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EQ.1 y = cos x EQ.2 y = sin x
Solve the intersection points between the two equations.
`y = y`
`sin x = cos x`
Divide both sides by cos x.
`(sin x)/(cosx) = 1`
`tan x = 1`
Refer to Unit Circle Chart or Table of Trigonometric Function for Special Angle to determine the value of x.
So at the interval `0<=x<=2pi` values of x are,
`x = pi/4 and (5pi)/4`
Then substitute these values to either EQ.1 or EQ.2.
`y = cos x = cos (pi/4) = sqrt(2)/2`
`y = cos x = cos ((5pi)/4) = -sqrt2/2`
So the two equations intersect at points `(pi/4, sqrt2/2)` and `((5pi)/4, sqrt2/2)` .
The graph of the two equations is:
(Note: The color of the graph of EQ.1 is red, while EQ.2 is purple.)
Then, use the formula for area between two curves.
`A = int y_U - y_L dx`
Use the x-coordinate of the intersection points as the limits of the integral.
`A= int_(pi/4)^((5pi)/4) (sin x - cosx) dx = -cosx - sinx |_(pi/4)^((5pi)/4)`
`A= -cos x - sin x |_(pi/4)^((5pi)/4) =(-cos ((5pi)/4) - sin((5pi)/4)) - (-cos (pi/4) - sin (pi/4))`
`A = (-(-sqrt2/2 )- (-sqrt2/2)) - (-sqrt2/2 - sqrt2/2)` `= (sqrt2/2 + sqrt2/2) - (-sqrt2/2 - sqrt2/2)`
`A = sqrt2 - (-sqrt2) = sqrt2+sqrt2 = 2sqrt2`
Answer: Area bounded by two curves is `2sqrt2` .
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