Worked examples
The question
Question: find the average value of f(x) = 2x + 1 on [0, 4].
f(x) = 2x + 1
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 4 minus 0: 4.
= 14 − 0∫ from 0 to 4 of (2x + 1) 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 x² + x.
∫ from 0 to 4 of (2x + 1) dx = (x2 + x) from 0 to 4
Put in 4, then take away the same with 0.
= (42 + 4) − (02 + 0)
That is 20 minus 0: 20.
= 20 − 0 = 20
3 Divide by the width
Divide by the width, 4.
The average value is 5.
average = 204 = 5
Careful!
20 is the area.
To get the average height, divide it by the width, 4.
Careful: not average = 20 ✗
4 Check
Let's check: a rectangle 5 high and 4 wide has area 20: the same as the integral.
Right!
5 × 4 = 20 = 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 = 20 ✗
▶ Watch it solved
The question
Question: find the average value of f(x) = 2x + 1 on [1, 4], and the number c where f(c) is the average.
f(x) = 2x + 1
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 4 minus 1: 3.
= 14 − 1∫ from 1 to 4 of (2x + 1) 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 x² + x.
∫ from 1 to 4 of (2x + 1) dx = (x2 + x) from 1 to 4
Put in 4, then take away the same with 1.
= (42 + 4) − (12 + 1)
That is 20 minus 2: 18.
= 20 − 2 = 18
3 Divide by the width
Divide by the width, 3.
The average value is 6.
average = 183 = 6
Careful!
18 is the area.
To get the average height, divide it by the width, 3.
Careful: not average = 18 ✗
4 Where f(c) is the average
Now find c where f is exactly the average.
f(c) = 6
2c + 1 = 6
Take away 1 on both sides.
2c = 5
Divide by 2: c = 5/2.
c = 52
5 Check
Let's check: a rectangle 6 high and 3 wide has area 18: the same as the integral.
Right!
6 × 3 = 18 = 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.
Then: where f(c) is the average.
Last: check.
Answer Careful: not average = 18 ✗
▶ Watch it solved
The question
Question: find the average value of f(x) = x² on [0, 6], and the number c where f(c) is the average.
f(x) = x2
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 6 minus 0: 6.
= 16 − 0∫ from 0 to 6 of x2 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 x³ over 3.
∫ from 0 to 6 of x2 dx = (x33) from 0 to 6
Put in 6, then take away the same with 0.
= (633) − (033)
That is 72 minus 0: 72.
= 72 − 0 = 72
3 Divide by the width
Divide by the width, 6.
The average value is 12.
average = 726 = 12
Careful!
72 is the area.
To get the average height, divide it by the width, 6.
Careful: not average = 72 ✗
4 Where f(c) is the average
Now find c where f is exactly the average.
f(c) = 12
c2 = 12
Take the square root: c is plus or minus √12, about 3.46.
c = ±√12 ≈ ±3.46
Only c between 0 and 6 counts: c = √12.
c = √12 ≈ 3.46
5 Check
Let's check: a rectangle 12 high and 6 wide has area 72: the same as the integral.
Right!
12 × 6 = 72 = area
Let's look back.
First: the formula.
Then: the integral.
Then: divide by the width.
Then: where f(c) is the average.
Last: check.
Answer Careful: not average = 72 ✗
▶ Watch it solved
