Question

The area formed by triangular shaped region bounded by the curves $y = \sin x , y = \cos x$ and $x = 0$ is

• A. $\sqrt { 2 } - 1$
• B. 1
• C. $\sqrt { 2 }$
• D. $1 + \sqrt { 2 }$

Answer 1

Answer: (A)

$\sqrt { 2 } - 1$

Answer 2

Given required area has been shown in the figure.

$x = \frac { \pi } { 4 }$ is the point of intersection of both curve

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

$= [ \sin x + \cos x ] _ { 0 } ^ { \pi / 4 }$ $= \left[ \frac { 1 } { \sqrt { 2 } } + \frac { 1 } { \sqrt { 2 } } - 1 \right]$

=$\frac { 2 } { \sqrt { 2 } } - 1 = \sqrt { 2 } - 1$.