Question

Let y = f(x) be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If f(x) satisfies xf'(x) = x² + f(x) – 2, then the area bounded by f(x) with x-axis between ordinates x = 0 and x = 3 is equal to ______.

पर्याय

• 6.00

• 7.00

• 8.00

• 9.00

Answer 1

Let y = f(x) be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If f(x) satisfies xf'(x) = x² + f(x) – 2, then the area bounded by f(x) with x-axis between ordinates x = 0 and x = 3 is equal to 6.00.

Explanation:

`(dy)/(dx) + (-1/x)y = (x - 2/x)` (Linear differential equation)

⇒ f(x) = (x – 1)² + 1

∴ Required area = `int_0^3((x - 1)^2 + 1)dx` = 3 + 3 = 6