`int_0^(pi/6) (a sin (x) + b cos(x))\ dx` is equal to
- `((2 - sqrt 3)a - b)/2`
- `(b - (2 - sqrt 3) a)/2`
- `((2 - sqrt 3)a + b)/2`
- `((2 - sqrt 3) b - a)/2`
- `((2 - sqrt 3) b + a)/2`
Aussie Maths & Science Teachers: Save your time with SmarterEd
`int_0^(pi/6) (a sin (x) + b cos(x))\ dx` is equal to
`C`
`int_0^(pi/6) (a sin (x) + b cos(x))\ dx`
`= [-a cos(x) + b sin(x)]_0^(pi/6)`
`= [-a ⋅ sqrt 3/2 + b/2 – (-a + 0)]`
`= (2a – sqrt 3 a + b)/2`
`= ((2 – sqrt 3) a + b)/2`
`=> C`
Evaluate `int_0^(pi/2) sin (x/2)\ dx`. (3 marks)
`2\ – sqrt2`
`int_0^(pi/2) sin (x/2)\ dx`
`= [-2cos (x/2)]_0^(pi/2)`
`= -2 [ cos (pi/4) – cos 0]`
`= -2 [ 1/sqrt2 – 1]`
`= -2/sqrt2 + 2`
`= 2 – sqrt2`
If `f′(x) = 2cos(x) - sin(2x)` and `f(pi/2) = 1/2`, find `f(x)`. (3 marks)
`2sinx + 1/2cos(2x) – 1`
`f(x)` | `= int(2cosx – sin2x)dx` |
`= 2sinx + 1/2cos(2x) + c` |
`text(Substitute)\ \ f(pi/2) = 1/2:`
`1/2` | `= 2sin(pi/2) + 1/2cos(pi) + c` |
`1/2` | `= 2 – 1/2 + c` |
`c` | `= −1` |
`:. f(x)` | `= 2sinx + 1/2cos(2x) – 1` |