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

Answer: (C)

$\int_{0}^{1} (1 - \sqrt{1 - y^2}) \,dy$

Answer 2

The correct answer is (C) : $\int_{0}^{1} (1 - \sqrt{1 - y^2}) \,dy$
$\int_{0}^{2} \sqrt{2x} \,dx - \int_{0}^{2} \sqrt{1 - (x - 1)^2} \,dx = \int_{0}^{2} \left(1 - \frac{y^2}{2}\right) \,dy - \int_{0}^{1} \sqrt{1 - y^2} \,dy + 1 + l$
$⇒$ $\frac{8}{3} - 2\int_{0}^{1} \sqrt{1 - y^2} \,dy = \frac{2}{3} + 1 - \int_{0}^{1} \sqrt{1 - y^2} \,dy + l$
$⇒$ $I = 1 - \int_{0}^{1} \sqrt{1 - y^2} \,dy$
$⇒$ $I = \int_{0}^{1} (1 - \sqrt{1 - y^2}) \,dy$