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Find the area enclosed by the given curves.y=-sin(x), y=sin(2x), 0<- x <- pi ...
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1) It is important for both functions what is the domain of definition.
0<- x <- pi => `0 gt x gt pi`
Notice that x covers the third and the fourth quadrants, therefore the function sine is negative and the function cosine is negative between `pi` and `(3pi)/2 ` or positive between `(3pi)/2` and `2pi` =0.
The area enclosed by the curves y=sin and y=sin 2x is the difference between them. Since the area is positive, you need to know if sin x is larger then sin 2x or sin 2x is larger than sin x.
Take a value of x between `pi` and `(3pi)/2` .
Put `x = pi+ pi/6=gt sin(7pi)/6 = sin(pi+ pi/6) = -sin (pi/6) = -1/2` cos (pi + pi/6) = `-cos (pi/6) = -sqrt 3/2`
`sin 2x = 2 sin x*cos x =gt sin 2*(7pi)/6 = (2sqrt 3)/4 = sqrt 3/2 = sin pi/3`
These values of sin x and sin 2x show that sin x < sin 2x, if x in `[pi;(3pi)/2].`
Area = `int_pi^(3pi/2) (sin 2x - sin x) dx = int_pi^(3pi/2)sin 2x dx - int_pi^(3pi/2) sin x dx`
`Area = - (cos 2x)/2 + cos x = (-(cos 3pi)/2 + (cos pi)/2) + cos (3pi /2) - cos pi`
Area = `1 - 1/2 +0 + 1 = 3/2`
Take a value of x between `(3pi)/2` and `2pi` . Put `x = (3pi)/2+ pi/6=gt sin(10pi)/6 = sin((3pi)/2+ pi/6) = -cos (pi/6) = -sqrt 3/2`
`cos ((3pi)/2+ pi/6) =sin (pi/6) = 1/2`
`sin 2x = 2 sin x*cos x = -sqrt 3/2`
` x = (3pi)/2+ pi/4 = (7pi)/4 =gt sin ((3pi)/2+ pi/3) = -cos pi/3= -1/2`
`cos ((3pi)/2+ pi/3) = sqrt3/2`
`` `sin 2x = -sqrt3/2`
All these values of sin x and sin 2x show that sin 2x < sin x, if `x in [pi;(3pi)/2].`
Area =`int_(3pi/2)^(2pi) (sin x - sin 2x) dx = int_(3pi/2)^(2pi) sin x dx - int_(3pi/2)^(2pi) sin 2x dx`
`Area = - cos 2pi + cos (3pi)/2 + (cos 4pi)/2 - (cos 3pi)/2`
`Area = -1 + 0 + 1/2+1/2 = 0`
ANSWER: The area enclosed by the curves sin x and sin 2x is `A = 3/2.`
Posted by sciencesolve on December 1, 2011 at 10:56 PM (Answer #1)
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