A continuous function y = f(x) defined for x 0 satisfies f(1) = 16 . The area bounded by the curve y = t^2 f(t) , the t -axis, and the vertical lines t = 1 and t = x (where x 1 ) is given by the expression x^3 f(x) - 16 . Then the value of f(4) is equal to

Step-by-step solution

The area bounded by the curve from t = 1 to t = x can be written as a definite integral: ₁^x t^2 f(t) dt = x^3 f(x) - 16 Differentiating both sides with respect to x using the Leibniz rule and the product rule gives: x^2 f(x) = 3x^2 f(x) + x^3 f^ (x) Rearranging the terms, we get: x^3 f^ (x) + 2x^2 f(x) = 0 Since x 1 , we can divide by x^2 : x f^ (x) + 2f(x) = 0 f^ (x) f(x) = - 2 x Integrating both sides with respect to x : f^ (x) f(x) dx = - 2 x dx f(x) = -2 x + C f(x) = K x^2 Using the given condition f(1) = 16 :