## Why is the area under one hump of a sine curve exactly 2?

I was talking with a student recently who told me that he always found the fact that $\int_0^{\pi} \sin x \, dx = 2$ amazing. “How is it that the area under one hump of the sine curve comes out exactly 2?” He asked me if there is an easy way to see that, or is it something you just have to discover by doing the computation.

Imagine a particle travelling counterclockwise around a unit circle with unit speed.  Let’s say that it starts moving from the rightmost point of the circle to the leftmost point. (Orient so that the vertical axis is truly vertical.) It’s high noon and the sun is casting a shadow of the particle straight down on the ground. The shadow only moves in the horizontal direction. What is the shadow’s speed when the particle is at $\theta$? It’s $\sin \theta$. (Since we’re interested in distance traveled, we measure speed as a positive quantity.)

##### If the particle has speed 1, its shadow has speed $\sin \theta$.

The integral of speed is distance.

So the integral of the sine function over one hump is the total distance that the shadow travels, which is exactly the diameter of the circle.

In fact, with just a little tweaking, this setup also makes it clear that the indefinite integral of sine is the negative of cosine.

We're a math club for girls.
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### 21 Responses to Why is the area under one hump of a sine curve exactly 2?

1. Deleance Blakes says:

Nice observation, although i think there’s no logical answer why the number is magically 2. I mean consider any problem that involves seemingly a lot of mathematics, but comes to a simple answer–Like Euler’s Identity, we can show many ways to compute that outcome but the fact it’s as simple as it is, remains a mystery within math.

2. Gareth says:

I did my maths in the 60’s and it was only a few nights ago that I thought these same thoughts of why exactly 2 … there must be an explanation , it could be geometry ?

3. Leong Qi Jin says:

May I know why is the speed sine theta? I only know it because it is cos x differentiated. Is there any other way to explain it?

• girlsangle says:

Thank you for this question. It’s an important one. The speed of the shadow is found using trigonometry; no differentiation is needed. By construction, the ball going around is moving at a speed of 1 radian per second. The speed of the shadow is the magnitude of the component of the velocity in the horizontal direction, and this magnitude (for the part of the ball’s journey from the right side of the circle to the left) is sine theta. (See, for example, this YouTube video.)

• Leong Qi Jin says:

Thank you! Everything became clear when I realised theta above the velocity vector is equal to theta from the centre.

4. Shiv says:

y = sinx

( integral sin x from 0 to pi/2) area under the curve. = 1 sq unit
integral sinx from 0 to pi =2*1 = 2 sq units as symmetrical.

So you can prove its 2 by finding area under the curve

• girlsangle says:

Right. The student knew how to compute this using integration, but was surprised that the answer comes out to be an integer and wanted an explanation for the fact. Here, I’m offering an interpretation of the integral where one would be able to see the answer is an integer (the diameter of a unit circle) without having to do the integral.

5. Harikumar.P says:

For unit circle the area is pi but when integrate to find the area under sine curve it become 4. How it become changed ?

• girlsangle says:

Right, the area of a unit circle is pi. The sine function does give the height above the horizontal axis of points on the unit circle, but its argument is the angle that the radial line to that point makes with the positive horizontal axis, and so when you integrate sin t with respect to t, you’re not actually computing the area of the unit circle. If you want to express the area of the unit circle as an integral with respect to the horizontal coordinate, you would need to integrate over the function that gives the height of each point on the circle as a function of the horizontal coordinate (and then double the result to get the full circle area and not just the semicircular area), and not the angle. With respect to the horizontal coordinate x, the appropriate function would be

f(x) = (1 – x^2)^0.5.

• Harikumar.P says:

Thank you for clearing my misunderstanding between integration with respect to dx and d theta..Can we can able to find the area of the unit circle in terms of integration with respect to d theta in any other way

• girlsangle says:

Yes. What you have to do is look at the section of the area of the circle that is selected between two radial lines that are separated by a very small angle $\Delta t$. To first order in $\Delta t$, that area is $R^2/2 \Delta t$, where $R$ is the radius of the circle. We add these values up and take the limit as $\Delta t$ tends to 0 and this becomes $\int R^2/2 dt$. In general, if you can parametrize a loop about the origin in polar coordinates by $r = f(\theta)$, then the area is $\int f(t)^2/2 d\theta$. This paper by by Kiryl Tsishchanka explains it in more detail: https://cims.nyu.edu/~kiryl/Calculus/Section_9.4–Areas_and_Lengths_in_Polar_Coordinates/Areas_and_Lengths_in_Polar_Coordinates.pdf

• Harikumar.P says:

Vary much thanks for the reply as well as the provided link.- it is very useful.I am even now in confusion because If we cut the unit circle into two parts and placed side by side above and below the x-axis is it not formed a sine curve ? if that is the case then what is the total area under the sine curve with respect to d theta ?

• girlsangle says:

That curve is not the sine curve. If you look just the half of the circle above the x-axis, what you are looking at is essentially the graph of the function that gives the height of points on a circle as a function of their x-coordinates. However, the graph of the sine function gives the height of the points on the circle as a function of the angle, which, in this case, is the same as the arclength between the rightmost point of the circle and the point you are taking the sine of. I’ve attached a picture that shows the two circle halves and the graph of the sine function superimposed.

6. Harikumar.P says:

Very Very much thanks for the Clear clarification .and the picture attached showing the difference between the two.Once again thanks you.

7. ttweddi says:

How to tweak this setup to show indefinite integral of sin is cos? Can you please reply? This question kills my brain.

• girlsangle says:

Well, technically, the indefinite integral of sine is the negative of cosine (plus a constant). If that’s what you mean, thank you for the correction! I’ve fixed it. But if you’re meaning that you also don’t see how to show that, here’s what you can do: Instead of allowing the ball to travel halfway around the circle, have it travel only to the point that makes an angle $\alpha$ with the positive horizontal axis. The distance the shadow travels is now given by the integral of the sine function from 0 to $\alpha$, but this distance can also be seen directly from the corresponding figure to be $1 - \cos \alpha$. (If $\alpha$ is not between 0 and $\pi$, then you have to account for signs with negative speeds corresponding to movement to the right.)

8. William Rose says:

This is amazing. I modified your argument slightly to avoid using speed/distance. See here:

9. John G. Aiken, PhD in mathematics, LSU, Baton Rouge, 1972. says:

I spent 90 minutes searching the web for how to compute the area of the sine curve from 0 to pi. Nothing even close until I found this page. Thank you!

10. Mij says:

There are so many relationships that aren’t intuitive. To a lot of people, it’s not intuitive that the circumference of a circle divided by the diameter will always be the same value, and an American state governor, finding Pi an inconvenient number, once passed a by-law that in his state, the circumference of a circle would be 3 x the diameter. I have always found it amazing that the area of a sphere is the same as the area of a cylinder that the sphere would stand in (one whose diameter and height are equal to the diameter of the sphere).

• girlsangle says:

It is fascinating that though mathematical theorems are tautologies, they can be startling, surprising, unintuitive, and even unbelievable. I think it’s a great goal to think and think until such mathematical facts become “obvious”. Intuition can be changed through contemplation and practice.