Strictly Non-negative Region

Calculus Level 1

I'm going to prove that $\displaystyle \int_0^\pi \sin x \, dx = 0$ .

Use the substituion $y = \sin x$ :

when $x = 0$ , $y=0$ ,
when $x = \pi$ , $y=0$ ,
$dx = \dfrac{dy}{\cos x}$ .

So $\displaystyle \int_0^\pi \sin x \, dx = \int_0^0 y \cdot \dfrac{dy}{\cos x}$ .

Since the lower limit and the upper limit of the right hand side integral are equal, thus the integral is equal to 0.

Is my working correct?

Inspired by Rishabh Cool .

Yes, it is correct No, it is incorrect

1 solution

Rishabh Jain
Apr 9, 2016

The above working is not coreect since $\sin x$ is a non negative function in $(0,\pi)$ area bounded by curve $\sin x$ and $x-$ axis must be positive..and therefore $\displaystyle \int_0^\pi \sin x \, dx \neq 0$

Rather, $\displaystyle\int_0^{\pi}\sin x dx=-\cos x|_0^{\pi}=2\neq 0$

Also writing $dx=\dfrac{dy}{\cos x}$ is not permitted in $(0,\pi)$ since $\dfrac{dy}{\cos x}$ becomes undefined at $x=\frac{\pi}{2}$ and therefore this way of doing substitution is incorrect and therefore the working is incorrect...

You are missing out the question of "Is my working correct?" and "If no, why is it wrong"?

Calvin Lin Staff - 5 years, 2 months ago

Is it okay now???

Rishabh Jain - 5 years, 2 months ago

Not really. There is more at play here, and you are most likely missing out the subtlely of the Calculus arguments here.

At the heart, the question is "Why does the change of variables substitution fail in this particular case, when it seems to work everywhere else?" That hasn't been satisfactorily answered as yet.

E.g. Yes, $\frac{y}{ \cos x }$ is undefined at $x = \frac{ \pi}{2}$ , but if we look at the limits of $y$ , does that matter?

Calvin Lin Staff - 5 years, 2 months ago

I’m confuse all the time

Brayden Sarabia - 6 months, 4 weeks ago

Strictly Non-negative Region

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