Integration 1 · Integration: undoing differentiation

Module 35 of 46Lesson 1 of 9 · Integration 1 · Integration: undoing differentiation

Integration, areas & kinematics · Essential understanding 1

Integration: undoing differentiation

IB AA SL 5.5with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: Integration is differentiation run backwards. If you know the gradient function, integrating gives you back the original function, plus a mystery constant + C.

By the end of this lesson you can:

explain integration as anti-differentiation and say why we write + C
integrate $ax^n$ and sums of such terms using $\dfrac{a}{n+1}x^{n+1}$
rewrite brackets and fractions as powers of x before integrating
check any answer by differentiating it

💡 AA Paper 1 has no calculator

In AA, Paper 1 is calculator-free, so the by-hand working is what earns the marks there. Use the TI-Nspire steps for Paper 2, and to check your hand answers.

🧑🏾 Imagine this…

A car's speedometer tells you how fast you are going, but not where you are. If you know the speed all the way, you can work out the distance travelled. Differentiating goes from position to speed. Integrating goes the other way, from speed back to position. It is a rewind button.

Anti-derivative

a function whose derivative is the one you were given.

x^2

is an anti-derivative of

2x

Integrate

find the anti-derivative. Also called anti-differentiate.

Indefinite integral

the answer with a constant, written

\int f(x)\,dx

. The

dx

says the variable is x.

Constant of integration, C

a number we cannot know yet.

x^2+3

x^2-7

and

x^2

all differentiate to

2x

, so all of them are correct anti-derivatives.

★ The power rule for integration

$\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ , for $n \neq -1$ .
Add 1 to the power, then divide by the new power. (Differentiating does the opposite: multiply by the power, then take 1 off.)
A number in front stays: $\displaystyle\int ax^n\,dx = \dfrac{a}{n+1}x^{n+1} + C$ . A constant: $\displaystyle\int k\,dx = kx + C$ .
Integrate a sum term by term, and write + C once, at the end.
This rule is in the IB formula booklet. Negative powers work too: $\int x^{-2}dx = \dfrac{x^{-1}}{-1}+C$ .

⚠️ Mistakes that lose marks

Forgetting + C. Without it an indefinite integral loses a mark.
No product or quotient rule for integrals. $\int 2x(x+4)\,dx$ is not $\int 2x\,dx \times \int (x+4)\,dx$ . Expand the brackets first.
n = −1 is not allowed in the power rule (you would divide by zero). You will not meet $\int \dfrac{1}{x}dx$ at SL.
Dividing by the old power. Divide by the new power (after adding 1).

💡 The best habit

You can always check an integration by differentiating your answer. If you get back what you started with, you are right. Do this in the exam if you have 20 seconds to spare.

🧪 Try it yourselfWhy there is a + C

The curve is y = x² + c. Slide c and watch the curve move up and down. The tangent is at x = 1: its gradient stays 2 every time, because every curve y = x² + c has the same derivative 2x. One gradient function, many possible curves.

Worked examples

Example 1.1Integrate term by term

Find (a)

\int 8x^3\,dx

(b)

\int (6x^2 - 10x + 7)\,dx

(c)

\int 9\,dx

🧠 Think it through

Add 1 to each power, divide by the new power, keep the number in front.
A plain number like 7 is $7x^0$ , so it becomes $7x$ .
Write + C once at the end.

✍️ By hand

(a)

\int 8x^3\,dx = \dfrac{8}{4}x^4 + C = 2x^4 + C

(b)

\int (6x^2 - 10x + 7)\,dx = \dfrac{6}{3}x^3 - \dfrac{10}{2}x^2 + 7x + C = 2x^3 - 5x^2 + 7x + C

(c)

\int 9\,dx = 9x + C

🔢 On your TI-Nspire CX II

The non-CAS TI-Nspire CX II cannot give indefinite integrals, so this part is by hand. But it can check your answer: differentiate with the derivative template at a test value.
On a Calculator page type the derivative of your (b) answer at x = 2 and compare with the original at x = 2: 6*2^2-10*2+7 should equal the slope of your answer.
Both give 11, so the answer is consistent

✅ Answer (a) 2x⁴ + C (b) 2x³ − 5x² + 7x + C (c) 9x + C

Example 1.2Negative powers

Find

\int \left(4x^{-3} + 6x^{-2}\right)dx

, and write the answer without negative powers.

🧠 Think it through

The rule works for negative powers too. Add 1 to −3 to get −2. Add 1 to −2 to get −1.
Divide by the new power, which is negative, so watch the signs.
At the end, $x^{-2}$ means $\dfrac{1}{x^2}$ .

