Find the area of the region bounded above by y=2cosx and above by y=secx,-π/4≤x≤π/4

Question

Find the area of the region bounded above by y=2cosx and above by y=secx,-π/4≤x≤π/4

Answer 1

To find the area of the region bounded above by the curves y = 2cosx and y = secx over the interval -π/4 ≤ x ≤ π/4, we need to set up an integral.

First, let's find the points of intersection between the two curves by setting them equal to each other:

2cosx = secx

To simplify this equation, we can rewrite secx as 1/cosx:

2cosx = 1/cosx

Multiplying both sides by cosx, we get:

2cos^2x = 1

Rearranging this equation, we have:

2cos^2x - 1 = 0

Using the identity cos^2x - sin^2x = 1, we can rewrite the equation as:

cos^2x - sin^2x = 0

Factoring the left side, we get:

(cosx + sinx)(cosx - sinx) = 0

Setting each factor equal to zero, we have:

cosx + sinx = 0 --> sinx = -cosx --> tanx = -1 --> x = -π/4 or 3π/4

cosx - sinx = 0 --> sinx = cosx --> tanx = 1 --> x = π/4 or 5π/4

Now that we have the intersection points, we can set up the integral considering the area between the curves:

∫[from -π/4 to π/4] (2cosx - secx) dx

To find the area, we need to evaluate this integral. Let's break it down into two integrals:

∫[from -π/4 to π/4] 2cosx dx - ∫[from -π/4 to π/4] secx dx

The integral of 2cosx is:

[2sinx] from -π/4 to π/4

Substituting the limits of integration, we get:

2sin(π/4) - 2sin(-π/4)

Since sin(-θ) = -sin(θ), this simplifies to:

2sin(π/4) + 2sin(π/4)

Using the value of sin(π/4) = √2/2, we have:

2(√2/2) + 2(√2/2) = 2√2

The integral of secx is a bit trickier. We can rewrite it as:

∫[from -π/4 to π/4] (1/cosx) dx

This can be evaluated using the natural logarithm function. The integral becomes:

[ln|secx + tanx|] from -π/4 to π/4

Substituting the limits of integration, we get:

ln|sec(π/4) + tan(π/4)| - ln|sec(-π/4) + tan(-π/4)|

Since sec(θ) = 1/cos(θ) and tan(θ) = sin(θ)/cos(θ), we can simplify this to:

ln|1 + 1| - ln|1 - 1| = ln(2)

Now, let's substitute the values back into the original expression:

2√2 - ln(2)

Therefore, the area of the region bounded above by the curves y = 2cosx and y = secx over the interval -π/4 ≤ x ≤ π/4 is given by the value 2√2 - ln(2).

Answer 2

To find the area of the region bounded above by the curves y = 2cos(x) and y = sec(x) over the interval -π/4 ≤ x ≤ π/4, we need to calculate the definite integral of the difference between the two functions within that interval.

The first step is to set up the integral. The area, A, can be expressed as:

A = ∫[a,b] (f(x) - g(x)) dx,

where f(x) = 2cos(x) and g(x) = sec(x), and [a, b] represents the interval of integration, -π/4 ≤ x ≤ π/4.

Next, we can rewrite the functions f(x) and g(x) in terms of the same trigonometric function to simplify the integration.

Recall that sec(x) is the reciprocal of cosine, so we can express g(x) as:

g(x) = sec(x)
= 1/cos(x).

Now, let's substitute these functions into the integral:

A = ∫[-π/4, π/4] (f(x) - g(x)) dx
= ∫[-π/4, π/4] (2cos(x) - 1/cos(x)) dx.

To evaluate this integral, we can split it into two separate integrals:

A = ∫[-π/4, π/4] (2cos(x) - 1/cos(x)) dx
= ∫[-π/4, π/4] 2cos(x) dx - ∫[-π/4, π/4] 1/cos(x) dx.

The integral of 2cos(x) can be easily evaluated as:

∫ 2cos(x) dx = 2 sin(x) + C1,

where C1 is the constant of integration.

Now, let's focus on the integral of 1/cos(x). We can rewrite the integrand using the trigonometric identity sec(x) = 1/cos(x):

∫ 1/cos(x) dx = ∫ sec(x) dx.

To evaluate this integral, we recall that the derivative of the secant function is the secant function times the tangent function: d/dx(sec(x)) = sec(x) * tan(x). With that in mind, we can rewrite the integral as:

∫ sec(x) dx = ln |sec(x) + tan(x)| + C2,

where C2 is another constant of integration.

Now, let's substitute these results back into the integral:

A = ∫[-π/4, π/4] (2cos(x) - 1/cos(x)) dx
= [2 sin(x) - ln |sec(x) + tan(x)|] evaluated from -π/4 to π/4.

To find the area, we evaluate the expression at the upper limit of the interval (π/4) and subtract it from the expression evaluated at the lower limit (-π/4):

A = [2 sin(π/4) - ln |sec(π/4) + tan(π/4)|] - [2 sin(-π/4) - ln |sec(-π/4) + tan(-π/4)|].

To simplify further, we can substitute the values of sin(π/4), sec(π/4), tan(π/4), sin(-π/4), sec(-π/4), and tan(-π/4):

A = [2 (1/√2) - ln |(1 + 1/√2)/(1/√2)|] - [2 (-1/√2) - ln |(1 - 1/√2)/(1/√2)|].

Simplifying the expression within the logarithms and evaluating the square roots, we get:

A = [√2 - ln(2 + √2)] - [-√2 - ln(2 - √2)]
= √2 - ln(2 + √2) + √2 + ln(2 - √2)
= 2√2 + ln(2 - √2) - ln(2 + √2).

Therefore, the area of the region bounded above by y = 2cos(x) and y = sec(x) over the interval -π/4 ≤ x ≤ π/4 is 2√2 + ln(2 - √2) - ln(2 + √2).

Answer 3

Assuming you meant the region below y= 2cos(x) and above y=sec(x):