How do you find the exact value of the area in the first quadrant enclosed by graph of y=sinx and y=cosx?

Question

How do you find the exact value of the area in the first quadrant enclosed by graph of  y=sinx and y=cosx?

Christopher Agresta · Accepted Answer

#2sqrt(2)-2#

The area enclosed by the graphs #y=sinx# and #y=cosx# is shaded in the above graph.

By symmetry this area is #2A#, where

# A= int_0^(pi/4) cosx - sinx \ dx #
# \ \ \ = [sinx +cosx]_0^(pi/4) #
# \ \ \ = { (sin(pi/4)+cos(pi/4)) - (sin 0 + cos 0)} #
# \ \ \ = (1/2sqrt(2)+1/2sqrt(2)) - (0+1)#
# \ \ \ = sqrt(2)-1 #

Hence shaded are is #2A=2sqrt(2)-2#

Samantha Adkison · Answer

undefined To find the exact value of the area in the first quadrant enclosed by the graphs of $ y = \sin(x) $ and $ y = \cos(x) $, follow these steps:

1. Identify the points of intersection between the two curves by setting them equal to each other and solving for $ x $.
2. Determine the area bounded by the curves within the interval where $ \sin(x) $ is above $ \cos(x) $ (which occurs when $ \sin(x) > \cos(x) $) and within the interval where $ \cos(x) $ is above $ \sin(x) $ (which occurs when $ \sin(x) < \cos(x) $).
3. Integrate $ \sin(x) - \cos(x) $ over each of these intervals to find the total area enclosed by the curves in the first quadrant.