Question

Question #a297a

Answer 1

The area you're searching is nothing but the following integral:

`\int_0^\pi 2 \sin (x) + \sin (2x)\ dx`

The integral of the sum is the sum of the integrals:

`\int_0^\pi 2 sin (x)\ dx+ \int_0^\pi \sin(2x)\ dx`

We can factor out constant:

2 `\int_0^\pi sin (x)\ dx+ \int_0^\pi \sin(2x)\ dx`

For the first integral, we have that since `d/{dx} (-\cos(x)==\sin(x)`, then `\int \sin(x)=-\cos(x)`. The second integral is quite similar: since `d/{dx} (-\cos(2x))=2\sin(2x)`, we have that `\int \sin(2x)=-1/2 \cos(2x)`

Evaluating the integral, we have

`-2[\cos(x)]_0^\pi -1/2 [\cos(2x)]_0^\pi`, which equals to
`-2[\cos(\pi)-\cos(0)] -1/2[\cos(2\pi)-\cos(0)]`
and since `\cos(0)=\cos(2\pi)=1`, `\cos(\pi)=-1`, the result is 4.

Two little post scriptums:

1. You get the "greater or equal" simbol by writing the command \ge, written among #
2. You could have observed that the second integral was zero, because `\sin(2x)` is a "squeezed" version of the sine function, and it has period `\pi` instead of `2\pi`, so you were integrating the function over its period, and the result is zero, just like when you integrate sine or cosine functions from `0` to `2\pi`