Verified Solution • 2025–26 Session moderate

QJEEM13Mab92d: JEE Main 2013 Mathematics - Class 12 JEEM13Mab92d Solution: If then is equal to

If ∫f(x)dx=ψ(x),\int {f\left( x \right)dx = \psi \left( x \right),} then ∫x5f(x3)dx\int {{x^5}f\left( {{x^3}} \right)dx} is equal to

Multiple Choice Options

  • A 13[x3ψ(x3)−∫x2ψ(x3)dx]+C{1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C
  • B 13x3ψ(x3)−3∫x3ψ(x3)dx+C{1 \over 3}{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^3}\psi \left( {{x^3}} \right)dx} + C
  • C 13x3ψ(x3)−∫x2ψ(x3)dx+C{1 \over 3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}\psi \left( {{x^3}} \right)dx} + C
  • D 13[x3ψ(x3)−∫x3ψ(x3)dx]+C{1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - \int {{x^3}\psi \left( {{x^3}} \right)dx} } \right] + C

Step-by-Step Verified Solution

Let ∫f(x)dx=ψ(x)\int {f\left( x \right)dx = \psi \left( x \right)}

Let I=∫x5f(x3)dxI = \int {{x^5}} f\left( {{x^3}} \right)dx

put x3=t⇒3x2dx=dt{x^3} = t \Rightarrow 3{x^2}dx = dt

I=13∫3.x2.x3.f(x3).dxI = {1 \over 3}\int {3.{x^2}} .{x^3}.f\left( {{x^3}} \right).dx

=13∫tf(t)dt= {1 \over 3}\int {tf} \left( t \right)dt

=13[t∫f(t)dt−∫f(t)dt]= {1 \over 3}\left[ {t\int {f\left( t \right)dt - \int {f\left( t \right)dt} } } \right]

=13[tψ(t)−∫ψ(t)dt]= {1 \over 3}\left[ {t\psi \left( t \right) - \int {\psi \left( t \right)dt} } \right]

=13[x3ψ(x3)−3∫x2ψ(x3)dx]+C= {1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C

=13x3ψ(x3)−∫x2ψ(x3)dx+C= {1 \over 3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}} \psi \left( {{x^3}} \right)dx + C

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