NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice
Let ∫ x 3 + x 2 + x 12 x 3 + 15 x 2 + 20 x d x = f x where f 1 = 47 30 . If f 2 2 is equal to p 225 , then the value of p is equal to
Correct answer
196
Step-by-step solution
In the given integral, multiplying numerator and denominator by x , we get, f x = ∫ x 4 + x 3 + x 2 12 x 5 + 15 x 4 + 20 x 3 d x Now substitute 12 x 5 + 15 x 4 + 20 x 3 = t 2 ⇒ 60 x 4 + x 3 + x 2 d x = 2 t d t ⇒ f x = 1 30 ∫ t d t t = 1 30 12 x 5 + 15 x 4 + 20 x 3 + C As f 1 = 47 30 ⇒ C = 0 So, f 2 = 784 30 ⇒ f 2 2 = 784 900 = 196 225 Hence, p = 196