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Indefinite Integrals question

2019 · 10 Jan · Shift 2 · Q43
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Indefinite Integrals question

2019 · 10 Jan · Shift 2 · Q43

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If ∫ \int \,∫ x5.e −-− 4x3 dx = 148{1 \over {48}}481​ e −-− 4x3 f(x) + C, where C is a constant of inegration, then f(x) is equal to -
  1. A
    −-− 2x3 −-− 1
  2. B
    −-− 2x3 + 1
  3. C
    4x3 + 1
  4. D
    −-− 4x3 −-− 1
View written solutionFree

Correct answer: D

  1. Interpret the integrand carefully

The expression in the question is intended as

∫x5e−4x3 dx=148e−4x3f(x)+C.\int x^5 e^{-4x^3}\,dx = \frac{1}{48} e^{-4x^3} f(x) + C.∫x5e−4x3dx=481​e−4x3f(x)+C.

We need to find f(x)f(x)f(x).

  1. Use substitution

Let

t=x3  ⟹  dt=3x2 dx.t = x^3 \implies dt = 3x^2\,dx.t=x3⟹dt=3x2dx.

But a more direct way is to notice that

ddx(e−4x3)=−12x2e−4x3.\frac{d}{dx}\left(e^{-4x^3}\right) = -12x^2 e^{-4x^3}.dxd​(e−4x3)=−12x2e−4x3.

Since the integrand contains x5=x3⋅x2x^5 = x^3\cdot x^2x5=x3⋅x2, write

∫x5e−4x3 dx=∫x3⋅x2e−4x3 dx.\int x^5 e^{-4x^3}\,dx = \int x^3 \cdot x^2 e^{-4x^3}\,dx.∫x5e−4x3dx=∫x3⋅x2e−4x3dx.

Now put

u=−4x3  ⟹  du=−12x2 dx,u = -4x^3 \implies du = -12x^2\,dx,u=−4x3⟹du=−12x2dx, so

x2 dx=−112du,x3=−u4.x^2\,dx = -\frac{1}{12}du, \qquad x^3 = -\frac{u}{4}.x2dx=−121​du,x3=−4u​.

Thus,

= \int \left(-\frac{u}{4}\right)e^u\left(-\frac{1}{12}\right)du = \frac{1}{48}\int u e^u\,du.$$ --- 3. **Integrate $\int u e^u\,du$** Using integration by parts, $$\int u e^u\,du = e^u(u-1) + C.$$ Hence, $$\int x^5 e^{-4x^3}\,dx = \frac{1}{48} e^u(u-1) + C.$$ Substitute back $u=-4x^3$: $$\int x^5 e^{-4x^3}\,dx = \frac{1}{48} e^{-4x^3}(-4x^3-1) + C.$$ So, $$f(x) = -4x^3 - 1.$$ --- 4. **Match with options** The correct option is $$\boxed{\text{D: } -4x^3 - 1}.$$ --- 5. **Compare with stored correct answer** Stored correct answer: $\text{D}$ Derived answer: $\text{D}$ They agree.
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