I was thinking about the graph of the curve $\sin(x)$. I know that we can generate the graph of $\sin(x)$ by plotting the heights given on the unit circle for various angle measures. Following this line of reasoning, I reasoned that we could find the are under a sine curve by simply looking at the area of that part of the unit circle.
However, if we try (integrating using the anti-derivative) $\int_{0}^{\pi/2} \sin x \,dx$, we get the answer of $1$. A glance at the unit circle would suggest that the answer would be $\pi/4$, as this is the area of the quarter-circle.
What went wrong?
from Hot Weekly Questions - Mathematics Stack Exchange
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