If f x = ∫ 5 x 8 + 7 x 6 x 2 + 1 + 2 x 7 2 d x , x ≥ 0 , and f 0 = 0 , then the value of f ( 1 ) is

Question

If f x = ∫ 5 x 8 + 7 x 6 x 2 + 1 + 2 x 7 2 d x , x ≥ 0 , and f 0 = 0 , then the value of f ( 1 ) is, - 1 4, 1 2, 1 4, - 1 2

MARKS App · Accepted Answer

We have, f x = ∫ 5 x 8 + 7 x 6 x 2 + 1 + 2 x 7 2 d x = ∫ 5 x 8 + 7 x 6 x 14 1 x 5 + 1 x 7 + 2 2 d x = ∫ 5 x 6 + 7 x 8 1 x 5 + 1 x 7 + 2 2 d x Put, 1 x 5 + 1 x 7 + 2 = t ⇒ - 5 x 6 - 7 x 8 d x = d t ∴ f x = - ∫ d t t 2 = 1 t + c , where c is the constant of integration. = x 7 x 2 + 1 + 2 x 7 + c Now, f 0 = 0 ⇒ c = 0 ∴ f 1 = 1 1 2 + 2 + 2.1 7 = 1 4

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