Steply

Double integral of 6 over 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 - 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 6/0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x.

∫ ∫ 6 dA, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x

1 Draw the region

Draw the grid.

Draw y = 1 − x

and y = 0.

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 0 to 1 − x.

Outside, x goes from 0 to 1.

∫ ∫ 6 dA = ∫ from 0 to 1 of ∫ from 0 to 1 − x of 6 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 1 − x of 6 dy = 6y from 0 to 1 − x

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

∫ from 0 to 1 − x of 6 dy = 6(1 − x) − 6(0)

The bottom part is 0.

So it is 6 − 6x.

∫ from 0 to 1 − x of 6 dy = 6 − 6x

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 (6 − 6x) dx = [6x − 3x2] from 0 to 1

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

∫ from 0 to 1 of (6 − 6x) dx = (6(1) − 3(1)2) − (6(0) − 3(0)2)

The bottom part is 0.

So it is 3.

∫ from 0 to 1 of (6 − 6x) dx = 3

5 Check

Let's check if it makes sense.

The region's area is 1/2.

On it, f goes from 6 to 6.

area = 1/2, f from 6 to 6

So the answer must be between 3 and 3.

3 is.

Right!

3 ≤ 3 ≤ 3

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 (6 − 6x) dx = 3

Your turn

Try this one yourself, then check:

Find the double integral of ∫ ∫ 2 dA, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x
Show the answer
1

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