Question

What is the integral of sin(x) dx from 0 to 2pi?

Answer 1

Using the definition of the integral and the fact that `sinx` is an odd function, from `0` to `2pi`, with equal area under the curve at `[0, pi]` and above the curve at `[pi, 2pi]`, the integral is `0`.

This holds true for any time `sinx` is evaluated with an integral across a domain where it is symmetrically above and below the x-axis.

`int_0^(2pi) sinxdx = [-cosx]|_(0)^(2pi)`

`= -cos2pi - (-cos0)`

`= -1 - (- 1) = 0`

Answer 2

An alternative way to do this starting from the limit definition is:

`int_(a)^(b) f(x)dx = lim_(n->oo) sum_(i=1)^N f(x_i^"*")Deltax`

where:

• `n` is the number of rectangles used to approximate the integral, i.e. the area between the curve and the x-axis.
• `i` is the index of each rectangle in `[0,2pi]`.
• `N` is the index of the final rectangle in `[0,2pi]`.
• `f(x_i^"*")` is the height of each given rectangle in `[0,2pi]`, which varies as `sin(x)`.
• `Deltax` is the width of each given rectangle in `[0,2pi]`, which converges to `0` as `n->oo`.

If we use the midpoint-rectangular approximation method (MRAM), we choose a convenient interval `Deltax` such that we can find a midpoint for each rectangle of dimension `Deltax xx f(x_i^"*")`, where the midpoint of the `i`th rectangle is defined as

`M_i = x_(i-1)+(x_i - x_(i-1))/2`.

Let us choose `Deltax = pi/2` such that `x = {0,pi/2,pi,(3pi)/2,2pi}` for `[0,2pi]` and `n = {1,2,3,4}`.

Then each rectangle's width is `pi/2`, and:

• Rectangle `1` spans `[0,pi/2]`.
• Rectangle `2` spans `[pi/2,pi]`.
• Rectangle `3` spans `[pi,(3pi)/2]`.
• Rectangle `4` spans `[(3pi)/2,2pi]`.

Each corresponds to an `f(x_i^"*")` that gives you the height of the `i`th rectangle as

`f(x_1^"*") ~~ f(M_1) = sin(x_0+(x_1-x_0)/2) = sin(pi/4) = sqrt2/2,`

`f(x_2^"*") ~~ f(M_2) = sin(x_1+(x_2-x_1)/2) = sin((3pi)/4) = sqrt2/2,`

`f(x_3^"*") ~~ f(M_3) = sin(x_2+(x_3-x_2)/2) = sin((5pi)/4) = -sqrt2/2,`

`f(x_4^"*") ~~ f(M_4) = sin(x_3+(x_4-x_3)/2) = sin((7pi)/4) = -sqrt2/2.`

In the end, what you get from MRAM is the following result:

`color(blue)(int_(0)^(2pi) sin(x)dx ~~ lim_(n->4) sum_(i=1)^4 sin(M_i)Deltax)`

`= (sin(pi/4) + sin((3pi)/4) + sin((5pi)/4) + sin((7pi)/4))*Deltax`

`= (sqrt2/2 + sqrt2/2 - sqrt2/2 - sqrt2/2) * pi/2`

`= color(blue)(0)`

Which is not surprising given that `sin(x)` in `[0,pi]` is equal to `-sin(x)` in `[pi,2pi]`, which means that

`int_(0)^(pi) sinxdx = -int_(pi)^(2pi) sinxdx`,

and thus:

`color(green)(int_(0)^(2pi) sinxdx)`

`= int_(0)^(pi) sinxdx + int_(pi)^(2pi) sinxdx`

`= color(green)(0)`,

regardless of the chosen method.

Note that RRAM, LRAM, and MRAM are approximations, so it was a coincidence that it gave the exact answer.