What is the area between y=sin2x and y=cosx?

What is the area between y=sin2x and y=cosx for -pi/2 </= x </= pi/6 ?
asked Jan 19, 2013 in CALCULUS

+1 vote

Given that

y = sin(2x) and y = cos(x) for  -π/2 < x < π/6

sin2x will have double the frequency than a normal sine and we shall first find out the intersection points between the

2curves

sin(2x) = cos(x)

Note : sin(2A) = 2sin(A)cos(A)

2sin(x)cos(x) = cos(x)

Divide each side by cosx

(2sin(x)cos(x)) / (cos(x)) = cos(x) / cos(x).

Simplify

2sin(x) = 1

Divide each side by 2.

(2sin(x)) / 2 = 1 / 2

Simplify

sin(x) = 1 / 2

sin(x) = sin(30)            ( sin 30 = 1/2 )

Cancel common terms.

x = 30 = pi/6

Now our region of area can be divided into 2 parts.

First that rises from 0 - π/6 along the sine curve and second along the cosine curve from π/6 - π/2

hence, total area (A) is

π/6            π/2
A = ∫sin(2x) dx + ∫cos(x) dx
0                π/6

π/6      π/2
A = -2[cos2x] + [sinx]                      [ ∫sin(2x)dx = -2cos(2x)  , ∫cos(x)dx = sin(x) ]
0          π/6

A = -2[cosπ/3 - cos0] + [sinπ/2 - sinπ/6]

Note : [ cosπ/3 = 1/2 = 0.5 , cos0 = 1  ,  sinπ/2 = 1  ,  sinπ/6 = 1/2 = 0.5 ]

A = -2[0.5 - 1] + [1 - 0.5]

Simplify

A = (-2)(0.5)-(-2)(1)+(1)-(0.5)

A = -1 + 2 + 1 - 0.5

A = 3 - 1.5

A = 1.5

There fore area A = 1.5

Area bounded between the two curves is 2.25 square units.

Let y = f (x ) = sin2x and y = g (x ) = cosx.

Formula for area bounded between the curves is as follows:

or

To find which function is greater, substiute x  value between the interval [-π/2 , π/6]

Let us take x = 0.

f (x ) = sin2(0) = sin(0) = 0.

g (x ) = cos(0) = 1.

Therefore g (x ) > f (x ).

[Since ʃsinax = - 1/a cosax ]

Therefore area bounded between the two curves is 2.25 square units.