Let f be a real valued continuous function defined on the positive real axis such that g(x)=∫₀ˣ t f(t) d t. If g(x³)=x⁶+x⁷, then value of Σᵣ₌₁¹⁵ f(r³) is :

Solution

Maths sir · 4 steps

Write g in terms of u = x³ first, then differentiate: Leibniz gives g'(u) = u f(u) directly.

1. Leibniz rule. Since g(x)=∫₀ˣ t f(t) dt with a constant lower limit, g'(x) = x f(x).
2. Rewrite g. Put u = x³, so x = u^(1/3). Then g(u) = u² + u^(7/3), which gives g'(u) = 2u + 7/3u^(4/3).

   u f(u) = 2u + 7/3u^(4/3) ⇒ f(u) = 2 + 7/3u^(1/3)

3. Add up. So f(r³) = 2 + 7r/3. Summing from 1 to 15 gives 2 × 15 + 7/3 × (15 × 16)/2 = 30 + 7/3 × 120 = 30 + 280.
4. So. Σᵣ₌₁¹⁵ f(r³) = 310, which is option (c).

The idea to keep. When g is given at a composite argument like x³, rename that argument as u first. Then Leibniz gives g'(u) = integrand at u, and you avoid any chain-rule slip.

Answer from the NTA final answer key. Solution by PrepWiser.