## 2008 - Higher Secondary (integration)

Zzzz
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Location: 22° 48' 0" N / 89° 33' 0" E

### 2008 - Higher Secondary (integration)

$f(x)$ 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...
Every logical solution to a problem has its own beauty.

tanvirab
Posts: 446
Joined: Tue Dec 07, 2010 2:08 am

### 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.

Zzzz
Posts: 172
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Location: 22° 48' 0" N / 89° 33' 0" E

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

Thanks .. Every logical solution to a problem has its own beauty.

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### 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 $a=1$ $\text {and use}$ $f(x)+f(1-x)=1$)
One one thing is neutral in the universe, that is $0$.

Avik Roy
Posts: 156
Joined: Tue Dec 07, 2010 2:07 am

### 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

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
One one thing is neutral in the universe, that is $0$.