Question Video: Finding the Average Value of a Function of Combining Two Trigonometric Functions in a Given Interval

Find the average value of 𝑓(π‘₯) = sin π‘₯ βˆ’ 5 sin 2π‘₯ on the interval [0, πœ‹/2].

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Video Transcript

Find the average value of 𝑓 of π‘₯ is equal to the sin of π‘₯ minus five sin of two π‘₯ on the closed interval from zero to πœ‹ by two.

We’re given a function 𝑓 of π‘₯ which is the difference between two trigonometric functions. And we’re asked to find the average value of 𝑓 of π‘₯ on the closed interval from zero to πœ‹ by two. To do this, let’s start by recalling what we mean by the average value of a function 𝑓 of π‘₯ on a closed interval from π‘Ž to 𝑏. 𝑓 average is one over 𝑏 minus π‘Ž times the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. There’s one important thing we need to notice about this definition however. We’re taking the definite integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. So 𝑓 of π‘₯ needs to be integrable on the entire domain of integration, in this case, the closed interval from π‘Ž to 𝑏.

In this case, we don’t need to worry too much about this because 𝑓 of π‘₯ is the difference between two trigonometric functions. We know this is continuous on its entire domain. And in this case, we can see its domain is all real values of π‘₯. So it’s continuous for all real values of π‘₯. And if it’s continuous for all real values of π‘₯, then it’s integrable on any interval. So in particular, it’s integrable on the closed interval from zero to πœ‹ by two.

All this does is justify that our definite integral in our formula makes sense. Now all we have to do is apply this to find our value of 𝑓 average. We’ll set 𝑓 of π‘₯ to be our function sin π‘₯ minus five sin of two π‘₯, π‘Ž to be zero, and 𝑏 to be πœ‹ by two. Now we substitute our values for π‘Ž and 𝑏 and our expression for 𝑓 of π‘₯ into our formula. This gives us 𝑓 average is equal to one over πœ‹ by two minus zero times the integral from zero to πœ‹ by two of the sin of π‘₯ minus five times the sin of two π‘₯ with respect to π‘₯.

And all that’s left to do is simplify and evaluate this expression. First, πœ‹ by two minus zero is equal to πœ‹ by two. So our coefficient simplifies to give us one divided by πœ‹ by two. However, instead of dividing by the fraction πœ‹ by two, we can instead multiplied by the reciprocal. And of course, the reciprocal of πœ‹ by two is two over πœ‹, so this gives us the following expression for 𝑓 average.

Now all that’s left to do is evaluate our definite integral. And we can evaluate this integral term by term by recalling one of our standard trigonometric integral results. For any real constant π‘Ž not equal to zero, the integral of the sin of π‘Žπ‘₯ with respect to π‘₯ is equal to negative the cos of π‘Žπ‘₯ divided by π‘Ž plus the constant of integration 𝐢. So we’ll evaluate the integral of each term separately. First, by setting π‘Ž equal to one, we know the integral of the sin of π‘Žπ‘₯ with respect to π‘₯ is equal to negative the cos of π‘₯. Next, we’re subtracting the integral of five sin of two π‘₯ with respect to π‘₯, so we’ll set π‘Ž to be equal to two, giving us the integral of this is negative five times the cos of two π‘₯ divided by two.

And of course, we’re subtracting this value. And we don’t add a constant of integration because this is a definite integral. So we need to evaluate this at the limits of integration. So we’ve now shown that 𝑓 average is equal to two over πœ‹ times negative cos of π‘₯ minus negative five times the cos of two π‘₯ over two evaluated at the limits of integration π‘₯ is equal to zero and π‘₯ is equal to πœ‹ by two.

Before we evaluate this at the limits of integration, there’s one piece of simplification we can do. We have negative one multiplied by negative five over two. Of course, this is just equal to positive five over two. This gives us the following expression. All that’s left to do now is evaluate this at the limits of integration.

Evaluating our antiderivative at the limits of integration, we get two over πœ‹ times negative the cos of πœ‹ by two plus five cos of two times πœ‹ by two all over two minus negative the cos of zero plus five times the cos of two times zero all divided by two. And we can just evaluate this expression directly. First, the cos of πœ‹ by two is equal to zero. Next, two πœ‹ simplifies to give us πœ‹. And the cos of πœ‹ is equal to negative one. And in our final two terms, we have the cos of zero. And we know the cos of zero is equal to one.

So this entire expression simplifies to give us two over πœ‹ times negative five over two plus one minus five over two. And if we calculate this expression, we see it’s equal to negative eight divided by πœ‹, which is our final answer.

Therefore, we were able to show the average value of the function 𝑓 of π‘₯ is equal to the sin of π‘₯ minus five times the sin of two π‘₯ on the closed interval from zero to πœ‹ by two is equal to negative eight divided by πœ‹.

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