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.

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