The curves y=sinx and y=cosx intersects twice on the interval (0,2pi). Find the area of the region bounded by the two curves between the points of intersection.
To find the area bounded by the curves \(y=sin(x)\) and \(y=cos(x)\) between the points of intersection, we need to determine the x-values of these points first.
The curves intersect when \(sin(x) = cos(x)\), or when \(tan(x) = 1\). We know that \(tan(x) = \frac{sin(x)}{cos(x)}\), so setting \(tan(x) = 1\) gives us \(\frac{sin(x)}{cos(x)} = 1\).
Multiplying both sides of the equation by \(cos(x)\) gives us \(sin(x) = cos(x)\), which is satisfied when \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\).
The area we want to find is the integral of the positive difference of the two curves between these x-values. To accomplish this, we'll define a new function \(f(x)\) as \(f(x) = sin(x) - cos(x)\).
The area, denoted as \(A\), can now be computed as:
\[ A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} f(x) dx \]
To evaluate this integral, we'll calculate the indefinite integral of \(f(x)\) and then evaluate it between the two points of intersection.
The indefinite integral of \(f(x)\) is:
\[ \int f(x) dx = -cos(x) - sin(x) + C \]
Plugging in the values for the definite integral, we get:
\[ A = \left[-cos(x) - sin(x)\right]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \]
Evaluating this expression gives us the final answer.
To find the area of the region bounded by the curves y = sinx and y = cosx between the points of intersection, we need to find the x-coordinates of the points where these curves intersect.
The curves intersect when sinx = cosx. We can rewrite this equation as sinx/cosx = 1. Using the trigonometric identity tanx = sinx/cosx, we have tanx = 1.
The value of x where tanx = 1 lies in the interval (0, 2pi). To find this value, we can take the inverse tangent of both sides: x = arctan(1).
The value of arctan(1) is pi/4. Therefore, the point of intersection in the given interval occurs at x = pi/4.
To calculate the area of the region, we integrate the difference between the two curves over this interval. We want to find the area under y = sinx minus the area under y = cosx.
Let's denote the area as A:
A = ∫[0, π/4] (sinx - cosx) dx
To evaluate this integral, we can use the method of integration by parts.
Let u = sinx and dv = dx. Then, du = cosx dx and v = x.
Using the formula for integration by parts, ∫u dv = uv - ∫v du, we can rewrite the integral as:
A = [sinx * x] from 0 to π/4 - ∫[0, π/4] x * cosx dx
Evaluating the first part of the expression at x = π/4 and x = 0:
A = (sin(π/4) * π/4) - (sin(0) * 0) - ∫[0, π/4] x * cosx dx
Simplifying the expression:
A = (1/√2 * π/4) - 0 - ∫[0, π/4] x * cosx dx
Evaluating the integral using integration by parts again:
A = (1/√2 * π/4) - x * sinx from 0 to π/4 + ∫[0, π/4] sinx dx
Simplifying further:
A = (1/√2 * π/4) - (π/4 * sin(π/4) - 0 * sin(0)) + [-cosx] from 0 to π/4
A = (1/√2 * π/4) - (π/4 * 1/√2 - 0) + (-cos(π/4) - (-cos(0)))
Evaluating the remaining expressions:
A = (1/√2 * π/4) - (π/4 * 1/√2) + (-1/√2 - (-1))
Simplifying and rationalizing the denominators:
A = (π - π/2 + 1 - √2) / 2√2
Therefore, the area of the region bounded by y = sinx and y = cosx between the points of intersection is (π - π/2 + 1 - √2) / 2√2 square units.