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 ______.

Options

• 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:

${\displaystyle \frac{dy}{dx}+(-\frac{1}{x})y=(x-\frac{2}{x})}$ (Linear differential equation)

⇒ f(x) = (x – 1)² + 1

∴ Required area = ${\displaystyle {\int }_0^3({(x-1)}^2+1)dx}$ = 3 + 3 = 6