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Indefinite Integrals question

2013 · Shift 0 · Q33
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  5. /2013 · Shift 0 · Q33

Indefinite Integrals question

2013 · Shift 0 · Q33

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If ∫f(x)dx=ψ(x),\int {f\left( x \right)dx = \psi \left( x \right),}∫f(x)dx=ψ(x), then ∫x5f(x3)dx\int {{x^5}f\left( {{x^3}} \right)dx}∫x5f(x3)dx is equal to
  1. 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] + C31​[x3ψ(x3)−∫x2ψ(x3)dx]+C
  2. 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} + C31​x3ψ(x3)−3∫x3ψ(x3)dx+C
  3. 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} + C31​x3ψ(x3)−∫x2ψ(x3)dx+C
  4. 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] + C31​[x3ψ(x3)−∫x3ψ(x3)dx]+C
View written solutionFree

Correct answer: C

  1. We are given ∫f(x) dx=ψ(x).\int f(x)\,dx = \psi(x).∫f(x)dx=ψ(x). Hence, ψ′(x)=f(x).\psi'(x)=f(x).ψ′(x)=f(x).

  2. We need to evaluate I=∫x5f(x3) dx.I=\int x^5 f(x^3)\,dx.I=∫x5f(x3)dx.

  3. Rewrite the factor x5x^5x5 as x5=x3⋅x2.x^5=x^3\cdot x^2.x5=x3⋅x2. So, I=∫x3 x2f(x3) dx.I=\int x^3\,x^2 f(x^3)\,dx.I=∫x3x2f(x3)dx.

  4. Now use the substitution t=x3  ⟹  dt=3x2 dx  ⟹  x2 dx=dt3.t=x^3 \implies dt=3x^2\,dx \implies x^2\,dx=\frac{dt}{3}.t=x3⟹dt=3x2dx⟹x2dx=3dt​.

    Therefore,

    =\frac13\int t f(t)\,dt.$$

  5. Now integrate by parts for ∫tf(t) dt.\int t f(t)\,dt.∫tf(t)dt. Take u=t,dv=f(t) dt.u=t, \qquad dv=f(t)\,dt.u=t,dv=f(t)dt. Then du=dt,v=ψ(t).du=dt, \qquad v=\psi(t).du=dt,v=ψ(t).

    So, ∫tf(t) dt=tψ(t)−∫ψ(t) dt.\int t f(t)\,dt = t\psi(t)-\int \psi(t)\,dt.∫tf(t)dt=tψ(t)−∫ψ(t)dt.

  6. Substitute back into III: I=13[tψ(t)−∫ψ(t) dt].I=\frac13\left[t\psi(t)-\int \psi(t)\,dt\right].I=31​[tψ(t)−∫ψ(t)dt].

  7. Replace t=x3t=x^3t=x3: I=13[x3ψ(x3)−∫ψ(x3) d(x3)].I=\frac13\left[x^3\psi(x^3)-\int \psi(x^3)\,d(x^3)\right].I=31​[x3ψ(x3)−∫ψ(x3)d(x3)]. Since d(x3)=3x2 dx,d(x^3)=3x^2\,dx,d(x3)=3x2dx, we get ∫ψ(x3) d(x3)=3∫x2ψ(x3) dx.\int \psi(x^3)\,d(x^3)=3\int x^2\psi(x^3)\,dx.∫ψ(x3)d(x3)=3∫x2ψ(x3)dx.

    Hence, I=13[x3ψ(x3)−3∫x2ψ(x3) dx].I=\frac13\left[x^3\psi(x^3)-3\int x^2\psi(x^3)\,dx\right].I=31​[x3ψ(x3)−3∫x2ψ(x3)dx].

  8. Simplifying, I=13x3ψ(x3)−∫x2ψ(x3) dx+C.I=\frac13 x^3\psi(x^3)-\int x^2\psi(x^3)\,dx + C.I=31​x3ψ(x3)−∫x2ψ(x3)dx+C.

  9. Compare with the options:

    • Option C: 13x3ψ(x3)−∫x2ψ(x3) dx+C\frac13 x^3\psi(x^3)-\int x^2\psi(x^3)\,dx + C31​x3ψ(x3)−∫x2ψ(x3)dx+C

    This matches exactly.

Therefore, the correct option is C.

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