The area enclosed by the curves y=x^2, y=x^3 , x=0 and x=p , where p>1 , is 1 / 6 . The p equals

Step-by-step solution

Given curves are y=x^2 and y=x^3 Also, x=0 and x=p, p>1 Now, intersecting point is (1,1) Required Area aligned & = ₀^1 (x^2-x^3 d x+ ₁^p x^3 )-x^2 d x & 1 6 = x^3 3 - . x^4 4 |₀ ^1+ x^4 4 - . x^3 3 |₁ ^p & 1 6 = ( 1 3 - 1 4 )+ ( p^4 4 - p^3 3 - 1 4 + 1 3 ) & 1 6 - 1 3 + 1 4 + 1 4 - 1 3 = 3 p^4-4 p^3 12 & . p^3(3 p-4 12 )=0 p^3(3 p-4)=0 & p=0 or 4 3 & Since, it is given that p>1 & p can not be zero. & Hence, p= 4 3 aligned