int_0^{2014}\frac 1 2 dx=1007$$ May 20, 2014 at 13:09
• Thanks. That's help me. But if I don't know? May 20, 2014 at 13:19
• 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) May 20, 2014 at 13:33
• 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 $$2\mathcal{I}=\int_0^{2014}\ dx\quad\Rightarrow\quad \mathcal{I}=1007.$$ May 20, 2014 at 17:24

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

• Why that get similar integral? May 20, 2014 at 13:03

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 $W=b/2$.