Calculus find the area of the region bounded by y = sin 2x and y = cos x for (-pi/2)<=x<=pi/6 my sol'n is 2.25

Answer 1

A = ∫(cos x - sin 2x) dx between lims -π/2 to π/6
A = [ sin x + (1/2).cos 2x ] (limits as given)
A = [(sin π/6 +(1/2).cos π/3] - [sin(- π/2) +(1/2).cos (-π)]
A = [1/2 + 1/4] - [- 1 - (1/2) ]
A = 3 / 4 + 3 / 2
A = 9/4
A = 2.25

Answer 2

Integrate (sin 2x - cos x) dx
=
-1/2 cos 2x - sin x

Now apply the limits from -pi/2 to 0, take the modulus of the answer, then add it to the answer you get when you apply the limits from 0 ti pi/6.

You need to do this, because the area to the left of the origin is below the x-axis.

Answer 3

if the area you're bearing on is the 1st area sure by using those curves, first discover the x values the place the curves first intersect sin2x = cosx 2sinxcosx = cosx 2sinxcosx - cosx = 0 cosx(2sinx - a million) = 0 cosx = 0 x = pi/2 [first fee] 2sinx - a million = 0 sinx = a million/2 x = pi/6 [first fee] consequently, indispensable is between pi/2 and pi/6 A = int(sin2x)dx - int(cosx)dx = -(a million/2)cos(2x) - sinx + C substitute in values for particular indispensable A = (-(a million/2)cos(pi) - sin(pi/2)) - (-(a million/2)cos(pi/3) - sin(pi/6)) = (a million/2 - a million) - (-a million/4 - a million/2) = 0.25 sq. units

Answer 4

Area between f(x) = sin(2x) and g(x) = cos(x), for

-pi/2 <= x <= pi/6

First, we equate f(x) to g(x) to find any points of intersection between -pi/2 and pi/6.

sin(2x) = cos(x)

By the double angle identity,

2sin(x)cos(x) = cos(x)
2sin(x)cos(x) - cos(x) = 0
cos(x) [ 2sin(x) - 1 ] = 0

cos(x) = 0
2sin(x) - 1 = 0

We care about solutions between -pi/2 and pi/6.

cos(x) = 0 when x = -pi/2.
2sin(x) - 1 = 0 when sin(x) = 1/2, which occurs at pi/6.

Since there aren't any points of intersection between -pi/2 and pi/6, all we have to do is determine the greater function, since

A = Integral (-pi/2 to pi/6, higher curve minus lower curve dx)

Test f(0): f(0) = sin(2*0) = sin(0) = 0
g(0) = cos(0) = 1

Therefore, g(x) is greater, and our area is

A = Integral ( -pi/2 to pi/6, [ g(x) - f(x) ] dx )

A = Integral ( -pi/2 to pi/6, [ cos(x) - sin(2x) ] dx )

A = [ sin(x) + (1/2)cos(2x) ] {evaluated from -pi/2 to pi/6}

A = [ sin(pi/6) + (1/2)cos(pi/3)] - [sin(-pi/2) + (1/2)cos(-pi/2)]

A = [ (1/2) + (1/2)(1/2) ] - [ -1 + (1/2)(0) ]

A = [ (1/2) + (1/4) ] - [-1 ]

A = [ (2/4) + (1/4) ] + (1)
A = [3/4] + 4/4
A = 7/4 = 1.75

Answer 5

wrong its 2. 025