Exams Joint Entrance Examination Mathematics Definite Integration Question 237 Question 237 of 246 Medium MCQ
Let f f f be a real valued continuous function defined on the positive real axis such that g ( x ) = ∫ 0 x t f ( t ) d t g(x)=\int\limits_0^x t f(t) d t g ( x ) = 0 ∫ x t f ( t ) d t . If g ( x 3 ) = x 6 + x 7 g\left(x^3\right)=x^6+x^7 g ( x 3 ) = x 6 + x 7 , then value of ∑ r = 1 15 f ( r 3 ) \sum\limits_{r=1}^{15} f\left(r^3\right) r = 1 ∑ 15 f ( r 3 ) is :
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\[ g ( x ) = x 2 + x 7 3 g ′ ( x ) = 2 x + 7 3 x 4 3 f ( x ) = g ′ ( x ) x f ( x ) = 2 + 7 3 x 1 3 f ( r 3 ) = 2 + 7 r 3 ∑ r = 1 15 ( 1 + 7 3 r ) = 310 \begin{aligned}
& g(x)=x 2+x^{\frac{7}{3}} \\
& g^{\prime}(x)=2 x+\frac{7}{3} x^{\frac{4}{3}} \\
& f(x)=\frac{g^{\prime}(x)}{x} \\
& f(x)=2+\frac{7}{3} x^{\frac{1}{3}} \\
& f\left(r^3\right)=2+\frac{7 r}{3} \\
& \sum_{r=1}^{15}\left(1+\frac{7}{3} r\right)=310
\end{aligned} g ( x ) = x 2 + x 3 7 g ′ ( x ) = 2 x + 3 7 x 3 4 f ( x ) = x g ′ ( x ) f ( x ) = 2 + 3 7 x 3 1 f ( r 3 ) = 2 + 3 7 r r = 1 ∑ 15 ( 1 + 3 7 r ) = 310 \] Answer the question to unlock