Why is the AUC of sine x from 0 to pi =2? I can calculate it using integration BUT it does not match intuitively expect looking at the unit circle Area of a sine function:
I think I was able to add my diagrams
I have intuitive problems with this issue (area of sine function):
 showing area of circle


*

*From a visual perspective using the unit circle:
The area of a circle = πr^2.
Then, in the case of the unit circle, the area would be π1^2=π=3.1416
Thus, the area of half of the unit circle would be ½ π=1.571

As an object at location (1,0) on the unit circle sweeps across to (-1,0), it has swept from 0 to π (a circumference of π).
So, the area of y=sin⁡x from 0 to π should be 1.571. But it isn’t, it is 2
I have superimposed a 1 x 1 grid over the unit circle. To the eye, it appears to me that the area of half a circle is less than 2...and estimated to be 1.5, which I close to my calculated 1.57
But when I examine the area under the curve for the sine function from 0 to pi:
I can see just by rough estimation that the AUC is approximately 2
I cannot, no matter how hard I have tried, reconcile these two different ways of looking at the matter.
 A: The point $(x,y)=(\cos\theta,\sin\theta)$ moves along the circle, and as $\theta$ goes from $0$ to $\pi$, $x$ goes from $1$ to $-1$.  If you want $x$ to increase from $-1$ to $1$, then you need $\theta$ decreasing from $\pi$ to $0$.  So the area of the semi-circle that you describe is
$$
\frac \pi 2 = \int_{x=-1}^{x=1} (\text{height of the curve above the $x$-axis})\cdot dx = \int_{x=-1}^{x=1} \sin\theta\,dx.
$$
Note that
$$
\frac\pi2 = \int_{x=-1}^{x=1} \sin\theta\,dx \ne\int_0^\pi \sin\theta\,d\theta.
$$
I.e. $\underbrace{\sin\theta\,dx}$ is different from $\underbrace{\sin\theta\,d\theta}$. One of them has $dx$ where the other has $d\theta$.
Since $x=\cos\theta$, we have $\underbrace{dx=-\sin\theta\,d\theta}$, and that's not the same as $d\theta$.  In fact
$$
\int_{x=-1}^{x=1} \sin\theta\,dx = \int_{\theta=\pi}^{\theta=0} \sin\theta \Big({-\sin\theta\,d\theta}\Big) = \int_0^\pi \sin^2\theta\,d\theta.
$$
Calculus instruction does not give sufficient attention to the meanings of things like $dx$ and $d\theta$: they are infinitely small increments of $x$ and $\theta$.  And $x$ and $\theta$ change at different rates.  Speaking of infinitely small increments is immensely useful as a heuristic, and that's why you see that done in physics and engineering courses, but it's not in accord with modern standards of logical rigor.  Logical rigor gets idolatrously worshipped. It should be used rather than worshipped.
The integral $\displaystyle\int_0^\pi \sin\theta\,d\theta = 2 \ne \frac \pi 2$ is the area under the graph of the sine function.  Look closely and carefully at what that graph looks like and what the various measurements are.
