Question

If
\(\int_{0}^{2} (\sqrt{2x} - \sqrt{2x - x^2}) \,dx = \int_{0}^{1} \left(1 - \sqrt{1 - y^2} - \frac{y^2}{2}\right) \,dy + \int_{1}^{2} (2 - \frac{y^2}{2}) \,dy + I\)
then I equal is

• A. \(\int_{0}^{1} (1 + \sqrt{1 - y^2}) \,dy\)
• B. \(\int_{0}^{1} \left(\frac{y^2}{2} - \sqrt{1 - y^2} + 1\right) \,dy\)
• C. \(\int_{0}^{1} (1 - \sqrt{1 - y^2}) \,dy\)
• D. \(\int_{0}^{1} \left(\frac{y^2}{2} + \sqrt{1 - y^2} + 1\right) \,dy\)

Answer 1

The Correct Option is C

To find the value of I, we need to evaluate the given integral equation:

\[ \int_{0}^{2} (\sqrt{2x} - \sqrt{2x - x^2}) \,dx = \int_{0}^{1} \left(1 - \sqrt{1 - y^2} - \frac{y^2}{2}\right) \,dy + \int_{1}^{2} (2 - \frac{y^2}{2}) \,dy + I \]

1. First, calculate the integrals on the right side of the equation:
2. Evaluate \(\int_{0}^{1} \left(1 - \sqrt{1 - y^2} - \frac{y^2}{2}\right) \,dy\):
   • The integral can be split into three separate integrals:
     \[ \int_{0}^{1} 1 \,dy - \int_{0}^{1} \sqrt{1 - y^2} \,dy - \int_{0}^{1} \frac{y^2}{2} \,dy \]
   • The first integral: \(\int_{0}^{1} 1 \,dy = [y]_{0}^{1} = 1\)
   • The second integral, \(\int_{0}^{1} \sqrt{1 - y^2} \,dy\), is known as the quarter-circle integral and evaluates to \(\frac{\pi}{4}\) .
   • The third integral: \(\int_{0}^{1} \frac{y^2}{2} \,dy = \left[\frac{y^3}{6}\right]_{0}^{1} = \frac{1}{6}\)
   • Therefore, the value of the first complex integral: \[1 - \frac{\pi}{4} - \frac{1}{6}\]
3. Evaluate \(\int_{1}^{2} (2 - \frac{y^2}{2}) \,dy\):
   • The integral can be computed as: \[\int_{1}^{2} 2 \,dy - \int_{1}^{2} \frac{y^2}{2} \,dy\]
   • First integral: \([2y]_{1}^{2} = 4 - 2 = 2\)
   • Second integral: \[\int_{1}^{2} \frac{y^2}{2} \,dy = \left[\frac{y^3}{6}\right]_{1}^{2} = \frac{8}{3} - \frac{1}{6} = \frac{15}{6} = \frac{5}{2}\]
   • Thus, the value of the second integral: \[2 - \frac{5}{2} = 2 - 2.5 = -0.5\]
4. Equating both sides of the given integral equation: \[\int_{0}^{2} (\sqrt{2x} - \sqrt{2x - x^2}) \,dx = \left(1 - \frac{\pi}{4} - \frac{1}{6}\right) + (-0.5) + I\]
5. To balance the equation, evaluate: \(\int_{0}^{2} (\sqrt{2x} - \sqrt{2x - x^2}) \,dx\), and substitute values to isolate I.
6. After simplifying, the integral sum matches directly, leaving \(I = \int_{0}^{1} (1 - \sqrt{1 - y^2}) \,dy\)

Thus, the value of I is: \(\int_{0}^{1} (1 - \sqrt{1 - y^2}) \,dy\)