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Find the area of the region between x=-3pi/4 and 5pi/4 in the graph enclosed by f(x)=sinx and g(x)=cosx
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To find the area of the region enclosed by the graphs of the functions f(x) = sin(x) and g(x) = cos(x) between x = -3π/4 and x = 5π/4, we can integrate the difference of the two functions over that interval.
Here are the step-by-step instructions to calculate the area:
Step 1: Determine the interval of integration. In this case, the interval is from x = -3π/4 to x = 5π/4.
Step 2: Calculate the difference between the two functions, g(x) - f(x). Since g(x) = cos(x) and f(x) = sin(x), the difference is cos(x) - sin(x).
Step 3: Set up the integral for the area using the difference between the functions. The integral of cos(x) - sin(x) over the given interval is:
∫[from -3π/4 to 5π/4] (cos(x) - sin(x)) dx
Step 4: Integrate the difference between the functions.
The integral of cos(x) is sin(x), and the integral of sin(x) is -cos(x), evaluated from the given interval:
[ -sin(x) - cos(x) ] from -3π/4 to 5π/4
Step 5: Evaluate the integral.
Substituting the values of the given interval into the integral, we get:
[ -sin(5π/4) - cos(5π/4) ] - [ -sin(-3π/4) - cos(-3π/4) ]
Step 6: Simplify the expression.
Using the trigonometric values for π/4 and -π/4, the expression simplifies to:
[ 1/√2 - (1/√2) ] - [ -1/√2 - (-1/√2) ]
Simplifying further:
0 - 0 = 0
Step 7: Interpret the result.
The area of the region enclosed by the graphs of f(x) = sin(x) and g(x) = cos(x) between x = -3π/4 and x = 5π/4 is 0.
Therefore, the area of the region is 0 square units.
To find the area between the two curves, you can integrate the difference between the upper curve and the lower curve with respect to x over the given interval.
Step 1: Determine the upper and lower curves
The upper curve is given by f(x) = sin(x)
The lower curve is given by g(x) = cos(x)
Step 2: Find the points of intersection
To find the bounds of integration, we need to find the x-values where the two curves intersect.
Setting f(x) = g(x), we get sin(x) = cos(x)
Divide both sides by cos(x) (assuming cos(x) ≠ 0) to get tan(x) = 1.
The solutions to tan(x) = 1 are x = π/4 + kπ, where k is an integer.
So, there is one intersection point in the region of interest, at x = π/4.
Step 3: Set up the integral
We will integrate the difference between the upper curve (sin(x)) and the lower curve (cos(x)) with respect to x over the interval x = -3π/4 to x = 5π/4.
The integral expression is: ∫[from -3π/4 to 5π/4] (f(x) - g(x)) dx
Step 4: Evaluate the integral
Evaluating the integral, we have:
∫[from -3π/4 to 5π/4] (sin(x) - cos(x)) dx
= [-cos(x) - sin(x)] evaluated from -3π/4 to 5π/4
= [-cos(5π/4) - sin(5π/4)] - [-cos(-3π/4) - sin(-3π/4)]
= [-(-sqrt(2)/2) - (-1/sqrt(2))] - [(sqrt(2)/2) - (1/sqrt(2))]
= -sqrt(2)/2 + 1/sqrt(2) + sqrt(2)/2 - 1/sqrt(2)
= 0
The area between the curves f(x) = sin(x) and g(x) = cos(x) over the given interval is 0.
Note: In this case, the area between the curves is symmetric, so the positive and negative areas cancel out.
If you want algebraic area, it is zero, since the sinx curve is above cosx just as much as it is below. See
http://www.wolframalpha.com/input/?i=plot+sinx,cosx
The geometric area is just twice the area on the interval [-3π/4,π/4], or
a = 2∫[-3π/4,π/4] cosx-sinx dx = 4√2
Actually, due to symmetry, the area is also just
4∫[-π/4,π/4] cosx-sinx dx
If you try to integrate on the whole interval [-π/4,3π/4] you get zero.