Worked examples
The question
Question: find the area between y = x + 2 and y = x².
y = x + 2
y = x2
1 Draw the region
Draw both curves.
They cross twice.
The area we want is between the crossings.
2 Where they meet
They meet where the two y's are equal.
x + 2 = x2
Move everything to one side.
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
3 The top curve
Test a number between −1 and 2: x = 1.
The first gives 3, the second 1.
So y = x + 2 is on top.
x = 1: 3 > 1
The area between them is shaded.
4 Write the integral
Area between = the integral of the top one minus the bottom one.
area = ∫ from −1 to 2 of ((x + 2) − x2) dx
Careful!
Top minus bottom!
The other way gives a minus answer, −9/2.
Careful: not bottom − top ✗
A minus in front of a bracket changes the sign of every part inside.
Then like terms.
= ∫ from −1 to 2 of (−x2 + x + 2) dx
5 The antiderivative
The antiderivative: add 1 to each power, divide by the new power.
F(x) = −x33 + x22 + 2x
6 Top minus bottom
area = F(2) − F(−1)
Put in 2, then take away the same with −1.
= (−233 + 222 + 2(2)) − (−(−1)33 + (−1)22 + 2(−1))
= 103 − (−76)
So the area is 9/2 square units.
= 92 square units
7 Check
Let's check F: its derivative is −x² + x + 2: top minus bottom.
And the area is positive.
Right!
F′(x) = −x2 + x + 2
Let's look back.
First: draw the region.
Then: where they meet.
Then: the top curve.
Then: write the integral.
Then: the antiderivative.
Then: top minus bottom.
Last: check.
▶ Watch it solved
The question
Question: find the area between y = 4 − x² and y = x² − 4.
y = 4 − x2
y = x2 − 4
1 Draw the region
Draw both curves.
They cross twice.
The area we want is between the crossings.
2 Where they meet
They meet where the two y's are equal.
4 − x2 = x2 − 4
Move everything to one side.
2x2 − 8 = 0
Factor it.
x² − 4 is a difference of two squares: x² minus 2 squared.
So it is (x − 2)(x + 2).
2(x + 2)(x − 2) = 0
So x = −2 or x = 2.
x = −2 or x = 2
3 The top curve
Test a number between −2 and 2: x = 0.
The first gives 4, the second −4.
So y = 4 − x² is on top.
x = 0: 4 > −4
The area between them is shaded.
4 Write the integral
Area between = the integral of the top one minus the bottom one.
area = ∫ from −2 to 2 of ((4 − x2) − (x2 − 4)) dx
Careful!
Top minus bottom!
The other way gives a minus answer, −64/3.
Careful: not bottom − top ✗
A minus in front of a bracket changes the sign of every part inside.
Then like terms.
= ∫ from −2 to 2 of (−2x2 + 8) dx
5 The antiderivative
The antiderivative: add 1 to each power, divide by the new power.
F(x) = −2 × x33 + 8x
Now we have F(x) = −2x³ over 3 + 8x.
= −2x33 + 8x
6 Top minus bottom
area = F(2) − F(−2)
Put in 2, then take away the same with −2.
= (−2(2)33 + 8(2)) − (−2(−2)33 + 8(−2))
= 323 − (−323)
So the area is 64/3 square units.
= 643 square units
7 Check
Let's check F: its derivative is −2x² + 8: top minus bottom.
And the area is positive.
Right!
F′(x) = −2x2 + 8
Let's look back.
First: draw the region.
Then: where they meet.
Then: the top curve.
Then: write the integral.
Then: the antiderivative.
Then: top minus bottom.
Last: check.
Answer = 643 square units
▶ Watch it solved
The question
Question: find the area between y = 4 − x² and y = x + 2.
y = 4 − x2
y = x + 2
1 Draw the region
Draw both curves.
They cross twice.
The area we want is between the crossings.
2 Where they meet
They meet where the two y's are equal.
4 − x2 = x + 2
Move everything to one side.
x2 + x − 2 = 0
Factor it.
We need two numbers that multiply to −2 and add to 1: 2 and −1.
(x + 2)(x − 1) = 0
So x = −2 or x = 1.
x = −2 or x = 1
3 The top curve
Test a number between −2 and 1: x = 0.
The first gives 4, the second 2.
So y = 4 − x² is on top.
x = 0: 4 > 2
The area between them is shaded.
4 Write the integral
Area between = the integral of the top one minus the bottom one.
area = ∫ from −2 to 1 of ((4 − x2) − (x + 2)) dx
Careful!
Top minus bottom!
The other way gives a minus answer, −9/2.
Careful: not bottom − top ✗
A minus in front of a bracket changes the sign of every part inside.
Then like terms.
= ∫ from −2 to 1 of (−x2 − x + 2) dx
5 The antiderivative
The antiderivative: add 1 to each power, divide by the new power.
F(x) = −x33 − x22 + 2x
6 Top minus bottom
area = F(1) − F(−2)
Put in 1, then take away the same with −2.
= (−133 − 122 + 2(1)) − (−(−2)33 − (−2)22 + 2(−2))
= 76 − (−103)
So the area is 9/2 square units.
= 92 square units
7 Check
Let's check F: its derivative is −x² − x + 2: top minus bottom.
And the area is positive.
Right!
F′(x) = −x2 − x + 2
Let's look back.
First: draw the region.
Then: where they meet.
Then: the top curve.
Then: write the integral.
Then: the antiderivative.
Then: top minus bottom.
Last: check.
▶ Watch it solved
