Let f be a real valued continuous function defined on the positive real axis such that g ( x ) = ∫ 0 x t f ( t ) dt . If g ( x 3 ) = x 6 + x 7 , then value of ∑ r = 1 15 f ( r 3 ) is :

Question

Let f be a real valued continuous function defined on the positive real axis such that g ( x ) = ∫ 0 x ​ t f ( t ) dt . If g ( x 3 ) = x 6 + x 7 , then value of ∑ r = 1 15 ​ f ( r 3 ) is :

TestHub · Accepted Answer

g ( x ) = x 2 + x 7/3 ​ g ′ ( x ) = 2 x + 3 7 ​ x 4/3 f ( x ) = x g ′ ( x ) ​ f ( x ) = 2 + 3 7 ​ x 1/3 f ( r 3 ) = 2 + 3 7 ​ r r = 1 ∑ 15 ​ ( 2 + 3 7 ​ r ) = 2 ( 15 ) + 3 7 ​ ( 2 15 ( 16 ) ​ ) ​ = 310

Mathematics - Definite Integration Question with Solution | TestHub

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