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

Find the average value of 𝑓(π‘₯) = π‘₯𝑒^(βˆ’2π‘₯Β²) on the interval [βˆ’2, 0].

05:54

Video Transcript

Find the average value of 𝑓 of π‘₯ is equal to π‘₯ times 𝑒 to the power of negative two π‘₯ squared on the closed interval from negative two to zero.

We’re given a function 𝑓 of π‘₯, and we’re asked to find the average value of 𝑓 of π‘₯ on the closed interval from negative two to zero. To do this, we first need to recall what we mean by the average value of a function 𝑓 of π‘₯ on a closed interval from π‘Ž to 𝑏. We say that this is equal to one over 𝑏 minus π‘Ž times the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. And we want to find the average value of π‘₯ times 𝑒 to the power of negative two π‘₯ squared on the closed interval from negative two to zero. So we’ll set our function 𝑓 of π‘₯ to be π‘₯ times 𝑒 to the power of negative two π‘₯ squared, our value of π‘Ž to be equal to negative two, and our value of 𝑏 equal to zero.

However, there is one small thing we need to worry about. We do need 𝑓 of π‘₯ to be integrable on the closed interval from π‘Ž to 𝑏. So in this case, we would need to show that our function 𝑓 of π‘₯ is integrable on the closed interval from negative two to zero. One way of doing this is to show that 𝑓 of π‘₯ is continuous on this interval, and in fact we can do this. We can see that 𝑓 of π‘₯ is π‘₯ multiplied by 𝑒 to the power of negative two π‘₯ squared. 𝑒 to the power of negative two π‘₯ squared is the composition of continuous functions. And then we multiply this by π‘₯, so 𝑓 of π‘₯ is the products and composition of continuous functions. And the products and composition of continuous functions will be continuous on its domain. In particular, we can see this is defined for all real values of π‘₯.

So 𝑓 of π‘₯ is continuous for all real values of π‘₯. And, in particular, this means it will be integrable on the closed interval from negative two to zero. Now we can find the average value of our function on this interval by substituting in our values for π‘Ž and 𝑏 and our expression for 𝑓 of π‘₯. We get it’s equal to one over zero minus negative two times the integral from negative two to zero of π‘₯ times 𝑒 to the power of negative two π‘₯ squared with respect to π‘₯. And we can simplify this. First, one over zero minus negative two is equal to one-half.

And at this point, there’s a couple of different ways we can evaluate this integral. The first way to evaluate this integral is to ask the question, what happens when we differentiate 𝑒 to the power of negative two π‘₯ squared with respect to π‘₯? This expression appears in our integrand. And by using the chain rule, we know for any differentiable function 𝑔 of π‘₯, the derivative of 𝑒 to the power of 𝑔 of π‘₯ with respect to π‘₯ is equal to 𝑔 prime of π‘₯ times 𝑒 to the power of 𝑔 of π‘₯. In our case, we can see that our function 𝑔 of π‘₯ will be negative two π‘₯ squared. And we know the derivative of this by using the power rule for differentiation is negative four π‘₯. So we get the derivative of 𝑒 to the power of negative two π‘₯ squared with respect to π‘₯ is equal to negative four π‘₯ times 𝑒 to the power of negative two π‘₯ squared.

And now we can see something interesting. π‘₯ times 𝑒 to the power of negative two π‘₯ squared is our integrand, so we could divide both sides of this equation through by negative four. Then, all we need to do is move negative one-quarter inside of our derivative. This gives us that the derivative of the following expression is equal to our integrand. But that tells us that this is an antiderivative of our integrand . So we found an antiderivative of our integrand and we could use this to evaluate our integral. However, it’s not always possible to find antiderivatives using this method. So we’ll go through a second method to help us evaluate this integral.

Instead, we’ll try and evaluate this integral by using substitution. We’ll want to use 𝑒 is equal to negative two π‘₯ squared. And the reason we choose 𝑒 is equal to negative two π‘₯ squared is if we differentiate both sides of this expression with respect to π‘₯, we get d𝑒 by dπ‘₯ is equal to negative four π‘₯. And we can see that π‘₯ appears in our integrand, and this is a scalar multiple of d𝑒 by dπ‘₯. So we’ll divide both sides of this equation through by negative four, giving us negative one-quarter d𝑒 by dπ‘₯ is equal to π‘₯. And of course, we know d𝑒 by dπ‘₯ is not a fraction. However, when we’re using integration by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement in terms of differentials. Negative one-quarter d𝑒 is equal to π‘₯dπ‘₯, and we can see π‘₯dπ‘₯ appears in our integral.

But remember, since we’re evaluating a definite integral by using substitution, we need to find the new limits of integration. To find the upper limit of integration, we’ll start by substituting π‘₯ is equal to zero into our expression for 𝑒. This gives us 𝑒 is equal to negative two times zero squared, which is of course equal to zero. We can do the same for our lower limit of integration. We need to substitute in π‘₯ is equal to negative two. We can see this gives us 𝑒 is equal to negative eight. We’re now ready to evaluate our integral by using substitution. First, we’ll change our lower limit to negative eight and our upper limit to zero. Next, remember that negative two π‘₯ squared is equal to 𝑒, so we can replace 𝑒 to the power of negative two π‘₯ squared with 𝑒 to the power of 𝑒.

Finally, by using our statements in terms of differentials, we can replace π‘₯dπ‘₯ with negative one-quarter d𝑒. And we can simplify this integral slightly. We’ll take the factor of negative one-quarter outside of our integral. This means we’ve shown that 𝑓 average is equal to negative one-eighth times the integral from negative eight to zero of 𝑒 to the power of 𝑒 with respect to 𝑒. And we know the integral of 𝑒 to the power of 𝑒 with respect to 𝑒 is equal to itself. So this is just equal to negative one-eighth times 𝑒 to the power of 𝑒 evaluated at the limits of integration, 𝑒 is equal to negative eight and 𝑒 is equal to zero.

All that’s left to do now is evaluate this at the limits of integration. Evaluating this at the limits of integration, we get negative one-eighth times 𝑒 to the zeroth power minus 𝑒 to the power of negative eight. And of course, we can simplify this. 𝑒 to the zeroth power is just equal to one. We’ll also use our laws of exponents to rewrite 𝑒 to the power of negative eight as one over 𝑒 to the power of eight. Then, all we need to do is distribute negative one-eighth over our parentheses, and this gives us our final answer. Therefore, we were able to show the average value of 𝑓 of π‘₯ is equal to π‘₯ times 𝑒 to the power of negative two π‘₯ squared on the closed interval from negative two to zero is equal to negative one-eighth plus one over eight times 𝑒 to the eighth power.

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