Dear Uncle Colin
How would you prove that the area under the curve $y=\sin(x)$ from $x = 0$ to $x=\pi$ was exactly 2?
‘Cause I Realised Calculus Lacked Explanation
Hi, CIRCLE, and thanks for your message!
The standard way is just to know that the integral((with respect to $x$)) of $\sin(x)$ is $-\cos(x)$, plug in the limits and dust your hands as if you’ve done something clever. But – as you suggest – that’s more of a trick than an explanation.
Let’s look instead at a unit circle, and consider a tiny sector of it. The angle at the centre is $\Delta\theta$, and the whole thing is raised at an angle $\theta$ from the positive $x$-axis.
If $\Delta \theta$ is small enough, the arc of the sector is effectively straight and has length $\Delta \theta$ (because the radius is 1). If we split that into a horizontal and vertical component, the horizontal displacement is $-\Delta \theta \sin(\theta)$.
Now. Let’s split up the upper half of the circle so it’s made entirely out of $n$ equal tiny sectors. Then we know that $\sum_1^n -\Delta \theta \sin(\theta_i) = -2$, where $\theta_i$ is the inclination of the $i$th sector. It’s going to vanish in a minute, so don’t worry about it.
In the limit as $n$ approaches infinity, the sum becomes an integral and we get $\int_0^\pi -\sin(\theta) d\theta = -2$, which – once you get rid of the minus signs – is exactly what you wanted to prove.
Hope that helps!
- Uncle Colin
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