Sketch the region bounded by the curves y=sec^2x and y=4  on the interval [-pi/2, pi/2].  Find the area of the sketched region.

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The region bounded by the curves y = sec^2 x and y = 4 is shown in the graph below:

The curves `y = sec^2x` and y = 4 intersect at the point where `sec^2x = 4`

=> `cos^2x = 1/4`

=> `cos x = (1/2)` and `cos x =...

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The region bounded by the curves y = sec^2 x and y = 4 is shown in the graph below:

The curves `y = sec^2x` and y = 4 intersect at the point where `sec^2x = 4`

=> `cos^2x = 1/4`

=> `cos x = (1/2)` and `cos x = -1/2`

`x = pi/3` and `x = -pi/3`

The area of the bounded area is `int_(-pi/3)^(pi/3) 4 - sec^2x dx`

=> `4x - tan x|_(-pi/3)^(pi/3)`

=> `4*2*pi/3 - 2*sqrt 3`

=> `8*pi/3 - 2*sqrt 3`

The required area is `8*pi/3 - 2*sqrt 3`

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