Given \[ \frac{d}{dx} F(x) = \frac{1}{\sqrt{2x - x^2}} \] and \( F(1) = 0 \), find \( F(x) \).

Question

Let \( \mathbb{R}_+ \) denote the set of all non-negative real numbers. Then the function \( f : \mathbb{R}_+ \to \mathbb{R}_+ \) defined as \( f(x) = x^2 + 1 \) is:

Answer 1

1. Perform the substitution \( u = 2x - x^2 \):\[{Let } u = 2x - x^2 \quad \Rightarrow \quad \frac{du}{dx} = 2 - 2x.\]2. Express the integral \( F(x) \) in terms of \( u \):\[F(x) = \int \frac{1}{\sqrt{2x - x^2}} \, dx = \int \frac{1}{\sqrt{u}} \cdot \frac{1}{2 - 2x} \, du.\]3. Evaluate the integral:Compute the antiderivative using the substitution and simplify. Then, apply the condition \( F(1) = 0 \) to determine any constants of integration.(Note: Further steps may involve integration techniques and handling of constants. Provide details if required.)
Final Answer: \( \boxed{F(x) = {The function of } u { that satisfies } F(1) = 0.} \)

Given \[ \frac{d}{dx} F(x) = \frac{1}{\sqrt{2x - x^2}} \] and \( F(1) = 0 \), find \( F(x) \).

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Solution and Explanation

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