$f(x)$ is a complicated nonlinear function. $f(x)+f(1-x)=1$. Evaluate $\int_0^1f(x)dx$.

Complicated nonlinear function কী?

সমাধানটাও কেউ একটু দিয়েন...

## 2008 - Higher Secondary (integration)

### 2008 - Higher Secondary (integration)

Every logical solution to a problem has its own beauty.

(

(

**Important**: Please make sure that you have read about the**Rules, Posting Permissions and Forum Language**)### Re: 2008 - Higher Secondary (integration)

non-linear means it's graph is not a straight line. Complicated means it's complicated.

That information is irrelevant.

The important information is the given equation. Divide the interval of integration into two halves.

That information is irrelevant.

The important information is the given equation. Divide the interval of integration into two halves.

### Re: 2008 - Higher Secondary (integration)

Thanks ..

Every logical solution to a problem has its own beauty.

(

(

**Important**: Please make sure that you have read about the**Rules, Posting Permissions and Forum Language**)### Re: 2008 - Higher Secondary (integration)

I don't know how to type integration.So let's agree to denote it as sum only

$\int_0^a f(x)dx=\int_0^a f(a-x)dx$,so it is $\frac 1 2$(set $a=1$ $\text {and use}$ $ f(x)+f(1-x)=1$)

$\int_0^a f(x)dx=\int_0^a f(a-x)dx$,so it is $\frac 1 2$(set $a=1$ $\text {and use}$ $ f(x)+f(1-x)=1$)

One one thing is neutral in the universe, that is $0$.

### Re: 2008 - Higher Secondary (integration)

Masum, you can get the code to write the correct latex code by double clicking on the integration in Zubayer's post. I hope you can now edit your code

"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

### Re: 2008 - Higher Secondary (integration)

But the time is over.help-mods

One one thing is neutral in the universe, that is $0$.