Worked examples
The question
Question: find the double integral of x over the region between y = x² and y = x.
∫ ∫ x dA, between y = x² and y = x
1 Draw the region
Draw the grid.
Draw y = x
and y = x².
The region is between them.
Where do they meet?
Where the two heights are equal.
x = x2
Bring everything to one side.
x − x2 = 0
Take out x.
x(1 − x) = 0
So they meet at x = 0 and x = 1.
x = 0 or x = 1
Which is on top?
Try x = 0.5: y = x is higher.
x = 0.5: x → 0.5, x² → 0.25
2 Find the limits
Take a thin upright strip.
On it, y goes from the bottom curve up to the top curve.
So inside, y goes from x² to x.
Outside, x goes from 0 to 1.
∫ ∫ x dA = ∫ from 0 to 1 of ∫ from x² to x of x dy dx
Careful!
The inside integral is dy, so its limits are the y limits.
They may have x in them.
The outside limits are only numbers.
Careful: dy inside: the y limits go inside
3 The inner integral
Inside first: integrate in y.
x stays fixed, like a plain number.
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from x² to x of x dy = xy from x² to x
Put in the top y = x, then take away the bottom y = x².
∫ from x² to x of x dy = x(x) − x(x2)
So it is x² − x³.
∫ from x² to x of x dy = x2 − x3
4 The outer integral
Now the outside integral, in x.
Its limits are numbers.
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from 0 to 1 of (x2 − x3) dx = [x33 − x44] from 0 to 1
Put in the top x = 1, then take away the bottom x = 0.
∫ from 0 to 1 of (x2 − x3) dx = ((1)33 − (1)44) − ((0)33 − (0)44)
The bottom part is 0.
So it is 1/12.
∫ from 0 to 1 of (x2 − x3) dx = 112
5 Check
Let's check if it makes sense.
The region's area is 1/6.
On it, f goes from 0 to 1.
area = 1/6, f from 0 to 1
So the answer must be between 0 and 0.167.
0.083 is.
Right!
0 ≤ 0.083 ≤ 0.167
Let's look back.
First: draw the region.
Then: find the limits.
Then: the inner integral.
Then: the outer integral.
Last: check.
Answer ∫ from 0 to 1 of (x2 − x3) dx = 112
▶ Watch it solved
The question
Question: find the double integral of 2x/0 ≤ x ≤ 2, x² ≤ y ≤ 2x.
∫ ∫ 2x dA, 0 ≤ x ≤ 2, x2 ≤ y ≤ 2x
1 Draw the region
Draw the grid.
Draw y = 2x
and y = x².
The region is between them.
2 Find the limits
Take a thin upright strip.
On it, y goes from the bottom curve up to the top curve.
So inside, y goes from x² to 2x.
Outside, x goes from 0 to 2.
∫ ∫ 2x dA = ∫ from 0 to 2 of ∫ from x² to 2x of 2x dy dx
Careful!
The inside integral is dy, so its limits are the y limits.
They may have x in them.
The outside limits are only numbers.
Careful: dy inside: the y limits go inside
3 The inner integral
Inside first: integrate in y.
x stays fixed, like a plain number.
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from x² to 2x of 2x dy = 2xy from x² to 2x
Put in the top y = 2x, then take away the bottom y = x².
∫ from x² to 2x of 2x dy = 2x(2x) − 2x(x2)
So it is 4x² − 2x³.
∫ from x² to 2x of 2x dy = 4x2 − 2x3
4 The outer integral
Now the outside integral, in x.
Its limits are numbers.
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from 0 to 2 of (4x2 − 2x3) dx = [4x33 − x42] from 0 to 2
Put in the top x = 2, then take away the bottom x = 0.
∫ from 0 to 2 of (4x2 − 2x3) dx = (4(2)33 − (2)42) − (4(0)33 − (0)42)
The bottom part is 0.
So it is 8/3.
