Let n be a positive integer and b > n − 1 , f ( x ) = a x b and f n ( x ) = f − 1 ( x ) , where f n ( x ) is the n th derivative of f ( x ) .
(1) Find the area of the region bounded by f ( x ) and f − 1 ( x ) as a function of n .
(2) Using n = 3 : If the area A of the region R above bounded by f ( x ) and f − 1 ( x ) = f 3 ( x ) can be expressed as A = ( b − c a ) ∗ ( a d a + d b ) d a ∗ ( b + d b + c ) , where a , b , c and d are coprime positive integers, find a + b + c + d .
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f ( x ) = a x b ⟹ f n ( x ) = a m x b − n where m = b ( b − 1 ) ( b − 2 ) ∗ ∗ ∗ ( b − n + 1 )
and f − 1 ( x ) = a b 1 1 x b 1
For x b − n = x b 1 ⟹ b − n = b 1 ⟹ b = 2 n + n 2 + 4 for b > n − 1 .
For a m = a b 1 1 = a b − n 1 ⟹ a b − n + 1 m = 1 ⟹ a = ( m 1 ) b − n + 1 1
⟹ f ( x ) = ( m 1 ) b − n + 1 1 ∗ x b
and
f n ( x ) = m b − n + 1 b − n x b − n = f − 1 ( x )
f ( x ) = f − 1 ( x ) ⟹ x b − n ( m b − n + 1 b − n − ( m 1 ) b − n + 1 1 x n )
x = 0 and for x ≥ 0 ⟹ x = m n 1 for case n even.
⟹ The area A = ∫ 0 m n 1 f − 1 ( x ) − f ( x ) d x = ( b − n + 1 1 ) m n 2 − ( b ( b − n + 1 ) 1 ) m n 2
= ( b − n + 1 1 ) ( m n 2 ) ( b b − 1 ) =
( b − n + 1 1 ) ( b ( b − 1 ) ( b − 2 ) ∗ ∗ ∗ ( b − n + 1 ) ) n 2 ( b b − 1 )
Note: You could have used the Gamma function if so desired.
For n = 3 we have:
b = 2 3 + 1 3 ⟹ The area A = ( 1 3 − 1 2 ) ( 2 9 + 3 1 3 ) 1 3 2 ( 1 3 + 3 1 3 + 1 ) =
( 1 3 − 1 2 ) ∗ ( 2 3 2 + 3 1 3 ) 3 2 ∗ ( 1 3 + 3 1 3 + 1 ) =
( b − c a ) ∗ ( a d a + d b ) d a ∗ ( b + d b + c ) ⟹ a + b + c + d = 1 9 .