Let $f: [0, 2 ] [0,1]$ be the function defined by $f(x)= ^2 x$ and let $g: [0, 2 ] [0, )$ be the function defined by $g(x)= x 2 -x^2 $. The value of 16 ^3 _0^ 2 f(x) g(x) d x is

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Correct answer: A. A

Now I_1= _0^ 2 f(x) g(x) d x= 1 2 _0^ 2 g(x) d x (it is explained in previous question solution) i.e. 1 2 _0^ 2 ( 4 )^2- (x- 4 )^2 dx Using a^2-x^2 = 1 2 (x a^2-x^2 +a^2 ^ -1 ( x a ) )+C 1 2 [ (x- 4 ) 2 x 2 -x^2 + ^2 2 2 ^ -1 ( x- 4 4 ) ]_0^ / 2 aligned & 1 2 [ (0+ ^3 64 )- (0+ ( - ^3 64 ) ) ] \ & 1 2 ^3 32 aligned Now 16 ^3 ^3 64 = 1 4 =0.25

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