This is essentially the same as finding the area of the curve between f(x) = sin(2x) and g(x) = 0 (because doing it in this fashion finds even the "negative" areas).
To find the area between f(x) = sin(2x) and g(x) = 0, we must first determine where they intersect, on the interval 0 to 2pi. This is done by equating them to each other.
f(x) = g(x)
sin(2x) = 0
2x = { 0, pi, 2pi, 3pi }
x = { 0, pi/2, pi, 3pi/2 }
So we have four intersections, which means we will have three different integrals to add up (one from 0 to pi/2, one from pi/2 to pi, one from pi to 3pi/2). On each interval, one curve will be higher than the other. All we have to do is test a single value in each region, and see which of f(x) or g(x) is higher on the interval. After all, the area A is equal to
A = Integral (a to b, [higher curve] - [lower curve] dx )
On the interval 0 to pi/2, test pi/4, plug into each function. Then
f(pi/4) = sin(2*pi/4) = sin(pi/2) = 1
g(pi/4) = 0,
so f(x) is higher on the curve.
A = Integral (0 to pi/2, [f(x) - g(x)] dx ) + ? + ? + ?
On the interval pi/2 to pi, test x = 3pi/4. Then
f(3pi/4) = sin(2*3pi/4) = sin(3pi/2) = -1
g(3pi/4) = 0,
so g(x) is higher.
A = Integral (0 to pi/2, [f(x) - g(x)] dx ) + Integral (pi/2 to pi, [g(x) - f(x)] dx ) + ? + ?
I'm not going to do the rest, but
from pi to 3pi/2, f(x) is higher.
From 3pi/2 to 2pi, g(x) is higher.
A = Integral (0 to pi/2, [f(x) - g(x)] dx ) + Integral (pi/2 to pi, [g(x) - f(x)] dx ) + Integral (pi to 3pi/2, [f(x) - g(x)] dx) + Integral (3pi/2 to 2pi, [g(x) - f(x)] dx )
Let's fill in the functions.
A = Integral (0 to pi/2, [sin(2x) - 0] dx ) + Integral (pi/2 to pi, [0 - sin(2x)] dx ) + Integral (pi to 3pi/2, [sin(2x) - 0] dx) + Integral (3pi/2 to 2pi, [0 - sin(2x)] dx )
A = Integral (0 to pi/2, [sin(2x)] dx ) + Integral (pi/2 to pi, [-sin(2x)] dx ) + Integral (pi to 3pi/2, [sin(2x)] dx) + Integral (3pi/2 to 2pi, [-sin(2x)] dx )
The negative sin(2x), we can factor the negative out of the integral, resulting in.
A = Integral (0 to pi/2, [sin(2x)] dx ) - Integral (pi/2 to pi, [sin(2x)] dx ) + Integral (pi to 3pi/2, [sin(2x)] dx) - Integral (3pi/2 to 2pi, [sin(2x)] dx )
Answers & Comments
Verified answer
From 0 to pi/2 : sin 2x > 0 ===> | sin 2x | = sin 2x
From pi/2 to pi : sin 2x < 0 ===> | sin 2x | = - sin 2x
From pi to 3pi/2 : sin 2x > 0 ===> |sin 2x | = sin 2x
From 3pi/2 to 2pi : sin 2x < 0 ===> | sin 2x | = - sin 2x
<=>
integral from 0 to 2pi | sin 2x | dx =
4 * integral from 0 to pi/2 sin 2x dx =
4 * [ - 1/2 * cos 2x ] from 0 to pi/2 =
4 * [ (- 1/2 * cos (2 * pi/2)) - (- 1/2 * cos (2 * 0)) ] =
4 * [ (- 1/2) * (- 1) - (- 1/2) * 1 ] =
4 * [ 1/2 + 1/2 ] =
4 * 1 =
4 area units
http://s636.photobucket.com/albums/uu88/stoupils/?...
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â« from 0 to 2pi Isin(2x)Idx
= 4 â« from 0 to (pi/2) sin(2x) dx
= 4 â« 2 sin(x) cos(x) dx (limit 0 to PI/2)
= 8 â« sin(x) d[sin(x)] (0 to PI/2)]
= 8 sin^2(x)/2
= 4 sin^2(x) [ 0 to PI/2]
= 4 sin^2(PI/2) - 4 sin^2(0)
= 4
â« (0 to 2pi, | sin(2x) | dx )
This is essentially the same as finding the area of the curve between f(x) = sin(2x) and g(x) = 0 (because doing it in this fashion finds even the "negative" areas).
