Area between y = x^2 - 9 and the x-axis from x = 0 to x = 4 Step by Step

Step by step

The question

Question: find the area between y = x² − 9 and the x-axis from x = 0 to x = 4.

y = x² − 9

1 Draw the region

Draw the curve first.

2 Where it crosses

Where does it cross the x-axis?

Where y is 0.

x² − 9 = 0

Factor it.

x² − 9 is a difference of two squares: x² minus 3 squared.

So it is (x − 3)(x + 3).

(x + 3)(x − 3) = 0

So x = −3 or x = 3.

x = −3 or x = 3

From 0 to 3 it is below the axis: shaded red.

From 3 to 4 the curve is above the axis: shaded green.

An integral counts area below the x-axis as negative.

But an area is never negative.

So we work out each piece on its own.

∫ below the x-axis → a minus number

3 The antiderivative

The antiderivative: add 1 to each power, divide by the new power.

F(x) = x³3 − 9x

4 Each piece

The first piece: from 0 to 3.

Top minus bottom.

∫ from 0 to 3 of f(x) dx = F(3) − F(0)

= (3³3 − 9(3)) − (0³3 − 9(0))

= −18 − 0

That is −18.

= −18

Now the second piece: from 3 to 4.

Top minus bottom.

∫ from 3 to 4 of f(x) dx = F(4) − F(3)

= (4³3 − 9(4)) − (3³3 − 9(3))

= −443 − (−18)

That is 10/3.

= 103

Careful!

Not one integral from 0 to 4!

The piece below the axis counts as negative, so the pieces cancel.

Careful: not area = ∫ from 0 to 4 of f(x) dx = −443 ✗

5 Add the areas

−18 is below the axis.

Its area is the size, without the minus: 18.

below: area = 18

Add the areas of the pieces.

area = 18 + 103

So the area is 64/3 square units.

= 643 square units

6 Check

Let's check the pieces: the whole integral from 0 to 4 is −44/3.

F(4) − F(0) = −443

The signed pieces add up to the same: −18 plus 10/3.

Right!

−18 + 103 = −443

Let's look back.

First: draw the region.

Then: where it crosses.

Then: the antiderivative.

Then: each piece.

Then: add the areas.

Last: check.

Answer