Area between y = x^2 - 9 and the x-axis Step by Step

Step by step

The question

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

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 −3 to 3 it is below the axis: shaded red.

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 integral: from −3 to 3.

Top minus bottom.

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

= (3³3 − 9(3)) − ((−3)³3 − 9(−3))

= −18 − 18

That is −36.

= −36

Careful!

An area is never negative.

The minus only says: below the axis.

Careful: not area = −36 ✗

5 Add the areas

−36 is below the axis.

Its area is the size, without the minus: 36.

below: area = 36

So the area is 36 square units.

area = 36 square units

6 Check

Let's check F: its derivative is x² − 9, the function we started with.

Right!

F′(x) = x² − 9

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