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

Find the average value of 𝑓(π‘₯) = sin π‘₯ cos⁴π‘₯ on the interval [0, πœ‹/2].

04:12

Video Transcript

Find the average value of the function 𝑓 of π‘₯ is equal to the sin of π‘₯ multiplied by the cos to the fourth power of π‘₯ on the closed interval zero to πœ‹ over two.

We start by recalling that the average value of a function over a closed interval from π‘Ž to 𝑏 is equal to one over 𝑏 minus π‘Ž multiplied by the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯, with respect to π‘₯. In our question, we have that 𝑓 of π‘₯ is equal to sin of π‘₯ multiplied by the cos to the fourth power of π‘₯. And we see that this is defined on the closed interval zero to πœ‹ over two, so we can have that π‘Ž is equal to zero and that 𝑏 is equal to πœ‹ over two.

Now, we can substitute this information into our formula to give us that the average value of the function 𝑓 of π‘₯ over the interval zero to πœ‹ over two is equal to one divided by πœ‹ over two minus zero multiplied by the integral from zero to πœ‹ over two of the sin of π‘₯ multiplied by the cos to the fourth power of π‘₯, with respect to π‘₯. We could start by noticing that one divided by πœ‹ over two minus zero can be simplified to just be two divided by πœ‹.

Now, to evaluate this integral, we need to notice that sin π‘₯ is a multiple of the derivative of the inner part of our composite function, the cos to the fourth power of π‘₯. What this tells us is that we can use integration by substitution to evaluate this integral. So, if we let 𝑒 be equal to the cos of π‘₯, we must have the derivative of 𝑒 with respect to π‘₯ will be equal to negative sin of π‘₯. We must be careful at this point since d𝑒 by dπ‘₯ is not actually a fraction. But when we are performing an integration by substitution, we can treat it a little bit like a fraction. This would give us that d𝑒 is equal to negative the sin of π‘₯ dπ‘₯.

We see that we now have a negative sin π‘₯ dπ‘₯, but our integrand has sin π‘₯ dπ‘₯. We can get around this by multiplying our integrand by negative one and multiplying the whole integral by negative one. We can then use the fact that d𝑒 is equal to negative the sin of π‘₯ dπ‘₯ to rewrite our integral as negative two over πœ‹ multiplied by the integral of the cos to the fourth power of π‘₯ d𝑒, where we don’t write the limits to our integral yet because we know an integration by substitution will change the value of our limits.

We now use our substitution that 𝑒 is equal to the cos of π‘₯ to rewrite our integrand of the cos to the fourth power of π‘₯ as just 𝑒 to the fourth power. We will now alter the limits of our integral by using substitution, 𝑒 is equal to the cos of π‘₯. We have when π‘₯ is equal to πœ‹ over two, 𝑒 is equal to the cos of πœ‹ over two, which we know is equal to zero. So, we can write the upper limit of our integral as zero.

Similarly, when π‘₯ is equal to zero, we will have that 𝑒 is equal to the cos of zero, which is equal to one, giving us that the lower limit of our integral is one. We can now evaluate this integral by using the power rule for integrals, which says that if 𝑛 is not equal to negative one, then the integral of π‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus our constant of integration, 𝑐. Where it’s worth noting that we can ignore the constant of integration in this case since we are working with a definite integral.

This gives us that the integral of 𝑒 to the fourth power is equal to 𝑒 to the four plus one divided by four plus one, which we can simplify to just be 𝑒 to the fifth power divided by five. We can then evaluate this at the limits of zero and one to give us negative two over πœ‹ multiplied by zero to the fifth power over five minus one to the fifth power over five. Which, if we calculate, will give us two divided by five πœ‹. Meaning that the average value of the function 𝑓 of π‘₯ is equal to sin of π‘₯ multiplied by cos to the fourth power of π‘₯ on the closed interval zero to πœ‹ over two is equal to two divided by five πœ‹.

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