Let I(x) = 6 2 x(1 - x)2 dx. If I(0) = 3, then I ( 12 ) is equal to
- 3
- 2 3
- 6 3
- 3 3
Answer
(4) 3 3
Solution
1 2 x = cosec2 x, the integral is I(x) = 6 cosec2 x(1 - x)2 dx
Put 1 - x = t, so cosec2 x dx = dt
I = 6t2 dt = - 6t + c = - 61 - x + c = 6 x x - x + c
I(0) = 0 + c = 3, so c = 3
12 = 2 + 3, we get 1 - 12 = -(1 + 3)
I ( 12 ) = 6 3 + 1 + 3 = 3( 3 - 1) + 3 = 3 3
Answer: (4)