Question Video: Finding the Average Value of a Function on a Given Interval Involving Using Integration by Substitution Mathematics

Video Transcript

Find the average value of the function 𝑓 of 𝑡 is equal to 𝑒 to the power of sin 𝑡 times the cos of 𝑡 on the closed interval from zero to 𝜋 by two.

The question gives us a function 𝑓 of 𝑡. And it wants us to find the average value of this function on the closed interval from zero to 𝜋 by two. Let’s start by recalling what the average value of a function on a closed interval is. We say if 𝑓 is continuous on the closed interval from 𝑎 to 𝑏, then the average value of 𝑓 on this interval is equal to one divided by 𝑏 minus 𝑎 times the integral from 𝑎 to 𝑏 of 𝑓 of 𝑥 with respect to 𝑥. In our case, we’re finding the average value of 𝑓 of 𝑡 is equal to 𝑒 to the power of sin 𝑡 times the cos of 𝑡 on the closed interval from zero to 𝜋 by two. So our function 𝑓 is 𝑒 to the sin 𝑡 times the cos of 𝑡. We have 𝑎 is equal to zero and 𝑏 is equal to 𝜋 by two.

The first thing we need to use this formula is that our function is continuous on this interval. And we can see this directly from our definition of 𝑓 of 𝑡. First, both the cos of 𝑡 and the sin of 𝑡 are continuous for all real values of 𝑡. Next, we also know the exponential function is continuous for all real values of 𝑡. So our function 𝑓 of 𝑡 is the composition of continuous functions and the product of continuous functions. So our function 𝑓 of 𝑡 is continuous for all real values of 𝑡. In particular, this means it’s continuous on the closed interval from zero to 𝜋 by two.

So we can use this formula to find the average value of our function on this interval. We get 𝑓 average is equal to one divided by 𝜋 by two minus zero times the integral from zero to 𝜋 by two of 𝑒 to the power of sin 𝑡 times the cos of 𝑡 with respect to 𝑡. And it’s worth noting it doesn’t matter if we use our variable of 𝑡 or 𝑥. But since the question uses 𝑡, we’ll continue to use 𝑡.

So to evaluate this, we need to find an integral of 𝑒 to the power of sin 𝑡 times the cos of 𝑡. There’s actually several different ways of evaluating this integral. For example, we could differentiate 𝑒 to the power of sin 𝑡 by using the chain rule. And if we did this, we know the derivative of the sin of 𝑡 is the cos of 𝑡 and the derivative of 𝑒 to the power of 𝑡 is just 𝑒 to the power of 𝑡. So the derivative of 𝑒 to the power of sin 𝑡 is 𝑒 to the power of sin 𝑡 times the cos of 𝑡. And this is actually equal to our integrand, so our integrand is a derivative.

In other words, we’ve shown 𝑒 to the power of sin 𝑡 is the antiderivative of our integrand. So 𝑓 average is equal to one divided by 𝜋 by two minus zero times 𝑒 to the power of sin 𝑡 evaluated at the limits of our integral, zero and 𝜋 by two. But finding antiderivatives in this manner can be difficult, so we’ll show a second method to evaluate this integral. We’ll do this by using a substitution. In particular, we want to notice that cos of 𝑡 is the derivative of the sin of 𝑡. This gives us our hint of using 𝑢 is equal to the sin of 𝑡 as our substitution.

We differentiate both sides of this substitution with respect to 𝑡. We get d𝑢 by d𝑡 is equal to the cos of 𝑡. And although d𝑢 by d𝑡 is not a fraction, when we’re using integration by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement d𝑢 is equal to the cos of 𝑡 d𝑡. And the cos of 𝑡 d𝑡 appears in our integrand. So we could just replace this with d𝑢.

However, before we do this, there’s one last thing we need to do. Remember, we’re evaluating a definite integral, so we need to calculate the new limits of our integral. Let’s start with the upper limit 𝑡 is equal to 𝜋 by two. We get that 𝑢 is then equal to the sin of 𝜋 by two, which is equal to one. And for our lower limit, when 𝑡 is equal to zero, we get that 𝑢 is equal to the sin of zero, which is equal to zero. So by using the substitution 𝑢 is equal to the sin of 𝑡, we’ve rewritten 𝑓 average as one divided by 𝜋 by two minus zero times the integral from zero to one of 𝑒 to the 𝑢 with respect to 𝑢. And this is now in a form which we can integrate. The integral of the exponential function is just equal to the exponential function plus a constant of integration 𝑐.

So let’s start simplifying this expression. First, one divided by 𝜋 by two minus zero is the same as one divided by 𝜋 by two. And, of course, one over 𝜋 by two is the reciprocal of 𝜋 by two, which is two over 𝜋. Next, the integral of 𝑒 to the power of 𝑢 is 𝑒 to the power of 𝑢 plus 𝑐. But this is a definite integral, so we don’t need our constant of integration. Therefore, 𝑓 average is equal to two over 𝜋 times 𝑒 to the power of 𝑢 evaluated at the limits of our integral, 𝑢 is equal to zero and 𝑢 is equal to one.

Now, all we need is to evaluate this at the limits of our integral. We get two over 𝜋 times 𝑒 to the first power minus 𝑒 to the zeroth power. And 𝑒 to the first power is equal to 𝑒, and 𝑒 to the zeroth power is equal to one. So this gives us two over 𝜋 times 𝑒 minus one. And we can distribute this over our parentheses and rearrange to get negative two over 𝜋 plus two 𝑒 over 𝜋. And this gives us our final answer.

Therefore, we’ve shown the average value of the function 𝑓 of 𝑡 is equal to 𝑒 to the power of sin 𝑡 times the cos of 𝑡 on the closed interval from zero to 𝜋 by two is equal to negative two over 𝜋 plus two 𝑒 over 𝜋.