If ∫ x5 e-4 x3 dx = (1/48) e-4x3 f(x) + C, where C is a constant of integration, then f(x) is equal to :

Question

If ∫ x5 e-4 x3 dx = (1/48) e-4x3 f(x) + C, where C is a constant of integration, then f(x) is equal to : (A)  - 4 x3 - 1  (B)  4 x3 + 1  (C)  - 2 x3 - 1  (D)  - 2 x3 + 1

Tardigrade Admin · Accepted Answer

∫ x5 .e-4x3 dx = (1/48) e-4x3f(x)+c Put x3=t 3x2 dx =dt ∫ x3.e-4x3.x2dx (1/3) ∫ t.e-4tdt (1/3) [t. (e-4t/-4) -∫ (e-4t/-4) dt] - (e-4t/48) [4t+1]+c (-e-4x3/48) [4x3 +1]+c ∴ f(x) = -1 - 4x3 (From the given options (1) is most suitable)

Q. If $\int x^{5} e^{- 4 x^{3}} d x = \frac{1}{48} e^{- 4 x^{3}} f (x) + C$ , where C is a constant of integration, then f(x) is equal to :

$- 4 x^{3} - 1$

$4 x^{3} + 1$

$- 2 x^{3} - 1$

$- 2 x^{3} + 1$

Q. If ∫x5e−4x3dx=481​e−4x3f(x)+C, where C is a constant of integration, then f(x) is equal to :

−4x3−1

4x3+1

−2x3−1

−2x3+1

∫x5.e−4x3dx=481​e−4x3f(x)+c Put x3=t 3x2dx=dt ∫x3.e−4x3.x2dx 31​∫t.e−4tdt 31​[t.−4e−4t​−∫−4e−4t​dt] −48e−4t​[4t+1]+c 48−e−4x3​[4x3+1]+c ∴f(x)=−1−4x3 (From the given options (1) is most suitable)

Q. If $\int x^{5} e^{- 4 x^{3}} d x = \frac{1}{48} e^{- 4 x^{3}} f (x) + C$ , where C is a constant of integration, then f(x) is equal to :

$- 4 x^{3} - 1$

$4 x^{3} + 1$

$- 2 x^{3} - 1$

$- 2 x^{3} + 1$