NTA Abhyas JEE Main2020MathematicsDefinite IntegrationPractice
If I 1 = ∫ 0 2 π sin 3 x d x and I 2 = ∫ 0 1 ln 1 x - 1 d x , then
Options
- AI 1 + I 2 > 0
- BI + I 2 < 0
- CI 1 < I 2
- DI 1 = I 2
Correct answer
D. I 1 = I 2
Step-by-step solution
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