Evaluate the definite integral.

Question

Evaluate the definite integral.

∫π/2 0 sin(2t)dt

π/2 is upper while 0 is lower if you know what I am saying

Answer 1

Hi Jay!

You can utilize u substitution on this.

du du

let u = 2t, then --- = 2 or --- = dt

dt 2

π

--

2 du

∫sin(u) ---

0 2

π

--

1 2

-- ∫ sin(u)du

2 0

π

t= ----

1 2

-- [ -cos(u)]

2 t=0

π

--

1 2

- ---[cos(2t)]

2 0

1

- ---[cos(2*π/2) - cos(2*0)]

2

1

- ---[cos(π) - cos(0)]

2

1

-- ---[-1 - 1]

2

1

-- -- [-2]

2

1