Find the exact length of the curve $y=x-x^2+{\sin -1}^x$ from $x=0$ to $x=1$.

• A. ${\int }_0^1\sqrt{1+{\left(1-2x\right)}^2} dx$
• B. ${\int }_0^1\sqrt{1+{\left(1-2x+\frac{1}{\sqrt{1-x^2}}\right)}^2} dx$
• C. ${\int }_0^1\sqrt{1+{\left(1+2x+\frac{1}{\sqrt{1-x^2}}\right)}^2} dx$
• D. ${\int }_0^1\sqrt{1+{\left(1-2x-\frac{1}{\sqrt{1-x^2}}\right)}^2} dx$

Answer: (B)

Verified step by step guidance

1. Step 1: Recall the formula for the arc length of a curve y = f(x) from x = a to x = b. The formula is given by: \( L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \). Here, we need to compute the derivative \( \frac{dy}{dx} \) for the given function y = x - x^2 + \sin^{-1} x.
2. Step 2: Differentiate y = x - x^2 + \sin^{-1} x with respect to x. Using the derivative rules, \( \frac{d}{dx}(x) = 1 \), \( \frac{d}{dx}(x^2) = 2x \), and \( \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \), we find \( \frac{dy}{dx} = 1 - 2x + \frac{1}{\sqrt{1 - x^2}} \).
3. Step 3: Substitute \( \frac{dy}{dx} \) into the arc length formula. This gives \( L = \int_0^1 \sqrt{1 + \left(1 - 2x + \frac{1}{\sqrt{1 - x^2}}\right)^2} \, dx \).
4. Step 4: Simplify the integrand if possible. Expand \( \left(1 - 2x + \frac{1}{\sqrt{1 - x^2}}\right)^2 \) and combine terms under the square root. This step involves algebraic manipulation, but the integral remains in its current form.
5. Step 5: Recognize that the integral \( \int_0^1 \sqrt{1 + \left(1 - 2x + \frac{1}{\sqrt{1 - x^2}}\right)^2} \, dx \) represents the exact length of the curve. Evaluate the integral using numerical methods or advanced techniques if required, but the setup is complete.