I'm a bit confused about the average equations with integration. I need to solve this problem, and I think I'm supposed to use substitution, but I can't get it to work. The problem looks like this:
z(t)=4tsin(t^2), [0,π^1/2]
Any input?
Let
u=t^2
du = 2tdt
∫z(t)dt
=∫4tsin(t^2)dt from 0 to sqrt(π)
=∫2sin(u)du from 0 to π
=-2cos(u) from 0 to π
=-2[-1 - (1)]
=4
To find the average of a function over an interval using integration, you'll need to follow these steps:
1. Determine the function you want to find the average of over the given interval. In this case, the function is z(t) = 4tsin(t^2).
2. Set up the integral for finding the average. The average of a function f(x) over an interval [a, b] is given by the formula:
Average = (1 / (b - a)) ∫[a,b] f(x) dx.
In our case, the interval is [0, π^(1/2)], and the function is z(t) = 4tsin(t^2). Therefore, the integral we need to set up becomes:
Average = (1 / (π^(1/2) - 0)) ∫[0,π^1/2] 4tsin(t^2) dt.
3. Apply the substitution method to simplify the integral. To do this, you need to choose a new variable u and find its derivative du/dt. In this case, let u = t^2, so du/dt = 2t.
4. Rewrite the integral in terms of u. By substituting u = t^2, the limits of integration also change. When t = 0, u = (0)^2 = 0, and when t = π^(1/2), u = (π^(1/2))^2 = π/2.
Therefore, the integral becomes:
Average = (1 / (π^(1/2) - 0)) ∫[0,π^1/2] 4tsin(t^2) dt = (1 / (π^(1/2) - 0)) ∫[0,π/2] 4sin(u) (1 / 2t) du.
Simplifying further, we get:
Average = (2 / π^(1/2)) ∫[0,π/2] 2sin(u) / t du.
5. Evaluate the integral. Now that the integral is simplified, you can compute it using standard integration techniques or a computer algebra system. In this case, the integral ∫ 2sin(u) / t du evaluates to -2cos(u).
Therefore, the average becomes:
Average = (2 / π^(1/2)) (-2cos(u)) evaluated from 0 to π/2.
6. Substitute the limits of integration back in. When u = 0, cos(u) = 1, and when u = π/2, cos(u) = 0.
Hence, the average of the function z(t) = 4tsin(t^2) over the interval [0, π^(1/2)] is:
Average = (2 / π^(1/2)) (-2cos(u)) evaluated from 0 to π/2 = (2 / π^(1/2)) (-2(0 - 1)) = (2 / π^(1/2)) * 2 = 4 / π^(1/2).
So, the average is 4 / π^(1/2).
Note: It's always a good practice to check your work and evaluate the average to see if it makes sense in the context of the problem and the given function.