$\int_{0}^{2\pi }\sin(x)dx = 0$
While the integral equals zero, the area is obviously positive.
When should i accept this as a solution and when not?
2 Answers
$\begingroup$The integral computes the signed area, which is $0$. Essentially, the "negative" when $x \in [\pi, 2\pi]$ cancels out the "positive" when $x \in [0, \pi]$.
If you want the area in a purely geometric sense, you want to integrate $\int_{0}^{2\pi} |\sin x| dx = 4$.
$\endgroup$ $\begingroup$Or integrate where it's above the x-axis, like: $$\int_0^\pi\sin(x) dx=-\cos (\pi)+\cos (0)=2$$ as $\sin $ goes from $0$ to $1$ and back down to $0$ on $[0,\pi] $.
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