What is the area of the region closed by the curve y=1/cos^2x, x axis and the lines x=pi/4 , x=pi/3?
We need to find the area of the region enclosed between the curve y = 1/(cos)^2 , the x-axis and the lines x = pi/4 and x = pi/3
To do this the definite integral of y = 1/(cos x)^2 has to be found between the limits x = pi/4 and x = pi/3.
1/(cos x)^2 = (sec x)^2
Int [ (sec x)^2 dx], x = pi/4 to x = pi/3
=> tan (pi/3) - tan (pi/4)
=> sqrt 3 - 1
The required are is sqrt 3 - 1
To determine the area of he region bounded by the given curve and lines, we'll have to compute the definite integral of y.
This integral will be evaluated using the Leibniz-Newton formula.
Int f(x)dx = F(b) - F(a), where x = a to x = b
Let y = f(x) = 1/(cos x)^2
We'll compute the indefinite integral, first:
Int dx/(cos x)^2 = tan x + C
We'll note the result F(x) = tan x + C
We'll determine F(a), for a = pi/3:
F(pi/3) = tan pi/3
F(pi/3) = sqrt 3
We'll determine F(b), for b = pi/4:
F(pi/4) = tan pi/4
F(pi/4) = 1
We'll evaluate the definite integral:
Int dx/(cos x)^2 = F(pi/3) - F(pi/4)
Int dx/(cos x)^2 = sqrt 3 - 1
The area of the region, bounded by the curve y=1/(cos x)^2, x axis and the lines x=pi/4 and x=pi/3, is: A = (sqrt 3 - 1) square units.