Grade 12 · Average function values · 204 of 600
Average function values · Multiplier a=3; Power p=1
Use the interval 0≤x≤1.28. Find the endpoint value, integral, and average function height. Givens: Multiplier a=3; Power p=1.
Try the question, use a hint when you need one, and compare your reasoning with the worked solution. The animation starts with this chapter’s values; changing its controls explores a new case.
01 · Make a prediction
Your practice question
Let f(x)=3x^1. Use the interval 0≤x≤1.28. Find the endpoint value, integral, and average function height.
Use this scratch space or work on paper. Your notes stay on this page and clear when you leave. Answers are for self-checking; they are not automatically graded.
A starting point
Divide the integral by the interval length 1.28; an average height is not the integral itself.
Work through the reasoning
Step 1
Identify the model and target
Let f(x)=3x^1. Use the interval 0≤x≤1.28. The governing relation is f(x)=axᵖ; average on [0,T]=aTᵖ/(p+1). Divide the integral by the interval length 1.28; an average height is not the integral itself.
Step 2
Substitute and calculate
The integral is 3(1.28)^(1+1)/(1+1)=2.4576. Dividing by interval length 1.28 gives average height 1.92; this is endpoint value 3.84 divided by 2.
Step 3
Check the mathematical meaning
Average height times 1.28 equals the integral 2.4576. Because this function is increasing and nonnegative, its average 1.92 lies between zero and the endpoint value 3.84. Interval length: 1.28; Endpoint value: 3.84; Integral: 2.458; Average height: 1.92. Decimal values are rounded, so use unrounded intermediate values.
The answer
The integral is 3(1.28)^(1+1)/(1+1)=2.4576. Dividing by interval length 1.28 gives average height 1.92; this is endpoint value 3.84 divided by 2. Average height times 1.28 equals the integral 2.4576. Because this function is increasing and nonnegative, its average 1.92 lies between zero and the endpoint value 3.84. Animation check: Interval length: 1.28; Endpoint value: 3.84; Integral: 2.458; Average height: 1.92. Decimal displays are rounded; retain the original parameters and exact π until the final step.
Compare the method as well as the result. A different valid method may reach the same answer. Keep exact values until the last step when the question asks for rounding.
02 · Explore the model
See the mathematical relationship move.
Use Play, Pause, and the timeline to inspect the construction. Reset restores the question’s original settings. The displayed assumptions describe where this model applies.
Watch the relationship
PausedHD animation studio
From experiment to screen.
Present a crisp Canvas scene, save a full-HD image, or capture your model as a silent video.
The mathematical idea
The average value of a function is the constant height with the same signed area over the interval. Dividing the integral by interval length supplies that height. For a positive power on [0,T], the average is a fixed fraction of the endpoint value, and that fraction depends on the power. Let f(x)=3x^1. Use the interval 0≤x≤1.28. The requested state occurs at 60% playback; use the exact target stated in the question for your calculation.
f(x)=axᵖ; average on [0,T]=aTᵖ/(p+1)
03 · Reflect and transfer
Explain what changes and why.
For a fixed endpoint, why is the average height exactly 1/(p+1) times the endpoint value?
This is a distinct guided scenario using a reusable mathematical model. Similar-looking diagrams can represent different given values and conclusions; they are not different mathematical theories.