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Math Snap
PROBLEM
Find the area of the region enclosed by the curves y=2cosx and y=2cos2x for 0≤x≤π.
STEP 1
1. We are given two functions: y=2cosx and y=2cos2x. 2. We need to find the area of the region enclosed by these curves from x=0 to x=π. 3. The area between two curves is found by integrating the difference of the functions over the given interval.
STEP 2
1. Find the points of intersection of the curves within the interval 0≤x≤π. 2. Determine the intervals where one function is above the other. 3. Set up the integral(s) to calculate the area between the curves. 4. Evaluate the integral(s) to find the total enclosed area.
STEP 3
Find the points of intersection by setting the functions equal: 2cosx=2cos2xSimplify: cosx=cos2xUsing the identity cos2x=2cos2x−1, we have: cosx=2cos2x−1Rearrange to form a quadratic equation: 2cos2x−cosx−1=0
STEP 4
Solve the quadratic equation: Let u=cosx, then: 2u2−u−1=0Use the quadratic formula: u=2a−b±b2−4acwhere a=2, b=−1, c=−1: u=41±1+8u=41±3Thus, u=1 or u=−21. Convert back to x: For cosx=1, x=0. For cosx=−21, x=32π.
STEP 5
Determine which function is above the other in each interval: 1. From x=0 to x=32π, compare y=2cosx and y=2cos2x. 2. From x=32π to x=π, compare y=2cosx and y=2cos2x.
STEP 6
Evaluate the functions at a test point in each interval: 1. For x=3π, 2cosx=2×21=1 and 2cos2x=2×(−21)=−1. So, 2cosx is above. 2. For x=π, 2cosx=2×(−1)=−2 and 2cos2x=2×1=2. So, 2cos2x is above.
STEP 7
Set up the integral for each interval: 1. From x=0 to x=32π: ∫032π(2cosx−2cos2x)dx2. From x=32π to x=π: ∫32ππ(2cos2x−2cosx)dx
STEP 8
Evaluate the integrals: 1. Evaluate ∫(2cosx−2cos2x)dx: ∫2cosxdx=2sinx∫−2cos2xdx=−sin2xEvaluate from 0 to 32π: [2sinx−sin2x]032π2. Evaluate ∫(2cos2x−2cosx)dx: ∫2cos2xdx=sin2x∫−2cosxdx=−2sinxEvaluate from 32π to π: [sin2x−2sinx]32ππ
STEP 9
Calculate the definite integrals: 1. From 0 to 32π: [2sinx−sin2x]032π=(2sin32π−sin34π)−(2sin0−sin0)=(2×23−(−23))=(3+23)=2332. From 32π to π: [sin2x−2sinx]32ππ=(sin2π−2sinπ)−(sin34π−2sin32π)=(0−0)−(−23−2×23)=(23+3)=233
SOLUTION
Add the areas from both intervals to find the total area: Total Area=233+233=33The area of the region enclosed by the curves is: 33