Steply

What is the area of the region between y = x^2 - 1 and the x-axis…

Step by step, with the reason for every step · AP Calculus AB Practice Exam 1 · AP Calculus AB

Step by step

The question

Question: find the area between y = x² − 1 and the x-axis from x = 0 to x = 3.

y = x2 − 1

There are four choices.

First we solve it, the normal way.

Then we pick.

1 Draw the region

Draw the curve first.

2 Where it crosses

Where does it cross the x-axis?

Where y is 0.

x2 − 1 = 0

Factor it.

1 is 1 times 1.

So x² − 1 is x² minus 1²: one square take away another square.

The rule: a² − b² = (a − b)(a + b).

So it is (x − 1)(x + 1).

1 = 1 × 1
= 1²
a² − b² = (a − b)(a + b)

So it factors like this.

(x + 1)(x − 1) = 0

So x = −1 or x = 1.

x = −1 or x = 1

From 0 to 1 it is below the axis: shaded red.

From 1 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 − x

4 Each piece

The first piece: from 0 to 1.

Top minus bottom.

∫ from 0 to 1 of f(x) dx = F(1) − F(0)

Put in 1, then take away the same with 0.

= (133 − 1) − (033 − 0)

First the powers and the times.

= (13 − 1) − (0 − 0)

Then each bracket.

1 is 3/3, so the same bottom.

The first is −2/3.

The second is 0.

= −23 − 0

That is −2/3.

= −23

Now the second piece: from 1 to 3.

Top minus bottom.

∫ from 1 to 3 of f(x) dx = F(3) − F(1)

Put in 3, then take away the same with 1.

= (333 − 3) − (133 − 1)

First the powers and the times.

= (9 − 3) − (13 − 1)

Then each bracket.

The first is 6.

1 is 3/3, so the same bottom.

The second is −2/3.

= 6 − (−23)

Taking away a minus is the same as adding.

− (−) = +

18 plus 2 is 20, over 3.

That is 20/3.

= 203

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 = 6 ✗

5 Add the areas

−2/3 is below the axis.

Its area is the size, without the minus: 2/3.

below: area = 23

Add the areas of the pieces.

area = 23 + 203

So the area is 22/3 square units.

= 223 square units

6 Check

Let's check the pieces: the whole integral from 0 to 3 is 6.

F(3) − F(0) = 6

The signed pieces add up to the same: −2/3 plus 20/3.

Right!

−23 + 203 = 6

7 Pick the choice

Our answer is area = 22/3 square units.

That is choice C.

C) 223

8 Why not the others

A is 6.

Wrong: ∫ from 0 to 3: the part under the axis takes area away.

That is the Careful mistake from before.

A) 6 ✗ ∫ from 0 to 3: the part under the axis takes area away

B is 20/3.

Wrong: only the part above the axis.

B) 203 ✗ only the part above the axis

D is 8.

Wrong: put 3 into x² − 1.

D) 8 ✗ put 3 into x^2 − 1

Let's look back.

First: draw the region.

Then: where it crosses.

Then: the antiderivative.

Then: each piece.

Then: add the areas.

Then: check.

Then: pick the choice.

Last: why not the others.

Answer
= 223 square units

Your turn

Try this one yourself, then check your answer:

Find the area between y = x2 − 1 and the x-axis from x = −1 to x = 3
Show the answer
area = 8 square units

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