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

Find the average value of π(π‘) = π^(sin π‘) cos π‘ on the interval [0, π/2].

05:12

### Video Transcript

Find the average value of the function π of π‘ is equal to π to the power of sin π‘ times the cos of π‘ on the closed interval from zero to π by two.

The question gives us a function π of π‘. And it wants us to find the average value of this function on the closed interval from zero to π by two. Letβs start by recalling what the average value of a function on a closed interval is. We say if π is continuous on the closed interval from π to π, then the average value of π on this interval is equal to one divided by π minus π times the integral from π to π of π of π₯ with respect to π₯. In our case, weβre finding the average value of π of π‘ is equal to π to the power of sin π‘ times the cos of π‘ on the closed interval from zero to π by two. So our function π is π to the sin π‘ times the cos of π‘. We have π is equal to zero and π 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 π of π‘. First, both the cos of π‘ and the sin of π‘ are continuous for all real values of π‘. Next, we also know the exponential function is continuous for all real values of π‘. So our function π of π‘ is the composition of continuous functions and the product of continuous functions. So our function π of π‘ is continuous for all real values of π‘. In particular, this means itβ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 π average is equal to one divided by π by two minus zero times the integral from zero to π by two of π to the power of sin π‘ times the cos of π‘ with respect to π‘. And itβs worth noting it doesnβt matter if we use our variable of π‘ or π₯. But since the question uses π‘, weβll continue to use π‘.

So to evaluate this, we need to find an integral of π to the power of sin π‘ times the cos of π‘. Thereβs actually several different ways of evaluating this integral. For example, we could differentiate π to the power of sin π‘ by using the chain rule. And if we did this, we know the derivative of the sin of π‘ is the cos of π‘ and the derivative of π to the power of π‘ is just π to the power of π‘. So the derivative of π to the power of sin π‘ is π to the power of sin π‘ times the cos of π‘. And this is actually equal to our integrand, so our integrand is a derivative.

In other words, weβve shown π to the power of sin π‘ is the antiderivative of our integrand. So π average is equal to one divided by π by two minus zero times π to the power of sin π‘ evaluated at the limits of our integral, zero and π by two. But finding antiderivatives in this manner can be difficult, so weβll show a second method to evaluate this integral. Weβll do this by using a substitution. In particular, we want to notice that cos of π‘ is the derivative of the sin of π‘. This gives us our hint of using π’ is equal to the sin of π‘ as our substitution.

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

However, before we do this, thereβs one last thing we need to do. Remember, weβre evaluating a definite integral, so we need to calculate the new limits of our integral. Letβs start with the upper limit π‘ is equal to π by two. We get that π’ is then equal to the sin of π by two, which is equal to one. And for our lower limit, when π‘ is equal to zero, we get that π’ is equal to the sin of zero, which is equal to zero. So by using the substitution π’ is equal to the sin of π‘, weβve rewritten π average as one divided by π by two minus zero times the integral from zero to one of π to the π’ with respect to π’. 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 π.

So letβ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 π to the power of π’ is π to the power of π’ plus π. But this is a definite integral, so we donβt need our constant of integration. Therefore, π average is equal to two over π times π to the power of π’ evaluated at the limits of our integral, π’ is equal to zero and π’ is equal to one.

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

Therefore, weβve shown the average value of the function π of π‘ is equal to π to the power of sin π‘ times the cos of π‘ on the closed interval from zero to π by two is equal to negative two over π plus two π over π.