If ∫ x 5 e - 4 x 3 d x = 1 48 e - 4 x 3 f x + C , where C is a constant of integration, then f x is equal to

Question

If ∫ x 5 e - 4 x 3 d x = 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, - 2 x 3 + 1, - 2 x 3 - 1, 4 x 3 + 1

MARKS App · Accepted Answer

We have, I = ∫ x 5 e - 4 x 3 d x Let, - 4 x 3 = t ⇒ - 12 x 2 d x = d t ⇒ I = ∫ - t 4 e t - d t 12 = 1 48 ∫ t e t d t = 1 48 t e t - ∫ e t d t = t - 1 e t 48 + C = - 4 x 3 - 1 e - 4 x 3 48 + C , where C is the constant of integration. Hence, f x = - 4 x 3 - 1

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