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(D) Given $g(x^3) = x^6 + x^7$. Let $u = x^3$,then $x = u^{1/3}$.Substituting this,we get $g(u) = (u^{1/3})^6 + (u^{1/3})^7 = u^2 + u^{7/3}$.By the Fu

Vedclass · Accepted Answer

$310$

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(D) Given $g(x^3) = x^6 + x^7$. Let $u = x^3$,then $x = u^{1/3}$.Substituting this,we get $g(u) = (u^{1/3})^6 + (u^{1/3})^7 = u^2 + u^{7/3}$.By the Fundamental Theorem of Calculus,$g'(x) = x f(x)$,so $f(x) = \frac{g'(x)}{x}$.Differentiating $g(x) = x^2 + x^{7/3}$ with respect to $x$,we get $g'(x) = 2x + \frac{7}{3}x^{4/3}$.Thus,$f(x) = \frac{2x + \frac{7}{3}x^{4/3}}{x} = 2 + \frac{7}{3}x^{1/3}$.Now,$f(r^3) = 2 + \frac{7}{3}(r^3)^{1/3} = 2 + \frac{7}{3}r$.We need to calculate $\sum_{r=1}^{15} f(r^3) = \sum_{r=1}^{15} (2 + \frac{7}{3}r) = \sum_{r=1}^{15} 2 + \frac{7}{3} \sum_{r=1}^{15} r$.$= (2 	imes 15) + \frac{7}{3} 	imes \frac{15 	imes 16}{2} = 30 + \frac{7}{3} 	imes 120 = 30 + 7 	imes 40 = 30 + 280 = 310$.

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 the value of $\sum_{r=1}^{15} f(r^3)$ is:

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