integrate |sinx| from 0 to 13 pi/3
13π/3 = 4π + 1/3 π
And of course |sinx| has all your sine curves above the x-axis
see:
https://www.wolframalpha.com/input/?i=y+%3D+%7Csinx%7C+from+0+to+13%CF%80%2F3
so we need
4∫ sinx dx from 0 to π + ∫ sinx dx from 0 to π/3
= 4[-cosx] from 0 to π + [-cosx] from 0 to π/3
= 4(-cosπ - (-cos0)) + (-cos π/3 - (-cos0) )
= 4(1 +1) + (-1/2 + 1)
= 8 + 1/2
= 8 1/2
= 8.5
https://www.wolframalpha.com/input/?i=%E2%88%AB+%7Csinx%7C+from+0+to+13%CF%80%2F3
To integrate |sin(x)| from 0 to 13π/3, we can split the integral into two parts based on the intervals where sin(x) is positive or negative.
First, let's find the intervals where sin(x) is positive and negative:
sin(x) is positive on the intervals [0, π/2] and [2π, 3π/2].
sin(x) is negative on the intervals [π/2, π] and [3π/2, 2π].
Now, let's integrate the function |sin(x)| in each interval separately:
1. Interval [0, π/2]:
∫ |sin(x)| dx = ∫ sin(x) dx (since sin(x) is positive in this interval)
= -cos(x) + C
2. Interval [π/2, π]:
∫ |sin(x)| dx = ∫ -sin(x) dx (since sin(x) is negative in this interval)
= cos(x) + C
3. Interval [π, 2π]:
∫ |sin(x)| dx = ∫ sin(x) dx (since sin(x) is positive in this interval)
= -cos(x) + C
4. Interval [2π, 3π/2]:
∫ |sin(x)| dx = ∫ -sin(x) dx (since sin(x) is negative in this interval)
= cos(x) + C
Now, we need to evaluate the definite integral from 0 to 13π/3:
∫[0 to 13π/3] |sin(x)| dx = [-cos(x)] [from 0 to π/2] + [cos(x)] [from π/2 to π] + [-cos(x)] [from π to 2π] + [cos(x)] [from 2π to 13π/3]
Plugging in the limits of integration, we get:
= [-cos(π/2) + cos(0)] + [cos(π) - cos(π/2)] + [-cos(2π) + cos(0)] + [cos(13π/3) - cos(2π)]
Simplifying further, we have:
= [-0 + 1] + [-1 - (-0)] + [-1 + 1] + [cos(13π/3) - 1]
= 1 + 1 + 0 + [cos(13π/3) - 1]
= 2 + cos(13π/3) - 1
Finally, noting that cos(13π/3) is equal to 1/2, we get:
= 2 + 1/2 - 1
= 3/2
Therefore, the integral of |sin(x)| from 0 to 13π/3 is equal to 3/2.
To integrate the absolute value of sine, |sin(x)| from 0 to 13π/3, we need to split the integral into two parts.
First, we need to find the values of x where |sin(x)| changes sign. Since the absolute value of sine is positive when sine is positive and negative when sine is negative, we need to find the values of x where sin(x) = 0.
We know that sin(x) = 0 when x is a multiple of π. So, we can set up the equation x = nπ, where n is an integer.
Next, we find the values of x within the given interval (0 to 13π/3) where |sin(x)| is positive and negative. Looking at the interval, we can see that sin(x) is positive for x ∈ (0, π) and (2π, 3π) and so on. Similarly, sin(x) is negative for x ∈ (π, 2π) and (3π, 4π) and so on.
Now we can set up the integral:
∫[0 to 13π/3] |sin(x)| dx = ∫[0 to π] sin(x) dx + ∫[π to 2π] -sin(x) dx + ∫[2π to 3π] sin(x) dx + ...
This pattern will continue until we reach the last interval of (12π to 13π/3). Note that the endpoints of each interval are the values we found before where |sin(x)| changes sign.
To calculate each integral, we can simply evaluate the antiderivative of sin(x), which is -cos(x). Using the Fundamental Theorem of Calculus, the expression becomes:
= [-cos(x)] [0 to π] + [-cos(x)] [π to 2π] + [-cos(x)] [2π to 3π] + ...
Simplifying this further, we have:
= [-cos(π) + cos(0)] + [cos(π) - cos(2π)] + [-cos(2π) + cos(3π)] + ...
Notice that for each interval, the cosine terms cancel each other out due to their alternating signs. Hence, we can simplify this expression to:
= -cos(0) = -1
Therefore, the value of the integral ∫[0 to 13π/3] |sin(x)| dx is -1.