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Area under y = 2x + 1 from x = 0 to x = 2

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

Step by step

The question

Question: find the area under y = 2x + 1 from x = 0 to x = 2.

y = 2x + 1

1 Draw the region

Draw the curve first.

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 (2x + 1) dx

2 The antiderivative

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

F(x) = 2 × x22 + x

Now we have F(x) = x² + x.

= x2 + x

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.

= (x2 + x) from 0 to 2

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

= (22 + 2) − (02 + 0)

Work out each bracket.

= 6 − 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 6 square units.

area = 6 square units

4 Check

Let's check with geometry.

Under a straight line the shape is a trapezoid: heights 1 and 5, width 2.

Its area is 6.

The same!

Right!

trapezoid: 1 + 52 × 2 = 6

Let's look back.

First: draw the region.

Then: the antiderivative.

Then: top minus bottom.

Last: check.

Answer
area = 6 square units

Your turn

Try this one yourself, then check:

Find the area under y = 2x + 3
Show the answer
area = 10 square units

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