#help ploxxxxx
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$|Sin2x|dx from 0 to pie$
infurnus
$\int_{x=0}^{\pi} |sin2\text{x}|d\text{x}$
gameswastedme
What's the area of sin curve from x= 0 to pi
2times
Cos2x/2
Integration from 0 to pie /2
I don't get you
reverse chain rule instead of multiply like you would in derivatives, you divide so it cancels. Example: derivative of sin(2x) = cos(2x) * 2. know to find the integral of cos(2x) * 2, you would keep the 2 coefficient, and integrate cos(2x) whihc is sin(2x) /2. This is the opposite of the chain rule since I divided sin(2x) by the derivative of the inside function. so the integral is 2sin(2x)/2 = sin(2x).
you can apply this to your question.
@haughty gyro there are two ways to do it, either draw the graph of mod(sin2x) and using its periodicity rewrite the limits
or evaluate the integral by separating it into two parts where sin2x is positive and negative
Se can Integrate sin 2x from 0 to π/2 then multiply by 2
Now to integrate sin 2x
Let 2x=u
du = 2 dx
dx=du/2
Now it becomes sin(u)/2 du
1/2 will jump out of the integration sign
So it will be sin(u)du
It will become -cos(u) + C
Or, - cos 2x /2 + c
This is what I did
Oh sorry didn't read that
Ans is 2
Noice
Using this method we get 1
Yus
Hmmmmm
Idk
You have to multiply by two cuz you're intergrating from 0 to pi/2
Mai 0 se pie
Tak karra
Tu kings rule ke agge wala
Rule lagaraa
Vo smaaj gyaa
You're Indian?
Yehh
Oh yeah
I haven't applied kings rule
You can't directly integrate when there is modulus function
What is kings rule?
12th
?
What country are you from??
Bhaiya Hindi samaj gya phir bhi nhi samjhe?
Eh?
Ayega terko
Kaise?
Kya?
Ohhhh
Definate
+CLOSE