# 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?