Pure 1 — Integration | Full Lesson

Pure Mathematics 1

Integration Lesson

Hello again. I am going to teach you everything you need to know about integration. I will assume you remember differentiation — because integration is its reverse. By the end, you will be finding areas, curves from gradients, and more. Let us begin.

What is Integration?

Integration is the reverse process of differentiation. If differentiation gives you the gradient (rate of change) of a function, integration takes you back to the original function. That is why it is also called the antiderivative.

The symbol for integration is a long, stretched S: ∫ — it stands for "sum" (because integration comes from summing up infinitely many tiny pieces).

Integration is the Reverse of Differentiation If dy/dx = f(x) then y = ∫ f(x) dx

Read "∫ f(x) dx" as "the integral of f of x with respect to x". The dx tells you which variable you are integrating with respect to.

Example 1: From Differentiation to Integration

You know that if y = x², then dy/dx = 2x (from differentiation).

So integration works the other way: if dy/dx = 2x, then y = x² + c.

(That + c is the constant of integration — we will talk about it soon.)

Think of differentiation as "down" (reduce the power) and integration as "up" (increase the power).

x² → (differentiate) → 2x

2x → (integrate) → x² + c

Integration undoes differentiation. They are inverse operations.

Teacher's Tip: If you can differentiate, you can already integrate — you just go backwards. Differentiation sends xⁿ → nxⁿ⁻¹. Integration sends xⁿ back: xⁿ → xⁿ⁺¹/(n+1).

Integrating Powers of x

The most important rule in integration is the power rule. Here it is:

The Power Rule for Integration ∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + c (for n ≠ −1)

In words: add 1 to the power, then divide by the new power. Then add the constant of integration c.

Positive Powers

Example 1: ∫ x³ dx

Add 1 to the power: 3 + 1 = 4

Divide by the new power: x⁴ / 4

Add the constant: + c

∫ x³ dx = x⁴/4 + c

Example 2: ∫ x dx (remember x = x¹)

∫ x¹ dx = x¹⁺¹ / (1 + 1) + c

= x²/2 + c

∫ x dx = x²/2 + c

Negative Powers

Example 3: ∫ x⁻² dx

Add 1 to the power: −2 + 1 = −1

Divide by the new power: x⁻¹ / (−1) = −x⁻¹

Add + c

∫ x⁻² dx = −x⁻¹ + c = −1/x + c

Fractional Powers

Example 4: ∫ √x dx (remember √x = x¹⸍²)

∫ x¹⸍² dx = x¹⸍²⁺¹ / (1/2 + 1) + c

= x³⸍² / (3/2) + c

= (2/3)x³⸍² + c

∫ √x dx = (2/3)x³⸍² + c

Integrating a Constant

Example 5: ∫ 5 dx (remember 5 = 5 × x⁰)

∫ 5 dx = 5 ∫ x⁰ dx = 5 × (x¹ / 1) + c

= 5x + c

∫ 5 dx = 5x + c

Teacher's Tip: The power rule works for every power except n = −1. ∫ x⁻¹ dx = ∫ 1/x dx = ln|x| + c (you will learn this later). For now, just remember: add 1, divide by new power.

Integrating term by term: When you have a sum, integrate each term separately:

∫ (f(x) + g(x)) dx = ∫ f(x) dx + ∫ g(x) dx

Example 6: ∫ (x² + 3x − 2) dx

∫ x² dx = x³/3

∫ 3x dx = 3∫ x dx = 3(x²/2) = 3x²/2

∫ (−2) dx = −2x

Add them together + c

∫ (x² + 3x − 2) dx = x³/3 + 3x²/2 − 2x + c

Indefinite Integrals

An indefinite integral is an integral without limits. The answer includes a constant of integration + c.

Indefinite Integral ∫ f(x) dx = F(x) + c

Why + c? Because when you differentiate a constant, you get zero. So if we only know the derivative, we cannot know the original constant — there are infinitely many possible original functions, all differing by a constant.

Example 1: Finding the Original Function

If f′(x) = 4x³, find f(x).

f(x) = ∫ 4x³ dx

= 4 ∫ x³ dx

= 4(x⁴/4) + c

= x⁴ + c

f(x) = x⁴ + c

Example 2: Given dy/dx = 6x² − 2x + 1, find y

y = ∫ (6x² − 2x + 1) dx

∫ 6x² dx = 6(x³/3) = 2x³

∫ (−2x) dx = −2(x²/2) = −x²

∫ 1 dx = x

y = 2x³ − x² + x + c

Teacher's Tip: Never forget the + c. It is easy to lose marks by forgetting it. If the question says "indefinite integral" or "find the original function", you must include + c.

