Area between y = 2x^2 - 8 and the x-axis Step by Step

Step by step

The question

Question: find the area between y = 2x² − 8 and the x-axis.

y = 2x² − 8

1 Draw the region

Draw the curve first.

2 Where it crosses

Where does it cross the x-axis?

Where y is 0.

2x² − 8 = 0

Factor it.

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

So it is (x − 2)(x + 2).

2(x + 2)(x − 2) = 0

So x = −2 or x = 2.

x = −2 or x = 2

From −2 to 2 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) = 2 × x³3 − 8x

Now we have F(x) = 2x³ over 3 − 8x.

= 2x³3 − 8x

4 Each piece

The integral: from −2 to 2.

Top minus bottom.

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

= (2(2)³3 − 8(2)) − (2(−2)³3 − 8(−2))

= −323 − 323

That is −64/3.

= −643

Careful!

An area is never negative.

The minus only says: below the axis.

Careful: not area = −643 ✗

5 Add the areas

−64/3 is below the axis.

Its area is the size, without the minus: 64/3.

below: area = 643

So the area is 64/3 square units.

area = 643 square units

6 Check

Let's check F: its derivative is 2x² − 8, the function we started with.

Right!

F′(x) = 2x² − 8

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