Question

Area enclosed by the curve (y – sin^-1x)² = x − x², is equal to

• A. $\frac { \pi } { 2 }$ sq. units
• B. $\frac { \pi } { 4 }$sq. units
• C. $\frac { \pi } { 8 }$sq. units
• D. None of these

Answer 1

Answer: (B)

$\frac { \pi } { 4 }$sq. units

Answer 2

Clearly x ≤ 1 and x – x² ≥ 0 ⇒ x ∈ [0, 1]. Curve has two branches whose equations are

y = sin^-1x ± $\sqrt { x - x ^ { 2 } }$ , and it is defined only in between [0, 1]. The area enclosed by the curve,

$\Delta = \int _ { 0 } ^ { 1 } \left( y _ { 1 } \sim y _ { 2 } \right) d x = 2 \int _ { 0 } ^ { 1 } \sqrt { x - x ^ { 2 } } d x$

$= 2 \int _ { 0 } ^ { 1 } \sqrt { \frac { 1 } { 4 } - \left( x - \frac { 1 } { 2 } \right) ^ { 2 } } d x$

$= 2 \left( \left. \frac { 2 x - 1 } { 4 } \sqrt { x - x ^ { 2 } } \right| _ { 0 } ^ { 1 } + \left. \frac { 1 } { 8 } \sin ^ { - 1 } ( 2 x - 1 ) \right| _ { 0 } ^ { 1 } \right)$

$= \frac { \pi } { 4 }$ sq. units