Welcome back to AP Calculus with Fiveable! This topic focuses on finding the average value of a continuous function using definite integrals.
The average value of a function will allow us to solve problems that involve the accumulation of change over an interval, which will later be used to understand more difficult topics of integration.
For questions that require the average value of a function, we are never given a finite number of data points. Therefore, we must use integration to determine what the average value is.
This idea is fairly simple once you memorize a key piece of information: if f is continuous on [a,b], then the average value of f on [a,b] is the following.
Average Value=b−a1∫abf(x)dx
Image Courtesy of ExamSolutions
Here are some steps to help break down the formula!
- Set up the integral so that you integrate f(x) from a to b with respect to x, which will calculate the area under the curve between these two limits.
- Then place the fraction in front of the integral, which is simply the reciprocal of the difference between a and b.
- Evaluating this expression allows you to get the average y-value of this function between [a,b]
If the formula still seems a little difficult to understand due to its notation, practice questions are the best way to better understand its use!
Consider the function f(x)=2x2−3x+5 on the interval [1,4]. Find the average value of this function on the interval.
In this case, a = 1 and b = 4. So we begin by subbing the 1 and 4 into both the denominator of the fraction in front of the integral and the limits of the integral.
Average Value=4−11∫14(2x2−3x+5)dx
Next, take the integral of f(x).
=31[32x3−23x2+5x]14
Finally, we can sub in the limits and evaluate.
31[(32(4)3−23(4)2+5(4))−(32(1)3−23(1)2+5(1))]=223
Time for you to practice some questions yourself! ⬇️
Give each of these problems a try before you move onto the solutions!
- What is the average value of 5x2+4 on the interval 0≤x≤6?
- What is the average value of x3−x2 on the interval 2≤x≤5?
- What is the average value of sin(x)+cos(x) on the interval 0≤x≤π?
Question 1 Solution
Average Value=6−01∫06(5x2+4)dx
=61[35x3+4x]06
61[(35(6)3+4(6))−(35(0)3+4(0))]=64
Question 2 Solution
Average Value=5−21∫25(x3−x2)dx
=31[41x4−31x3]25
31[(41(5)4−31(5)3)−(41(2)4−31(2)3)]=4151
Question 3 Solution
Average Value=π−01∫0πsin(x)+cos(x)dx
=π1[−cos(x)+sin(x)]0π
π1[(−cos(π)+sin(π))−(−cos(0)+sin(0))]=π2
Great job! This topic often shows up as part (a) of FRQs, so keep this in mind for the AP.