∫ from 0 to 2 of (4x2 − 2x3) dx = 83
5 Check
Let's check if it makes sense.
The region's area is 4/3.
On it, f goes from 0 to 4.
area = 4/3, f from 0 to 4
So the answer must be between 0 and 5.333.
2.667 is.
Right!
0 ≤ 2.667 ≤ 5.333
Let's look back.
First: draw the region.
Then: find the limits.
Then: the inner integral.
Then: the outer integral.
Last: check.
Answer ∫ from 0 to 2 of (4x2 − 2x3) dx = 83
▶ Watch it solved
The question
Question: find the double integral of x + 2y over the region between y = x² and y = 2 − x².
∫ ∫ (x + 2y) dA, between y = x² and y = 2 − x²
1 Draw the region
Draw the grid.
Draw y = 2 − x²
and y = x².
The region is between them.
Where do they meet?
Where the two heights are equal.
2 − x2 = x2
Bring everything to one side.
2 − 2x2 = 0
Move the number and divide: x² = 1.
x2 = 1
So they meet at x = −1 and x = 1.
x = −1 or x = 1
Which is on top?
Try x = 0: y = 2 − x² is higher.
x = 0: 2 − x² → 2, x² → 0
2 Find the limits
Take a thin upright strip.
On it, y goes from the bottom curve up to the top curve.
So inside, y goes from x² to 2 − x².
Outside, x goes from −1 to 1.
∫ ∫ (x + 2y) dA = ∫ from −1 to 1 of ∫ from x² to 2 − x² of (x + 2y) dy dx
Careful!
The inside integral is dy, so its limits are the y limits.
They may have x in them.
The outside limits are only numbers.
Careful: dy inside: the y limits go inside
3 The inner integral
Inside first: integrate in y.
x stays fixed, like a plain number.
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from x² to 2 − x² of (x + 2y) dy = [xy + y2] from x² to 2 − x²
Put in the top y = 2 − x², then take away the bottom y = x².
∫ from x² to 2 − x² of (x + 2y) dy = (x(2 − x2) + (2 − x2)2) − (x(x2) + (x2)2)
Work out each term on its own: multiply out the brackets.
∫ from x² to 2 − x² of (x + 2y) dy = ((2x − x3) + (4 − 4x2 + x4)) − (x3 + x4)
Add the like terms in each part.
∫ from x² to 2 − x² of (x + 2y) dy = (2x − x3 + 4 − 4x2 + x4) − (x3 + x4)
So it is 2x − 2x³ + 4 − 4x².
∫ from x² to 2 − x² of (x + 2y) dy = 2x − 2x3 + 4 − 4x2
4 The outer integral
Now the outside integral, in x.
Its limits are numbers.
Integrate each term: the power goes up by 1, and we divide by the new power.
∫ from −1 to 1 of (2x − 2x3 + 4 − 4x2) dx = [x2 − x42 + 4x − 4x33] from −1 to 1
Put in the top x = 1, then take away the bottom x = −1.
∫ from −1 to 1 of (2x − 2x3 + 4 − 4x2) dx = ((1)2 − (1)42 + 4(1) − 4(1)33) − ((−1)2 − (−1)42 + 4(−1) − 4(−1)33)
So it is 16/3.
∫ from −1 to 1 of (2x − 2x3 + 4 − 4x2) dx = 163
5 Check
Let's check if it makes sense.
The region's area is 8/3.
On it, f goes from −0.13 to 4.13.
area = 8/3, f from −0.13 to 4.13
So the answer must be between −0.333 and 11.
5.333 is.
Right!
−0.333 ≤ 5.333 ≤ 11
Let's look back.
First: draw the region.
Then: find the limits.
Then: the inner integral.
Then: the outer integral.
Last: check.
Answer ∫ from −1 to 1 of (2x − 2x3 + 4 − 4x2) dx = 163
▶ Watch it solved
