Let f be a real valued continuous function defined on the positive real axis such that g(x)=∫0xtf(t)dt. If g(x3)=x6+x7, then value of ∑r=115f(r3) is : [2025]
(4)
We have, g(x)=∫0xtf(t)dt
g(x)=x2+x7/3 ⇒ g'(x)=2x+73x4/3
As, f(x)=g'(x)x=2+73x1/3 ⇒ f(r3)=2+7r3
∴ ∑r=115(2+73r)=30+73[1+2+...+15]=310.