Let f be a real valued continuous function defined on the positive real axis such that g(x)=∫0x tf(t)dt. If gx3=x6+x7, then value of ∑r=115 fr3 is:

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

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270

310

320

340

answer is D.

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Detailed Solution

$\begin{array}{l} g (x) = x 2 + x^{\frac{7}{3}} \\ g^{'} (x) = 2 x + \frac{7}{3} x^{\frac{4}{3}} \\ f (x) = \frac{g^{'} (x)}{x} \\ f (x) = 2 + \frac{7}{3} x^{\frac{1}{3}} \\ f (r^{3}) = 2 + \frac{7 r}{3} \\ \sum_{r = 1}^{15} (1 + \frac{7}{3} r) = 310 \end{array}$