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

Step by step

The question

Question: find the area between y = 8 − x² and y = x².

y = 8 − x²

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.

8 − x² = x²

Move everything to one side.

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

3 The top curve

Test a number between −2 and 2: x = 0.

The first gives 8, the second 0.

So y = 8 − x² is on top.

x = 0: 8 > 0

The area between them is shaded.

4 Write the integral

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

area = ∫ from −2 to 2 of ((8 − x²) − x²) dx

Careful!

Top minus bottom!

The other way gives a minus answer, −64/3.

Careful: not bottom − top ✗

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

Then like terms.

= ∫ from −2 to 2 of (−2x² + 8) dx

5 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

6 Top minus bottom

area = F(2) − F(−2)

Put in 2, then take away the same with −2.

= (−2(2)³3 + 8(2)) − (−2(−2)³3 + 8(−2))

= 323 − (−323)

So the area is 64/3 square units.

= 643 square units

7 Check

Let's check F: its derivative is −2x² + 8: top minus bottom.

And the area is positive.

Right!

F′(x) = −2x² + 8

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.

Last: check.

Answer