### Related Formula Newton-Leibnitz Theorem for differentiation under integral sign: fracddx left( int_0^x tf(t) dt right) = xf(x)$\frac{d}{dx} \left( \int_{0}^{x} tf(t) dt \right) = xf(x)$ ### Core Logic Given: g(x) = int_0^x tf(t) dt$g(x) = \int_{0}^{x} tf(t) dt$ Differentiating both sides with respect to x$x$: g'(x) = xf(x) implies f(x) = fracg'(x)x$g'(x) = xf(x) \implies f(x) = \frac{g'(x)}{x}$ We are given g(x^3) = x^6 + x^7$g(x^3) = x^6 + x^7$. Let y = x^3 implies x = y^1/3$y = x^3 \implies x = y^{1/3}$. Substituting this into the expression for g$g$: g(y) = (y^1/3)^6 + (y^1/3)^7 = y^2 + y^7/3$g(y) = (y^{1/3})^6 + (y^{1/3})^7 = y^2 + y^{7/3}$ Thus, replacing y$y$ back with x$x$: g(x) = x^2 + x^7/3$g(x) = x^2 + x^{7/3}$ ### Step 1: Differentiate g(x) to find f(x) g'(x) = 2x + frac73x^4/3$g'(x) = 2x + \frac{7}{3}x^{4/3}$ Now find f(x)$f(x)$: f(x) = fracg'(x)x = frac2x + frac73x^4/3x = 2 + frac73x^1/3$f(x) = \frac{g'(x)}{x} = \frac{2x + \frac{7}{3}x^{4/3}}{x} = 2 + \frac{7}{3}x^{1/3}$ ### Step 2: Evaluate the Summation We need to find sum_r=1^15 f(r^3)$\sum_{r=1}^{15} f(r^3)$: f(r^3) = 2 + frac73(r^3)^1/3 = 2 + frac73r$f(r^3) = 2 + \frac{7}{3}(r^3)^{1/3} = 2 + \frac{7}{3}r$ Now, compute the summation from r=1$r=1$ to 15$15$: sum_r=1^15 f(r^3) = sum_r=1^15 left( 2 + frac73r right) = sum_r=1^15 2 + frac73sum_r=1^15 r$\sum_{r=1}^{15} f(r^3) = \sum_{r=1}^{15} \left( 2 + \frac{7}{3}r \right) = \sum_{r=1}^{15} 2 + \frac{7}{3}\sum_{r=1}^{15} r$ (2 times 15) + frac73 times frac15 times 162$(2 \times 15) + \frac{7}{3} \times \frac{15 \times 16}{2}$ 30 + frac73 times 120 = 30 + 7 times 40 = 30 + 280 = 310$30 + \frac{7}{3} \times 120 = 30 + 7 \times 40 = 30 + 280 = 310$ ### Pattern Recognition Converting g(x^3)$g(x^3)$ directly into a function of variable y=x^3$y=x^3$ prevents multi-layer chain rule complications when applying differentiation immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Definite Integration