Steply

Integrate ∫ from 0 to 1 of (2x + 1)^2 dx

Step by step, with the reason for every step · u-substitution · Calculus

Step by step

The question

Question: find the integral from 0 to 1 of (2x + 1)².

∫ from 0 to 1 of (2x + 1)2 dx =

1 Choose u

The inside: 2x + 1.

Call it u.

Its derivative is just a number, 2.

u = 2x + 1

2 Find du

The power rule: du/dx = 2.

dudx = 2

Multiply both sides by dx: du = 2 dx.

A small change in u is 2 times a small change in x.

du = 2 dx

We have just dx.

Divide both sides by 2: dx = du over 2.

dx = du2

3 Change the limits

The limits 0 and 1 are x values.

u needs its own: x = 0 gives u = 1.

x = 0 → u = 2(0) + 1 = 0 + 1 = 1

And x = 1 gives u = 3.

x = 1 → u = 2(1) + 1 = 2 + 1 = 3

4 Write it with u

2x + 1 becomes u, and dx becomes half of du.

A number in front can wait outside the integral.

The limits are the u values.

= 12∫ from 1 to 3 of u2 du

Careful!

Not the old limits!

0 and 1 are x values, but now we work with u.

Careful: not ∫ from 0 to 1 of 12u2 du ✗

5 Integrate

Integrate: add 1 to the power, divide by the new power.

2 plus 1 is 3.

= 12 × u33

Multiply the numbers: 1/2 times 1/3 is 1/6.

= [u36] from 1 to 3
12 × 13 = 16

6 Top minus bottom

Put in 3, then take away the same with 1.

= 336 − 136

Work out the powers.

= 276 − 16
= 266

So the integral is 13/3.

= 133

7 Check

Let's check without changing the limits: put x back in the antiderivative, and use the x limits 0 and 1.

in x: (2x + 1)36 from 0 to 1

A calculator gives 4.5 minus 0.1667: 4.3333.

The same number.

Right!

4.5 − 0.1667 = 4.3333

Let's look back.

First: choose u.

Then: find du.

Then: change the limits.

Then: write it with u.

Then: integrate.

Then: top minus bottom.

Last: check.

Answer
= 133

Your turn

Try this one yourself, then check:

∫ from 0 to 1 of (2x + 1)3 dx
Show the answer
10

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