Area between y = 2x - 6 and the x-axis from x = 1 to x = 5 Step by Step

Step by step

The question

Question: find the area between y = 2x − 6 and the x-axis from x = 1 to x = 5.

y = 2x − 6

1 Draw the region

Draw the curve first.

2 Where it crosses

Where does it cross the x-axis?

Where y is 0.

2x − 6 = 0

Factor it.

2(x − 3) = 0

So x = 3.

x = 3

From 1 to 3 it is below the axis: shaded red.

From 3 to 5 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) = 2 × x²2 − 6x

Now we have F(x) = x² − 6x.

= x² − 6x

4 Each piece

The first piece: from 1 to 3.

Top minus bottom.

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

= (3² − 6(3)) − (1² − 6(1))

= −9 − (−5)

That is −4.

= −4

Now the second piece: from 3 to 5.

Top minus bottom.

∫ from 3 to 5 of f(x) dx = F(5) − F(3)

= (5² − 6(5)) − (3² − 6(3))

= −5 − (−9)

That is 4.

= 4

Careful!

Not one integral from 1 to 5!

The piece below the axis counts as negative, so the pieces cancel.

Careful: not area = ∫ from 1 to 5 of f(x) dx = 0 ✗

5 Add the areas

−4 is below the axis.

Its area is the size, without the minus: 4.

below: area = 4

Add the areas of the pieces.

area = 4 + 4

So the area is 8 square units.

= 8 square units

6 Check

Let's check the pieces: the whole integral from 1 to 5 is 0.

F(5) − F(1) = 0

The signed pieces add up to the same: −4 plus 4.

Right!

−4 + 4 = 0

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