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Grade 12 · Average function values

Average function values: A scaled fourth power

Average function values: investigate a scaled fourth power with multiplier a = 2.5; power p = 4.

Predict what will happen, press Play, then pause and explain what changed. Every control also works without playback.

Watch the relationship

Paused
Average function values: A scaled fourth power. Interval length: 0.2. Endpoint value: 0.004. Integral: 0. Average height: 0.001The same area at a constant height040012yx → · labeled axes rescale to this model
The upper bound stays positive, avoiding division by zero. Powers are positive integers, and the average is a continuous-interval average rather than a mean of finitely many plotted samples.

Starts paused. Play once, pause anywhere, or use Step to inspect the mathematics. Playback stops when this panel leaves the screen.

Make it your experiment

Change one value. Notice what follows.

The controls adjust the model. Numbers below describe the current frame. Decimals are rounded.

Interval length
0.2
Endpoint value
0.004
Integral
0
Average height
0.001

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From experiment to screen.

Present a crisp Canvas scene, save a full-HD image, or capture your model as a silent video.

Understand what you are seeing

The idea behind the motion.

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. This investigation starts with Multiplier a = 2.5; Power p = 4. Predict the result before playing, then change one parameter while keeping the other fixed to test the reason for the change.

A relationship to keep

f(x)=axᵖ; average on [0,T]=aTᵖ/(p+1)

Read the symbols alongside the explanation. A diagram shows the relationship; the assumptions tell you when it applies.

  1. STEP 1

    Set up the mathematical model

    Choose a positive scale and an integer power, then identify the increasing curve. The starting case is “A scaled fourth power.”

  2. STEP 2

    Follow the changing quantity

    Move the right boundary from 0.2 to 2. Compare the shaded area with a rectangle of the displayed average height.

  3. STEP 3

    Explain and test the result

    Multiply average height by interval length to recover the integral. Explain why the endpoint value alone overstates the average for these increasing powers.

Your turn to explain

Make a prediction. Test your reasoning.

Keep Multiplier a = 2.5; Power p = 4. Pause the timeline at 60%. Given interval length = 1.28, calculate endpoint value, integral, average height. Show the substitution into the displayed formula.

Use the values specified in the question. Reset restores the initial values for this investigation.

Compare your explanation

The integral is 2.5(1.28)^(4+1)/(4+1)=1.717987. Dividing by interval length 1.28 gives average height 1.342177; this is endpoint value 6.710886 divided by 5. Results: Endpoint value: 6.711; Integral: 1.718; Average height: 1.342. Decimal values are rounded; retain the original parameters when checking.

Work through a full lesson

Connect the animation to a worked example and practice questions.

Grade 12 · Calculus

Turn an integral into an average height

Normalize accumulated area by interval length.