Question

The area of the region bounded by the lines x = 0, x = $\frac { \pi } { 2 }$ and ƒ(x) = sin x, g(x) = cos x is-

• A. 2 ($\sqrt { 2 }$ + 1)
• B. $\sqrt { 3 }$ – 1
• C. 2($\sqrt { 3 }$ – 1)
• D. 2 ($\sqrt { 2 }$– 1)

Answer 1

Answer: (D)

2 ($\sqrt { 2 }$– 1)

Answer 2

Required area =$\int _ { 0 } ^ { \pi / 2 } ( \sin x - \cos x ) d x$= $\int _ { 0 } ^ { \pi / 4 } ( \cos x - \sin x ) d x$+ $\int _ { \pi / 4 } ^ { \pi / 2 } ( \sin x - \cos x ) d x$

= $[ \sin x + \cos x ] _ { 0 } ^ { \pi / 4 }$ + $[ - \cos x - \sin x ] _ { \pi / 4 } ^ { \pi / 2 }$

= $\frac { 1 } { \sqrt { 2 } }$ + $\frac { 1 } { \sqrt { 2 } }$ – (0 + 1) – $\left\{ 1 - \left( \frac { 1 } { \sqrt { 2 } } + \frac { 1 } { \sqrt { 2 } } \right) \right\}$

= $\frac { 4 } { \sqrt { 2 } }$– 2 = 2$\sqrt { 2 }$– 2 = 2($\sqrt { 2 }$ – 1).