Find the area of the region. (Round your answer to three decimal places.)
between y = cos t and y = sin t for −π/2 ≤ t ≤ π/2
To get the algebraic (signed) area, just do
∫[-π/2,π/2] cos(t)-sin(t) dt = 2
But that value subtracts the area where cos < sin. If you want the physical area, you need to break up the region at π/4.
http://www.wolframalpha.com/input/?i=%E2%88%AB[-%CF%80%2F2%2C%CF%80%2F2]+%28cos%28t%29-sin%28t%29%29+dt
Would I then have to subtract the area sin(t)-cos(t)?
that would be correct, since then the sin curve is above the cos curve.
so would you subtract (-pi/2, pi/2) sin(t)-cos(t) on Wolframalpha?
I'm having trouble coming up with the value.
To find the area of the region between the curves y = cos(t) and y = sin(t) for −π/2 ≤ t ≤ π/2, we need to set up an integral that represents the area.
First, let's find the points of intersection between the curves. Set y = cos(t) equal to y = sin(t) and solve for t:
cos(t) = sin(t)
Divide both sides by sin(t):
cot(t) = 1
From this equation, we can see that t = π/4 is the only value on the interval −π/2 ≤ t ≤ π/2 that satisfies the equation.
Therefore, the area of the region can be obtained by integrating the difference between the two curves from t = -π/2 to t = π/4. The integral can be set up as follows:
A = ∫[−π/2, π/4] (sin(t) - cos(t)) dt
To evaluate this integral, you can use integration techniques such as substitution or integration by parts. However, since the bounds of integration are simple and the integrand is relatively straightforward, we can directly evaluate the integral:
A = [-cos(t) - sin(t)] from −π/2 to π/4
Substituting the bounds of integration:
A = [-cos(π/4) - sin(π/4)] - [-cos(-π/2) - sin(-π/2)]
Simplifying the expression inside the brackets:
A = [-√2/2 - √2/2] - [0 + 1]
A = -√2/2 - 1
Finally, rounding the answer to three decimal places:
A ≈ -1.707