Definite Integrals

A definite integral has limits. You evaluate it over a specific interval, and the result is a number (not a function). No + c needed.

Definite Integral ∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a)

You integrate f(x) to get F(x), then subtract F(a) from F(b): top limit minus bottom limit. The + c cancels out, so we do not write it.

Example 1: ∫₁³ x² dx

Step 1: Integrate: ∫ x² dx = x³/3

Step 2: Apply limits: [x³/3]₁³

Step 3: F(3) = 27/3 = 9

Step 4: F(1) = 1/3

Step 5: F(3) − F(1) = 9 − 1/3 = 27/3 − 1/3 = 26/3

∫₁³ x² dx = 26/3 ≈ 8.67

Example 2: ∫₀² (3x² + 2x) dx

Step 1: Integrate: ∫ (3x² + 2x) dx = x³ + x²

Step 2: Apply limits: [x³ + x²]₀²

Step 3: F(2) = 8 + 4 = 12

Step 4: F(0) = 0 + 0 = 0

Step 5: F(2) − F(0) = 12 − 0 = 12

∫₀² (3x² + 2x) dx = 12

Teacher's Tip: Definite integrals give a number. The + c disappears because it cancels out: (F(b) + c) − (F(a) + c) = F(b) − F(a). So never write + c in a definite integral.

Area Under a Curve

One of the most important uses of integration is finding the area under a curve. The area between the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is:

Area Under a Curve Area = ∫ₐᵇ y dx = ∫ₐᵇ f(x) dx

This gives the signed area — area above the x-axis is positive, area below is negative.

Example 1: Find the area under y = x² from x = 0 to x = 2

Area = ∫₀² x² dx

= [x³/3]₀²

= (8/3) − (0/3)

= 8/3

Area = 8/3 ≈ 2.67 square units

Dealing with Area Below the Axis

If the curve goes below the x-axis, the integral gives a negative value. To find the total area, take the absolute value of the negative part, or split the integral at the x-axis crossing.

Example 2: Find the area under y = x² − 4 from x = 0 to x = 3

The curve crosses the x-axis at x = 2 (where x² − 4 = 0).

Split into two parts: x = 0 to 2 (below axis) and x = 2 to 3 (above axis).

∫₀² (x² − 4) dx = [x³/3 − 4x]₀² = (8/3 − 8) − 0 = 8/3 − 24/3 = −16/3

Area below = |−16/3| = 16/3

∫₂³ (x² − 4) dx = [x³/3 − 4x]₂³ = (9 − 12) − (8/3 − 8) = −3 − (−16/3) = −3 + 16/3 = 7/3

Total area = 16/3 + 7/3 = 23/3

Total area = 23/3 ≈ 7.67 square units

Teacher's Tip: When in doubt, sketch the curve! If the area goes below the x-axis, split the integral at the x-axis crossing(s) and take absolute values for any negative parts.

Interactive Area Visualiser — y = x²

a = 0 b = 2

Shaded area = 0 square units

Area Between Two Curves

To find the area between two curves, subtract the lower curve from the upper curve, then integrate.

Area Between Two Curves Area = ∫ₐᵇ (y_upper − y_lower) dx

The limits a and b are the x-coordinates where the curves intersect (where y_upper = y_lower).

Example 1: Find the area between y = x and y = x² from x = 0 to x = 1

At x = 0: both are 0. At x = 1: both are 1. These are the intersection points.

Between them, y = x is above y = x² (since x > x² for 0 < x < 1).

Area = ∫₀¹ (x − x²) dx

= [x²/2 − x³/3]₀¹

= (1/2 − 1/3) − (0 − 0)

= 3/6 − 2/6 = 1/6

Area = 1/6 ≈ 0.167 square units

Example 2: Find the area between y = x² + 1 and y = 2x + 1

First find where they intersect: x² + 1 = 2x + 1 → x² − 2x = 0 → x(x − 2) = 0 → x = 0 or x = 2.

For 0 < x < 2, check which is higher at x = 1: y = 2(1) + 1 = 3 vs y = 1² + 1 = 2. So y = 2x + 1 is above.

Area = ∫₀² ((2x + 1) − (x² + 1)) dx

= ∫₀² (2x − x²) dx

= [x² − x³/3]₀²

= (4 − 8/3) − (0 − 0)

= 12/3 − 8/3 = 4/3

Area = 4/3 ≈ 1.333 square units

Teacher's Tip: Always figure out which curve is on top before integrating. Sketch the graphs or test a point between the intersection points. Upper minus lower, always.

