Steply

What is the area of the region between y = x and y = x^2?

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 = x and y = x².

y = x

There are four choices.

First we solve it, the normal way.

Then we pick.

y = x2

1 Draw the region

Draw both curves.

They cross twice.

The area we want is between the crossings.

2 Where they meet

They meet where the two y's are equal.

x = x2

We want 0 on one side.

So we take x away from both sides, and write the two sides the other way round.

x2 − x = 0

Factor it.

Every term has an x, so we take it out.

x(x − 1) = 0

So x = 0 or x = 1.

x = 0 or x = 1

3 The top curve

Test a number between 0 and 1: x = 1.

The first gives 1, the second 1.

So y = x is on top.

x = 1: 1 < 1

The area between them is shaded.

4 Write the integral

Area between = the integral of the top one minus the bottom one.

At each x, top minus bottom is the height of the gap between them.

area = ∫ from 0 to 1 of ((x) − x2) dx

Careful!

Top minus bottom!

The other way gives a minus answer, −1/6.

Careful: not bottom − top ✗

A minus in front of a bracket changes the sign of every part inside.

Then like terms.

= ∫ from 0 to 1 of (−x2 + x) dx

5 The antiderivative

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

F(x) = −x33 + x22

6 Top minus bottom

Top minus bottom: F(1) − F(0).

area = F(1) − F(0)

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

= (−133 + 122) − (−033 + 022)

First the powers and the times.

= (−13 + 12) − (0 + 0)

We count up in 3s and 2s.

In 3s: 3, 6.

In 2s: 2, 4, 6.

6 is the first number in both lists.

So 6 works for both.

In 3s: 3, 6
In 2s: 2, 4, 6

Then each bracket.

1/3 is 2/6, and 1/2 is 3/6, so the same bottom.

The first is 1/6.

The second is 0.

= 16 − 0

So the area is 1/6 square units.

= 16 square units

7 Check

Let's check F: its derivative is −x² + x: top minus bottom.

And the area is positive.

Right!

F′(x) = −x2 + x

8 Pick the choice

Our answer is area = 1/6 square units.

That is choice B.

B) 16

9 Why not the others

A is 1/2.

Wrong: only the area under y = x.

A) 12 ✗ only the area under y = x

C is 5/6.

Wrong: added the two areas: take one away.

C) 56 ✗ added the two areas: take one away

D is 1/3.

Wrong: only the area under y = x².

D) 13 ✗ only the area under y = x^2

Let's look back.

First: draw the region.

Then: where they meet.

Then: the top curve.

Then: write the integral.

Then: the antiderivative.

Then: top minus bottom.

Then: check.

Then: pick the choice.

Last: why not the others.

Answer
= 16 square units

Your turn

Try this one yourself, then check your answer:

Find the area between y = 1 and y = x²
Show the answer
area = 43 square units

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