NTA Abhyas JEE Main2020MathematicsDefinite IntegrationPractice
If Ι 1 = ∫ - 1 2 x s i n x 1 - x d x and Ι 2 = ∫ - 1 2 s i n x 1 - x d x , then Ι 1 Ι 2 is equal to
Options
- A2
- B1 2
- C1
- D1 3
Correct answer
B. 1 2
Step-by-step solution
Ι 1 = ∫ - 1 2 x s i n x ( 1 - x ) d x Ι 1 = ∫ - 1 2 1 - x s i n 1 - x x d x Ι 1 = ∫ - 1 2 s i n x 1 - x d x - ∫ - 1 2 x s i n 1 - x x d x Ι 1 = Ι 2 - Ι 1 2 Ι 1 = Ι 2 Ι 1 Ι 2 = 1 2