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Let ∫ e x ⋅ x 2 d x = f x e x + C (where, C is the constant of integration). The range of f x as x ∈ R is a , ∞ . The value of a 4 is

Correct answer

0.25

Step-by-step solution

As we know, ∫ e x f x + f ' x d x = e x ⋅ f x + C , Thus, ∫ e x ⋅ x 2 d x = ∫ e x x 2 + 2 x - 2 x - 2 + 2 d x = ∫ e x x 2 - 2 x + 2 + e x 2 x - 2 d x = e x x 2 - 2 x + 2 + C i.e. f x = x 2 - 2 x + 2 = x - 1 2 + 1 Hence, range is 1 , ∞ ⇒ a = 1 ∴ a 4 = 0.2 5

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