Finding the Equation of a Curve from Its Gradient

If you are given the gradient function dy/dx (the derivative) and a point that the curve passes through, you can find the original equation of the curve.

The process is:

1. Integrate dy/dx to get y = F(x) + c

2. Substitute the given point (x, y) to find the value of c

3. Write the complete equation of the curve

Finding the Curve from Its Gradient y = ∫ (dy/dx) dx → use point to find c

Example 1: The gradient of a curve is dy/dx = 4x³ − 6x. The curve passes through (1, 3). Find its equation.

Step 1: Integrate

y = ∫ (4x³ − 6x) dx

= x⁴ − 3x² + c

Step 2: Use the point (1, 3):

3 = (1)⁴ − 3(1)² + c

3 = 1 − 3 + c

3 = −2 + c

c = 5

Step 3: Write the equation:

y = x⁴ − 3x² + 5

Example 2: A curve has gradient dy/dx = 2x − 5 and passes through (3, −4). Find its equation.

Step 1: Integrate

y = ∫ (2x − 5) dx

= x² − 5x + c

Step 2: Use the point (3, −4):

−4 = (3)² − 5(3) + c

−4 = 9 − 15 + c

−4 = −6 + c

c = 2

Step 3: Write the equation:

y = x² − 5x + 2

Teacher's Tip: This type of question is very common in exams. The steps never change: integrate → substitute the point → find c → write the final equation. Practise it until it is automatic.

Practice — Step by Step

Solve each problem one step at a time. Type your answer for each step and click Check Step. You must get each step correct before moving to the next.

Mastery Kit

The integral of e to the x is e to the x. That is the beauty of calculus — the function that is its own derivative and its own integral.

— Calculus folklore

⚡ Quick Hacks

Power rule shortcut: ∫ xⁿ dx = xⁿ⁺¹/(n+1). Say it: "add one, divide by new one." Do this in your head for simple powers.
Area above/below axis: If the curve goes below the x-axis, the integral is negative. Split at the roots and take |negative area| + positive area.
+ c memory trick: Indefinite = "I" for "I need + c". Definite = "D" for "Done — no c". If there are limits, no + c.
Checking integration: Differentiate your answer. If you get back the original function, you integrated correctly. This is the best check.
Upper − lower: For area between curves, always do top curve minus bottom curve. Test one x-value between intersections to see which is higher.

📌 Key Notes

Integration reverses differentiation: If you know how to differentiate, you know how to integrate — just go backwards.
Every indefinite integral needs + c: The constant of integration represents the family of functions that share the same derivative.
Definite integrals give numbers: ∫ₐᵇ f(x) dx evaluates to a single number, never a function.
The power rule fails at n = −1: ∫ x⁻¹ dx = ln|x| + c (you will learn this separately).
Area is absolute: Total area considers regions below the axis as positive. Use splitting and absolute values.

⚠️ Common Mistakes

Forgetting + c: In an indefinite integral, missing the + c costs you a mark every time. Always add it.
Adding + c to a definite integral: Definite integrals do not need + c — it cancels out anyway.
Wrong order in F(b) − F(a): It is top limit minus bottom limit, not the other way around. F(b) − F(a), always.
Forgetting to split areas: If part of the curve is below the x-axis, the integral gives a negative number. You must split and take absolute values for total area.
Wrong curve as upper: When finding area between two curves, ensure you do (upper − lower). Sketch or test a point to check.

🧠 Integration Cheat Sheet

Power rule: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + c (n ≠ −1)

∫ k dx = kx + c | ∫ k·f(x) dx = k∫ f(x) dx

∫ (f±g) dx = ∫ f dx ± ∫ g dx

Definite: ∫ₐᵇ f(x) dx = F(b) − F(a) (no + c)

Area under curve: ∫ₐᵇ y dx

Area between curves: ∫ₐᵇ (y_upper − y_lower) dx

Find curve from gradient: y = ∫ (dy/dx) dx, then find c

💡

The Golden Rule of Integration

Differentiate your answer to check it. If you get back the original function, you are correct. This simple check catches almost every mistake — use it every single time.

You Have Completed the Integration Lesson

Here is what you have learned:

✅ Integration is the reverse of differentiation (antiderivative)
✅ The power rule: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + c (n ≠ −1)
✅ Indefinite integrals have + c; definite integrals are evaluated with limits
✅ Area under a curve = ∫ₐᵇ y dx (split if below axis)
✅ Area between two curves = ∫ₐᵇ (upper − lower) dx
✅ Finding a curve from its gradient: integrate and use a point to find c

Well done. You now know integration — one of the most powerful tools in mathematics. You can find areas, recover original functions, and much more.