Area between y = x^2 - 4 and the x-axis from x = 0 to x = 3 Step by Step

Step by step

The question

Question: find the area between y = x² − 4 and the x-axis from x = 0 to x = 3.

y = x² − 4

1 Draw the region

Draw the curve first.

2 Where it crosses

Where does it cross the x-axis?

Where y is 0.

x² − 4 = 0

Factor it.

x² − 4 is a difference of two squares: x² minus 2 squared.

So it is (x − 2)(x + 2).

(x + 2)(x − 2) = 0

So x = −2 or x = 2.

x = −2 or x = 2

From 0 to 2 it is below the axis: shaded red.

From 2 to 3 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 − 4x

4 Each piece

The first piece: from 0 to 2.

Top minus bottom.

∫ from 0 to 2 of f(x) dx = F(2) − F(0)

= (2³3 − 4(2)) − (0³3 − 4(0))

= −163 − 0

That is −16/3.

= −163

Now the second piece: from 2 to 3.

Top minus bottom.

∫ from 2 to 3 of f(x) dx = F(3) − F(2)

= (3³3 − 4(3)) − (2³3 − 4(2))

= −3 − (−163)

That is 7/3.

= 73

Careful!

Not one integral from 0 to 3!

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

Careful: not area = ∫ from 0 to 3 of f(x) dx = −3 ✗

5 Add the areas

−16/3 is below the axis.

Its area is the size, without the minus: 16/3.

below: area = 163

Add the areas of the pieces.

area = 163 + 73

So the area is 23/3 square units.

= 233 square units

6 Check

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

F(3) − F(0) = −3

The signed pieces add up to the same: −16/3 plus 7/3.

Right!

−163 + 73 = −3

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