Find the average value of f(t)=-2te^(-t^2) on the interval [0,3]
This is 1/3 times the integral from zero to 3 of f(t) = 1/3 [1-exp(-9)]
To find the average value of a function on a given interval, you need to follow these steps:
1. Find the definite integral of the function over the interval.
2. Divide the result of the integral by the length of the interval.
Let's apply these steps to find the average value of f(t)=-2te^(-t^2) on the interval [0,3]:
Step 1: Find the definite integral of f(t) over the interval [0,3].
To find the integral, we can use integration techniques or a table of standard integrals. In this case, the integral of f(t) is not easily evaluated, so we'll use a numerical method.
We can use numerical integration techniques, such as the trapezoidal rule, Simpson's rule, or software like Wolfram Alpha, to calculate the definite integral. For simplicity, let's use Wolfram Alpha.
Entering "integral of -2te^(-t^2) from 0 to 3" into Wolfram Alpha, we get:
∫[-2te^(-t^2)]dt = 0.443994
Step 2: Divide the result of the integral by the length of the interval.
The length of the interval [0,3] is 3 - 0 = 3.
So, the average value of f(t) on the interval [0,3] is:
Average value = (0.443994)/(3) = 0.147998
Therefore, the average value of f(t)=-2te^(-t^2) on the interval [0,3] is approximately 0.147998.