Worked examples
The question
Question: find the area between y = x² − 4x and the x-axis.
y = x2 − 4x
1 Draw the region
Draw the curve first.
2 Where it crosses
Where does it cross the x-axis?
Where y is 0.
x2 − 4x = 0
Factor it.
We need two numbers that multiply to 0 and add to −4: 0 and −4.
x(x − 4) = 0
So x = 0 or x = 4.
x = 0 or x = 4
From 0 to 4 it is below the axis: shaded red.
An integral counts area below the x-axis as negative.
But an area is never negative.
So we work out each piece on its own.
∫ below the x-axis → a minus number
3 The antiderivative
The antiderivative: add 1 to each power, divide by the new power.
F(x) = x33 − 4 × x22
Now we have F(x) = x³ over 3 − 2 x².
= x33 − 2x2
4 Each piece
The integral: from 0 to 4.
Top minus bottom.
∫ from 0 to 4 of f(x) dx = F(4) − F(0)
= (433 − 2(4)2) − (033 − 2(0)2)
= −323 − 0
That is −32/3.
= −323
Careful!
An area is never negative.
The minus only says: below the axis.
Careful: not area = −323 ✗
5 Add the areas
−32/3 is below the axis.
Its area is the size, without the minus: 32/3.
below: area = 323
So the area is 32/3 square units.
area = 323 square units
6 Check
Let's check F: its derivative is x² − 4x, the function we started with.
Right!
F′(x) = x2 − 4x
Let's look back.
First: draw the region.
Then: where it crosses.
Then: the antiderivative.
Then: each piece.
Then: add the areas.
Last: check.
Answer area = 323 square units
▶ Watch it solved
The question
Question: find the area between y = x − 3 and the x-axis from x = 0 to x = 3.
y = x − 3
1 Draw the region
Draw the curve first.
From 0 to 3 it is below the axis: shaded red.
An integral counts area below the x-axis as negative.
But an area is never negative.
So we work out each piece on its own.
∫ below the x-axis → a minus number
2 The antiderivative
The antiderivative: add 1 to each power, divide by the new power.
F(x) = x22 − 3x
3 Each piece
The integral: from 0 to 3.
Top minus bottom.
∫ from 0 to 3 of f(x) dx = F(3) − F(0)
= (322 − 3(3)) − (022 − 3(0))
= −92 − 0
That is −9/2.
= −92
Careful!
An area is never negative.
The minus only says: below the axis.
Careful: not area = −92 ✗
4 Add the areas
−9/2 is below the axis.
Its area is the size, without the minus: 9/2.
below: area = 92
So the area is 9/2 square units.
area = 92 square units
5 Check
Let's check F: its derivative is x − 3, the function we started with.
Right!
F′(x) = x − 3
Let's look back.
First: draw the region.
Then: the antiderivative.
Then: each piece.
Then: add the areas.
Last: check.
Answer area = 92 square units
▶ Watch it solved
The question
Question: find the area between y = x² − x − 2 and the x-axis from x = 0 to x = 3.
y = x2 − x − 2
1 Draw the region
Draw the curve first.
2 Where it crosses
Where does it cross the x-axis?
Where y is 0.
x2 − x − 2 = 0
Factor it.
We need two numbers that multiply to −2 and add to −1: 1 and −2.
(x + 1)(x − 2) = 0
So x = −1 or x = 2.
x = −1 or x = 2
From 0 to 2 it is below the axis: shaded red.
From 2 to 3 the curve is above the axis: shaded green.
An integral counts area below the x-axis as negative.
But an area is never negative.
So we work out each piece on its own.
∫ below the x-axis → a minus number
3 The antiderivative
The antiderivative: add 1 to each power, divide by the new power.
F(x) = x33 − x22 − 2x
4 Each piece
The first piece: from 0 to 2.
Top minus bottom.
∫ from 0 to 2 of f(x) dx = F(2) − F(0)
= (233 − 222 − 2(2)) − (033 − 022 − 2(0))
= −103 − 0
That is −10/3.
= −103
Now the second piece: from 2 to 3.
Top minus bottom.
∫ from 2 to 3 of f(x) dx = F(3) − F(2)
= (333 − 322 − 2(3)) − (233 − 222 − 2(2))
= −32 − (−103)
That is 11/6.
= 116
Careful!
Not one integral from 0 to 3!
The piece below the axis counts as negative, so the pieces cancel.
Careful: not area = ∫ from 0 to 3 of f(x) dx = −32 ✗
5 Add the areas
−10/3 is below the axis.
Its area is the size, without the minus: 10/3.
below: area = 103
Add the areas of the pieces.
area = 103 + 116
So the area is 31/6 square units.
= 316 square units
6 Check
Let's check the pieces: the whole integral from 0 to 3 is −3/2.
F(3) − F(0) = −32
The signed pieces add up to the same: −10/3 plus 11/6.
Right!
−103 + 116 = −32
Let's look back.
First: draw the region.
Then: where it crosses.
Then: the antiderivative.
Then: each piece.
Then: add the areas.
Last: check.
Answer = 316 square units
▶ Watch it solved
