# 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?

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