Double integral of x + 1 over 0 ≤ x ≤ 2, 0 ≤ y ≤ 4 Step by Step

Step by step

The question

Question: find the double integral of x + 1/0 ≤ x ≤ 2, 0 ≤ y ≤ 4.

∫ ∫ (x + 1) dA, 0 ≤ x ≤ 2, 0 ≤ y ≤ 4

1 Draw it

The rectangle lies on the floor: x from 0 to 2, y from 0 to 4.

Above it is the surface z = x + 1.

The double integral is the volume of this solid.

2 Find the limits

Write it as two integrals.

Inside: y from 0 to 4.

Outside: x from 0 to 2.

∫ ∫ (x + 1) dA = ∫ from 0 to 2 of ∫ from 0 to 4 of (x + 1) 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 0 to 4 of (x + 1) dy = [xy + y] from 0 to 4

Put in the top y = 4, then take away the bottom y = 0.

∫ from 0 to 4 of (x + 1) dy = (x(4) + (4)) − (x(0) + (0))

The bottom part is 0.

So it is 4x + 4.

∫ from 0 to 4 of (x + 1) dy = 4x + 4

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 (4x + 4) dx = [2x² + 4x] from 0 to 2

Put in the top x = 2, then take away the bottom x = 0.

∫ from 0 to 2 of (4x + 4) dx = (2(2)² + 4(2)) − (2(0)² + 4(0))

The bottom part is 0.

So it is 16.

∫ from 0 to 2 of (4x + 4) dx = 16

5 Check

Let's check in the other order.

x first: 4.

∫ from 0 to 2 of (x + 1) dx = 4

Then y: 16.

The same answer.

Right!

∫ from 0 to 4 of 4 dy = 16

Let's look back.

First: draw it.

Then: find the limits.

Then: the inner integral.

Then: the outer integral.

Last: check.

Answer