NTA Abhyas JEE Main2020MathematicsDefinite IntegrationPractice
The value of the integral ∫ 0 4 x 2 x 2 - 4 x + 8 d x is equal to
Correct answer
4
Step-by-step solution
Let I = ∫ 0 4 x 2 x x - 4 + 8 d x Using ∫ a b f x d x = ∫ a b f a + b - x d x , we get, I = ∫ 0 4 4 - x 2 4 - x - x + 8 d x Adding both the equations, we get, 2 I = ∫ 0 4 2 x 2 - 4 x + 8 x 2 - 4 x + 8 d x = 8 ⇒ I = 4