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Average value of f(x) = x^3 on [0, 2]

Step by step, with the reason for every step · Average value of a function · Calculus 1

Step by step

The question

Question: find the average value of f(x) = x³ on [0, 2].

f(x) = x3

1 The formula

The average value: the area under the curve, divided by the width of the interval.

average = 1b − a∫ from a to b of f(x) dx

Here the width is 2 minus 0: 2.

= 12 − 0∫ from 0 to 2 of x3 dx

Think of it as the height of a rectangle with the same area as under the curve.

average height × width = area

2 The integral

The antiderivative: add 1 to each power, divide by the new power.

Now we have x⁴ over 4.

∫ from 0 to 2 of x3 dx = (x44) from 0 to 2

Put in 2, then take away the same with 0.

= (244) − (044)

That is 4 minus 0: 4.

= 4 − 0 = 4

3 Divide by the width

Divide by the width, 2.

The average value is 2.

average = 42 = 2

Careful!

4 is the area.

To get the average height, divide it by the width, 2.

Careful: not average = 4 ✗

4 Check

Let's check: a rectangle 2 high and 2 wide has area 4: the same as the integral.

Right!

2 × 2 = 4 = area

In a picture:

the area under the curve,

and the rectangle at the average height.

The same area.

Let's look back.

First: the formula.

Then: the integral.

Then: divide by the width.

Last: check.

Answer
Careful: not average = 4 ✗

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