Visual SolutionPYQ 2024 · Jan Shift 2Tricky
Question
Let f:(0,)Rf : (0,\infty) \to \mathbb{R} and F(x)=0xtf(t)dtF(x) = \int_0^x t f(t) \, dt. If F(x2)=x4+x5F(x^2) = x^4 + x^5, then r=112f(r2)\sum_{r=1}^{12} f(r^2) is equal to:
Solution Path
From F(u)=u2+u5/2F(u) = u^2 + u^{5/2}, differentiate to get f(u)=2+52uf(u) = 2 + \frac{5}{2}\sqrt{u}. Then f(r2)=2+5r2f(r^2) = 2 + \frac{5r}{2}, sum =219= 219.
01Question Setup
1/4
F(x)=0xtf(t)dtF(x) = \int_0^x t\,f(t)\,dt and F(x2)=x4+x5F(x^2) = x^4 + x^5. Find r=112f(r2)\sum_{r=1}^{12} f(r^2).
f(r2)=  ?\sum f(r^2) = \;?
02Find f(t)
2/4
Let u=x2u = x^2: F(u)=u2+u5/2F(u) = u^2 + u^{5/2}. Since F(u)=uf(u)F'(u) = u\,f(u), we get f(u)=2+52uf(u) = 2 + \frac{5}{2}\sqrt{u}.
f(u)=2+52uf(u) = 2 + \frac{5}{2}\sqrt{u}
03Compute SumKEY INSIGHT
3/4
f(r2)=2+5r2f(r^2) = 2 + \frac{5r}{2}. Sum =24+5278=24+195=219= 24 + \frac{5}{2} \cdot 78 = 24 + 195 = 219.
=219\sum = 219
04Final Answer
4/4
r=112f(r2)=219\sum_{r=1}^{12} f(r^2) = 219.
219\boxed{219}
Concepts from this question1 concepts unlocked

Substitution Method

EASY

Replace a composite expression with a single variable to simplify the integrand

f(g(x))g(x)dx=f(t)dt,t=g(x)\int f(g(x))\,g'(x)\,dx = \int f(t)\,dt, \quad t = g(x)

Transforms complex integrands into standard forms that can be integrated directly

Trigonometric integralsRational functionsPower rule integrals
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