Video Transcript
What is the approximate average value of the function 𝑓 of 𝑥 equals six 𝑥 squared over three 𝑥 cubed plus five on the closed interval from two to six?
Let’s recall the formula for the average value of a function. If 𝑓 of 𝑥 is a continuous function on the closed interval from 𝑎 to 𝑏, then the average value of the function on the interval is given by the formula one over 𝑏 minus 𝑎 times by the integral from 𝑎 to 𝑏 of 𝑓 with respect to 𝑥. Before we start applying the formula, let’s check that the function 𝑓, given to us in the question, is indeed continuous.
A function 𝑓 of 𝑥 is continuous at 𝑥 equals 𝑐 if all of the following conditions exist. Number one, 𝑓 of 𝑐 exists. Number two, the limit as 𝑥 tends to 𝑐 𝑓 of 𝑥 exists. And number three, the limit as 𝑥 tends to 𝑐 of 𝑓 of 𝑥 equals 𝑓 evaluated at 𝑐. A function is continuous on an interval if it is continuous at every point in that interval. So in order to check that the function given to us in the question is continuous on the closed interval from two to six. We need to check that the three conditions for continuity hold for every point in the closed interval from two to six. Let 𝑓 of 𝑥 equal six 𝑥 squared over three 𝑥 cubed plus five and 𝑐 be an arbitrary point in the closed interval from two to six.
Number one, does 𝑓 of 𝑐 exist? Well, the functions six 𝑥 squared and three 𝑥 cubed plus five can both be evaluated at 𝑐. So their quotient 𝑓 can also be evaluated at 𝑐. Apart from when the denominator three 𝑥 cubed plus five is equal to zero as something divided by zero is undefined. Solving the equation three 𝑥 cubed plus five equals zero, we obtain a value of 𝑥 equal to the cube root of negative five over three, which, using a calculator, is equal to negative 1.19 to two decimal places. So 𝑓 of 𝑐 exists unless 𝑐 is equal to the cube root of negative five over three. However, the cube root of negative five over three is approximately equal to negative 1.19. And so it is not contained in the closed interval from two to six. Therefore, 𝑐 can never take that value. And so 𝑓 of 𝑐 does indeed exist. Number two, does the limit as 𝑥 tends to 𝑐 of 𝑓 of 𝑥 exist? Well, using direct substitution, the limit as 𝑥 tends to 𝑐 of 𝑓 of 𝑥 equals 𝑓 evaluated at 𝑐, which is defined, as we saw in the first condition. So this settles the continuity conditions two and three together.
Now that we are satisfied that 𝑓 is continuous on the closed interval from two to six, let’s apply the formula for the average value of a function. We have that 𝑎 equals two and 𝑏 equals six. So the coefficient outside of the integral becomes one over six minus two, which is just equal to one over four. Substituting in 𝑎 equals two and 𝑏 equals six for the limits of integration, we obtain that the average value of the function 𝑓 given to us in the question on the closed interval from two to six is one over four times by the integral from two to six of six 𝑥 squared over three 𝑥 cubed plus five with respect to 𝑥.
Let’s evaluate this integral using integration by substitution. Let 𝑢 of 𝑥 equal three 𝑥 cubed plus five. Then, the derivative of 𝑢 with respect to 𝑥, which is denoted by d𝑢 by d𝑥, is equal to nine 𝑥 squared. Here, we have used the fact that the derivative of a sum of functions is the sum of their derivatives as well as the standard formula for differentiating terms of the form 𝑎𝑥 to the power of 𝑏. Where we multiply the coefficient 𝑎 by the exponent 𝑛 and decreased the exponent by one and the fact that the derivative of a constant is zero. Now, we can rewrite d𝑥 as one over nine 𝑥 squared d𝑢. Substituting this as well as the function 𝑢 into our integral, we obtain that, without the limits, our integral is equal to one over four times by the integral of six 𝑥 squared over 𝑢 times by one over nine 𝑥 squared with respect to 𝑢.
We will come back to the limits of integration. Before that, let’s simplify the expression inside the integral. We can cancel the 𝑥 squared terms and take the six over nine outside the integral as it is a constant. One over four multiplied by six over nine is one over six. Now, let’s work out the limits of integration in terms of 𝑢. 𝑢 evaluated at two is equal to three times two cubed plus five, which is equal to 29. 𝑢 evaluated at six is equal to three times six cubed plus five, which is equal to 653. So the average value required in the question is given by one over six times by the integral from 29 to 653 of one over 𝑢 with respect to 𝑢.
Recall that the indefinite integral of one over 𝑥 with respect to 𝑥 is the natural logarithm of the absolute value of 𝑥 plus the constant of integration 𝑐. So the integral in question is equal to one over six times by the natural logarithm of the absolute value of 𝑢, evaluated from 29 to 653. This is equal to one over six times by the natural logarithm of 653 minus the natural logarithm of 29. Using a calculator, this is equal to 0.519 to three decimal places.
So the average value of the function 𝑓 of 𝑥 equals six 𝑥 squared over three 𝑥 cubed plus five on the closed interval from two to six is approximately 0.519.
