Find the area of the region enclosed between y = 2sin (x) and y = 2cos (x) from x = 0 to x = pi/4.
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Since cos x > sin x on the interval,
Integrate 2cos x - 2sin x on [0,pi/4]
2sin x + 2cos x [0,pi/4]
at x=pi/4, 2(1/√2 + 1/√2) = 2*2/√2 = 2√2
at x=0, 2*0 + 2*1 = 2
area = 2√2 - 2
To find the area of the region enclosed between the curves y = 2sin(x) and y = 2cos(x) from x = 0 to x = pi/4, we need to integrate the difference between the two functions over the given interval. Here are the step-by-step instructions:
Step 1: Find the x-values where the two curves intersect.
To find the intersection points, set the two functions equal to each other:
2sin(x) = 2cos(x)
Divide both sides by 2:
sin(x) = cos(x)
Divide both sides by cos(x):
tan(x) = 1
Using the inverse tangent function, we get:
x = pi/4
So the curves intersect at x = pi/4.
Step 2: Set up the integral for finding the area.
We need to integrate the difference between the two functions from x = 0 to x = pi/4. Since the function 2sin(x) is above the function 2cos(x) in that interval, we subtract the equation for 2cos(x) from the equation for 2sin(x):
Area = ∫[0, pi/4] (2sin(x) - 2cos(x)) dx
Step 3: Integrate the difference.
We can integrate each term separately:
Integral of 2sin(x) dx = -2cos(x)
Integral of 2cos(x) dx = 2sin(x)
Step 4: Evaluate the definite integral.
Now we can evaluate the definite integral over the given interval:
Area = [-2cos(x)] from 0 to pi/4 - [2sin(x)] from 0 to pi/4
= [-2cos(pi/4) - -2cos(0)] - [2sin(pi/4) - 2sin(0)]
= [-2√2/2 - -2] - [2/√2 - 0]
= [-√2 - -2] - [√2]
= 2 - √2 - √2
= 2 - 2√2
Therefore, the area of the region enclosed between y = 2sin(x) and y = 2cos(x) from x = 0 to x = pi/4 is 2 - 2√2.
To find the area of the region enclosed between the two curves, y = 2sin(x) and y = 2cos(x), we need to determine the points of intersection between these two curves.
Setting the two equations equal to each other, we have:
2sin(x) = 2cos(x)
Dividing both sides by 2, we get:
sin(x) = cos(x)
Now, we can use the trigonometric identity sin(x) = cos(π/2 - x) to rewrite the equation:
cos(π/2 - x) = cos(x)
According to the cosine function, two angles are equal if their differences are multiples of 2π. So we set up the following equation:
π/2 - x = x + 2kπ (where k is an integer)
Simplifying this equation, we have:
π/2 = 2x + 2kπ
Now, solve for x:
2x = π/2 - 2kπ
x = (π/4 - kπ)
To find the points of intersection within the given interval [0, π/4], we substitute values of k:
When k = 0, x = (π/4 - 0π) = π/4
When k = 1, x = (π/4 - π) = -3π/4 (outside the interval, so we discard it)
Now, we have the lower and upper limits of integration, 0 and π/4 respectively.
To find the area, we integrate the difference of the curves' equations between these limits:
Area = ∫[0, π/4] (2sin(x) - 2cos(x)) dx
Next, we integrate the term (2sin(x) - 2cos(x)) with respect to x:
Area = -2cos(x) - 2sin(x) |[0, π/4]
Evaluating the integral at the upper limit (π/4) and subtracting the integral at the lower limit (0):
Area = [-2cos(π/4) - 2sin(π/4)] - [-2cos(0) - 2sin(0)]
Area = [-2(√2/2) - 2(√2/2)] - [-2(1) - 2(0)]
Area = [-√2 - √2] - [-2]
Area = -2√2 + 2
Therefore, the area of the region enclosed between y = 2sin(x) and y = 2cos(x) from x = 0 to x = π/4 is -2√2 + 2.