integral- area?

Question

An area is difined by curve y = sin(2x) and x-axel. Calculate size of the area between 0 and 3pi/4

choose the right alternativeb. 0,5 a.e.

c. 1,5 a.e.

d. 2,5 a.e.

e. 3,5 a.e.

Answer 1

Integral is in 2 parts, A1 and A2.
A1 = ∫ sin 2x dx from x = 0 to x = π/2
A2 = ∫ sin 2x dx from x = π/2 to x = 3π/4

A1 = - (1/2) cos 2x between 0 and π/2
A1 = - (1/2) [ cos π - cos 0 ]
A1 = (-1/2) (- 1 - 1)
A1 = 1

A2 = - (1/2) cos 2x between π/2 and 3π/4
A2 = (-1/2) (- cos π + cos 3π/4)
A2 = (-1/2) (1 + 1/√2 )
A2 = - 1/2 - 1 / (2√2)

Total area(numerical) = A1 + A2
= 1 + 1/2 + 1 / (2√2)
= 3/2 + 1/(2√2)

Answer 2

Integrate between 0 and 3π/4.

∫ sin(2x) = [-cos(2x) / 2]

Area = [-cos(2*3π/4) / 2] - [-cos(2*0) / 2]
=-1/2 * [cos(6π/4) - cos(0)]
=-1/2 * [0.9966196... - 1]
= 0.00338035...

The area above the x-axis is positive, the area below is negative. If you integrated between 0 and 2π, the answer would be zero.

Answer 3

integrate with limits as x=0 to x=3pi/4

Answer 4

~(thats my integral sign)

~(from 0 to 3pi/4) of ((sin (2x))-x)

=(-1/2 cos (2x)-((x^2)/2))) evaluated from 0 to 3pi/4

As my book would say "We leave the details of the integration as an exercise." =)