To find the area between f(x) = sin(2x) and g(x) = 0, we must first determine where they intersect, on the interval 0 to 2pi. This is done by equating them to each other.
f(x) = g(x)
sin(2x) = 0
2x = { 0, pi, 2pi, 3pi }
x = { 0, pi/2, pi, 3pi/2 }
So we have four intersections, which means we will have three different integrals to add up (one from 0 to pi/2, one from pi/2 to pi, one from pi to 3pi/2). On each interval, one curve will be higher than the other. All we have to do is test a single value in each region, and see which of f(x) or g(x) is higher on the interval. After all, the area A is equal to
A = Integral (a to b, [higher curve] - [lower curve] dx )
On the interval 0 to pi/2, test pi/4, plug into each function. Then
f(pi/4) = sin(2*pi/4) = sin(pi/2) = 1
g(pi/4) = 0,
so f(x) is higher on the curve.
A = Integral (0 to pi/2, [f(x) - g(x)] dx ) + ? + ? + ?
On the interval pi/2 to pi, test x = 3pi/4. Then
f(3pi/4) = sin(2*3pi/4) = sin(3pi/2) = -1
g(3pi/4) = 0,
so g(x) is higher.
A = Integral (0 to pi/2, [f(x) - g(x)] dx ) + Integral (pi/2 to pi, [g(x) - f(x)] dx ) + ? + ?
I'm not going to do the rest, but
from pi to 3pi/2, f(x) is higher.
From 3pi/2 to 2pi, g(x) is higher.
A = Integral (0 to pi/2, [f(x) - g(x)] dx ) + Integral (pi/2 to pi, [g(x) - f(x)] dx ) + Integral (pi to 3pi/2, [f(x) - g(x)] dx) + Integral (3pi/2 to 2pi, [g(x) - f(x)] dx )
Let's fill in the functions.
A = Integral (0 to pi/2, [sin(2x) - 0] dx ) + Integral (pi/2 to pi, [0 - sin(2x)] dx ) + Integral (pi to 3pi/2, [sin(2x) - 0] dx) + Integral (3pi/2 to 2pi, [0 - sin(2x)] dx )
A = Integral (0 to pi/2, [sin(2x)] dx ) + Integral (pi/2 to pi, [-sin(2x)] dx ) + Integral (pi to 3pi/2, [sin(2x)] dx) + Integral (3pi/2 to 2pi, [-sin(2x)] dx )
The negative sin(2x), we can factor the negative out of the integral, resulting in.
A = Integral (0 to pi/2, [sin(2x)] dx ) - Integral (pi/2 to pi, [sin(2x)] dx ) + Integral (pi to 3pi/2, [sin(2x)] dx) - Integral (3pi/2 to 2pi, [sin(2x)] dx )
Leading to the exact same integral in each.
The integral of sin(2x) is (-1/2)cos(2x).
A = [(-1/2)cos(2x)] {evaluated from 0 to pi/2} -
[(-1/2)cos(2x)] {evaluated from pi/2 to pi} +
[(-1/2)cos(2x)] {evaluated from pi to 3pi/2} -
[(-1/2)cos(2x)] {evaluated from 3pi/2 to 2pi}
A = [(-1/2)cos(2*pi/2) - (-1/2)cos(0)] -
[(-1/2)cos(2*pi) - (-1/2)cos(2*pi/2)] +
[(-1/2)cos(2*3pi/2) - (-1/2)cos(2*pi)] -
[(-1/2)cos(2*2pi) - (-1/2)cos(2*3pi/2)]
A = [(-1/2)cos(pi) - (-1/2)(1)] -
[(-1/2)cos(2pi) - (-1/2)cos(pi)] +
[(-1/2)cos(3pi) - (-1/2)cos(2pi)] -
[(-1/2)cos(4pi) - (-1/2)cos(3pi)]
A = [(-1/2)(-1) + (1/2)(1)] -
[(-1/2)(1) + (1/2)(-1)] +
[(-1/2)(-1) - (-1/2)(1)] -
[(-1/2)(1) - (-1/2)(-1)]
A = [(1/2) + (1/2)] - [(-1/2) + (-1/2)] + [(1/2) + (1/2)] - [(-1/2) + (-1/2)]
A = 1 - (-1) + 1 - (-1)
A = 1 + 1 + 1 + 1
A = 4