Find f (x), if f '(x) = 7/(1 − x^2)^0.5, f(1/2)= 1?

Question

please help!!!!!!!!!!!!!!!!!!!!

Answer 1

f ' (x) = 7 / (1-x^2)^0.5

[ Note : 0.5 = 1 / 2  ,A^0.5 = A^1/2  = sqrt A ]

f ' (x) = 7 / sqrt ( 1-x^2)

Apply integral each side

integral f ' (x) = intergal [ 7 / sqrt ( 1-x^2) ]

[ Note : integral f ' (x) = f (x) ]  , 

f (x) = intergal [ 7 / sqrt ( 1-x^2) ]

f (x) = 7 intergal [ 1 / sqrt ( 1-x^2) ]

[ Note : intergal [ 1 / sqrt ( 1-x^2) ] = sin^-1 x +c ]

f (x) = 7 (sin^-1 x) + c -------------- (1)

substitute x = 1/2 in the equation (1)

f (1/2) = 7 (sin^-1 (1/2)) + c

Given that f( 1/2 ) = 1

1 = 7 (sin^-1 (sin30)) + c                   sin 30 = 1/2

Note : (sin^-1 (sinA)) = A

1 = 7 (pi / 6) + c

Subtract 7 (pi / 6) from each side.

1 - 7 (pi / 6) = 7 (pi / 6) + c - 7 (pi / 6)

Simplify

1 - 7 (pi / 6) = c

c = 1 - 7 (pi / 6)

c = (6 - 7 pi) / 6                here  1 , 6 LCM in 6

There fore

c = (6 - 7 pi) / 6

f (x) = 7 (sin^-1 x) + c

Comments

good explanation :)