Steply

Double integral of x + y over 0 ≤ x ≤ 1, x^2 ≤ y ≤ x

Step by step, with the reason for every step · Double integrals over general regions · Calculus 3

Step by step

The question

Question: find the double integral of x + y/0 ≤ x ≤ 1, x² ≤ y ≤ x.

∫ ∫ (x + y) dA, 0 ≤ x ≤ 1, x2 ≤ y ≤ x

1 Draw the region

Draw the grid.

Draw y = x

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 x.

Outside, x goes from 0 to 1.

∫ ∫ (x + y) dA = ∫ from 0 to 1 of ∫ from x² to x of (x + y) 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 + y) dy = [xy + y22] from x² to x

Put in the top y = x, then take away the bottom y = x².

∫ from x² to x of (x + y) dy = (x(x) + (x)22) − (x(x2) + (x2)22)

Work out each term on its own: multiply out the brackets.

∫ from x² to x of (x + y) dy = (x2 + x22) − (x3 + x42)

Add the like terms in each part.

∫ from x² to x of (x + y) dy = (3x22) − (x3 + x42)

So it is 3x² over 2 − x³ − x⁴ over 2.

∫ from x² to x of (x + y) dy = 3x22 − x3 − x42

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 (3x22 − x3 − x42) dx = [x32 − x44 − x510] from 0 to 1

Put in the top x = 1, then take away the bottom x = 0.

∫ from 0 to 1 of (3x22 − x3 − x42) dx = ((1)32 − (1)44 − (1)510) − ((0)32 − (0)44 − (0)510)

The bottom part is 0.

So it is 3/20.

∫ from 0 to 1 of (3x22 − x3 − x42) dx = 320

5 Check

Let's check if it makes sense.

The region's area is 1/6.

On it, f goes from 0 to 2.

area = 1/6, f from 0 to 2

So the answer must be between 0 and 0.333.

0.15 is.

Right!

0 ≤ 0.15 ≤ 0.333

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 1 of (3x22 − x3 − x42) dx = 320

Your turn

Try this one yourself, then check:

Find the double integral of ∫ ∫ (2x + 4y) dA, 0 ≤ x ≤ 1, x2 ≤ y ≤ x
Show the answer
1330

More from Double integrals over general regions

The whole lesson: Double integrals over general regions →