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Area between y = -x^2 + 2x and the x-axis

Step by step, with the reason for every step · Definite integrals and area · Calculus

Step by step

The question

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

y = −x2 + 2x

1 Draw the region

Draw the curve first.

Where does it cross the x-axis?

Where y is 0.

−x² + 2x = 0

Factor it.

We need two numbers that multiply to 0 and add to −2: 0 and −2.

−x(x − 2) = 0

So it crosses at x = 0 and x = 2.

x = 0 or x = 2

The area we want is shaded: under the curve, above the x-axis, from 0 to 2.

This area is the integral from 0 to 2.

area = ∫ from 0 to 2 of (−x2 + 2x) dx

2 The antiderivative

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

F(x) = −x33 + 2 × x22

Now we have F(x) = −x³ over 3 + x².

= −x33 + x2

No + C here: in F(b) minus F(a), the two Cs cancel.

no + C here: it cancels

3 Top minus bottom

The rule: put in the top number, then take away the bottom number.

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

Write F(x) with 0 and 2 on the bar.

= (−x33 + x2) from 0 to 2

Put in 2 first, then take away the same with 0.

= (−233 + 22) − (−033 + 02)

Work out each bracket.

= 43 − 0

Careful!

Top first, then take away the bottom.

The other way gives the wrong sign.

Careful: not F(0) − F(2) ✗

So the area is 4/3 square units.

area = 43 square units

4 Check

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

Right!

F′(x) = −x2 + 2x

Let's look back.

First: draw the region.

Then: the antiderivative.

Then: top minus bottom.

Last: check.

Answer
area = 43 square units

Your turn

Try this one yourself, then check:

Find the area between y = 3 − 3x2
Show the answer
area = 4 square units

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