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?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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