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

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Definite Integrals

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 :

Answer: (D)

With

g (x) = \int_{0}^{x} t f (t) d t

, differentiating gives

g^{'} (x) = x f (x)

.

Given

g (x^{3}) = x^{6} + x^{7}

, differentiate both sides with respect to

x

g^{'} (x^{3}) \cdot 3 x^{2} = 6 x^{5} + 7 x^{6} ⟹ g^{'} (x^{3}) = 2 x^{3} + \frac{7}{3} x^{4}

Since

g^{'} (x^{3}) = x^{3} f (x^{3})

x^{3} f (x^{3}) = 2 x^{3} + \frac{7}{3} x^{4} ⟹ f (x^{3}) = 2 + \frac{7}{3} x

f (r^{3}) = 2 + \frac{7 r}{3}

, and

r = 1 \sum 15 f (r^{3}) = r = 1 \sum 15 (2 + \frac{7 r}{3}) = 30 + \frac{7}{3} \cdot \frac{15 \cdot 16}{2} = 30 + 280 = 310

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option