If I 1 = ∫ 0 2 π sin 3 ⁡ x d x and I 2 = ∫ 0 1 ln ⁡ 1 x - 1 d x , then

Question

If I 1 = ∫ 0 2 π sin 3 ⁡ x d x and I 2 = ∫ 0 1 ln ⁡ 1 x - 1 d x , then, I 1 + I 2 > 0, I + I 2 0, I 1 I 2, I 1 = I 2

MARKS App · Accepted Answer

I 2 = ∫ 0 1 ln ⁡ 1 - x - ln ⁡ x d x = ∫ 0 1 ln ⁡ 1 - x d x - ∫ 0 1 ln ⁡ x d x = ∫ 0 1 ln ⁡ x d x - ∫ 0 1 ln ⁡ x d x (using a + b - x property) ⇒ I 2 = 0 Also, I 1 = ∫ 0 2 π sin 3 ⁡ x d x = ∫ 0 2 π sin 3 ⁡ 2 π - x d x = - ∫ 0 2 π sin 3 ⁡ x d x ⇒ I 1 = - I 1 ⇒ I 1 = 0 Hence, I 1 = I 2

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