g(x)=x2+x7/3 g′(x)=2x+37x4/3f(x)=xg′(x)f(x)=2+37x1/3f(r3)=2+37rr=1∑15(2+37r)=2(15)+37(215(16)) =310
Let f be a real valued continuous function defined on the positive real axis such that g(x)=∫_0^x t f( t) dt. If g (x^3 )=x^6+x^7, then value of…
JEE Main 2025 — Mathematics Calculus
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=1∑15f(r3) is :
Held on 28 Jan 2025 · Verified 6 Jul 2026.
Options
- A
270
- B
340
- C
320
- D
310
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