Area problemFind the area located under the curve f(x)=(cos ^2 x)^-1, x axis and x=0, x=pi/4

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We require the area between the curve f(x)=(cos ^2 x)^-1, the x- axis and the lines x=0 and  x=pi/4.

This is the same as calculating the definite integral of f(x) between the limits x = 0 and x = pi/4

Int[ f(x) dx] = Int [ 1/(cos x)^2 dx]

=> Int[ (sec x)^2 dx]

=> tan x + C

Between x = 0 and x = pi/4, the area is:

tan (pi/4) - tan 0

=> 1 - 0

=> 1

The required area is 1.

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The area of the region bounded by the curve f(x) = 1/(cos x)^2, x axis and the lines =0 and x=pi/4 is the definite integral of the curve, evaluated using the Leibniz-Newton formula.

Int f(x)dx = F(b) - F(a), where x = a to x = b

We'll put 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 = 0:

F(0) = tan 0

F(0) = 0

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/4) - F(0)

Int dx/(cos x)^2 = 1 - 0

The area of the region bounded by the curve f(x) = 1/(cos x)^2, x axis and the lines =0 and x=pi/4 is A = 1 square units.

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