✍️ By hand

\int 4x^{-3}dx = \dfrac{4}{-2}x^{-2} = -2x^{-2}

\int 6x^{-2}dx = \dfrac{6}{-1}x^{-1} = -6x^{-1}

Total:

-2x^{-2} - 6x^{-1} + C = -\dfrac{2}{x^2} - \dfrac{6}{x} + C

✅ Answer

-\dfrac{2}{x^2} - \dfrac{6}{x} + C

Check: differentiate −2x⁻² to get +4x⁻³ ✓, and −6x⁻¹ to get +6x⁻² ✓.

Example 1.3Expand the brackets first

Find

\int (x+3)(2x-1)\,dx

🧠 Think it through

There is no rule for integrating a product, so multiply out first.
Collect like terms into a sum of powers of x.
Then integrate term by term.

✍️ By hand

(x+3)(2x-1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3

\int (2x^2 + 5x - 3)\,dx = \dfrac{2}{3}x^3 + \dfrac{5}{2}x^2 - 3x + C

✅ Answer

\dfrac{2}{3}x^3 + \dfrac{5}{2}x^2 - 3x + C

Fractions in the answer are normal. Do not turn them into decimals unless asked.

Example 1.4Divide before you integrate

Find

\displaystyle\int \dfrac{6x^4 - 4x^2 + 10}{2x^2}\,dx

🧠 Think it through

Split the fraction: divide every term on top by the bottom.
Use the law of indices $\dfrac{x^a}{x^b} = x^{a-b}$ on each term.
Then integrate the sum of powers.

✍️ By hand

\dfrac{6x^4}{2x^2} - \dfrac{4x^2}{2x^2} + \dfrac{10}{2x^2} = 3x^2 - 2 + 5x^{-2}

\int (3x^2 - 2 + 5x^{-2})\,dx = x^3 - 2x + \dfrac{5x^{-1}}{-1} + C = x^3 - 2x - \dfrac{5}{x} + C

✅ Answer

x^3 - 2x - \dfrac{5}{x} + C

Example 1.5Spot the mistakes

A student writes:

\displaystyle\int \left(6x^2 - \dfrac{4}{x^3}\right)dx = 2x^3 - 4x^{-2}

. Find the two mistakes, then write the correct answer.

🧠 Think it through

Write $\dfrac{4}{x^3}$ as $4x^{-3}$ . The second term is $-4x^{-3}$ .
Integrate it: add 1 to the power, then divide by the new power. Was that done?
Check the whole answer by differentiating it. What is missing at the end?

✍️ By hand

\int 6x^2\,dx = 2x^3

is correct.

\int -4x^{-3}\,dx = \dfrac{-4}{-2}x^{-2} = 2x^{-2}

. The student added 1 to the power but forgot to divide by the new power (−2), and so kept −4.

The student also left out + C.

Correct:

2x^3 + 2x^{-2} + C = 2x^3 + \dfrac{2}{x^2} + C

✅ Answer

2x^3 + \dfrac{2}{x^2} + C

Check: differentiate 2x⁻² to get −4x⁻³ ✓. The student's version gives +8x⁻³, which does not match the original.

Challenge questions

⭐ Challenge 1.6Find the functionPaper 1 style

A function f has gradient function

f'(x) = 5x^4 - \dfrac{3}{x^2} + 2

. Find an expression for

f(x)

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Write $\dfrac{3}{x^2}$ as $3x^{-2}$ .
Integrate each term. Do not forget + C.

✍️ By hand

f(x) = \int (5x^4 - 3x^{-2} + 2)\,dx

= x^5 - \dfrac{3x^{-1}}{-1} + 2x + C = x^5 + \dfrac{3}{x} + 2x + C

✅ Answer

f(x) = x^5 + \dfrac{3}{x} + 2x + C

⭐ Challenge 1.7Expand, integrate, checkPaper 2 style

Let

g'(x) = (2x+1)^2

and

h'(x) = \dfrac{x^3 + 6x^2 - 4}{x^2}

(a) Find $g(x)$ .
(b) Differentiate your answer to (a) to show that you get back $(2x+1)^2$ .
(c) Find $h(x)$ .

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Expand $(2x+1)^2$ first. It is not $4x^2+1$ .
To check, differentiate each term of your answer and compare.
For (c) divide every term by x², then integrate.

✍️ By hand

(a)

(2x+1)^2 = 4x^2 + 4x + 1

, so

g(x) = \dfrac{4}{3}x^3 + 2x^2 + x + C

(b)

g'(x) = \dfrac{4}{3}(3x^2) + 2(2x) + 1 = 4x^2 + 4x + 1 = (2x+1)^2

✓

(c)

\dfrac{x^3 + 6x^2 - 4}{x^2} = x + 6 - 4x^{-2}

, so

h(x) = \dfrac{x^2}{2} + 6x + \dfrac{4}{x} + C

✅ Answer (a)

\dfrac{4}{3}x^3 + 2x^2 + x + C

(b) shown (c)

\dfrac{x^2}{2} + 6x + \dfrac{4}{x} + C

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL Integration 1 · Integration: undoing differentiation

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

Integration: undoing differentiation

Worked examples

Challenge questions