Grade 11 · Pure Mathematics 1 · Cambridge A-Level 9709 · Age 16–17
Welcome to Integration Techniques!
Integration is the reverse process of differentiation — the second great pillar of calculus. In Cambridge A-Level Pure Mathematics 1, you build from basic indefinite integrals right through to definite integrals, trigonometric and exponential integrals, integration by substitution, and using integration to find curves from gradient functions and to compute areas.
Integrate polynomials using the power rule (including negative and fractional powers)
Recognise and handle the special case ∫(1/x) dx = ln|x| + C
Apply the constant of integration C and explain why it is essential
Evaluate definite integrals using the Fundamental Theorem of Calculus
Handle negative areas (curve below x-axis) and compute absolute area by splitting at roots
Integrate sin x, cos x, sec²x, eˣ directly
Apply the reverse chain rule for linear composites (ax + b)
Use integration by substitution u = g(x) for more complex integrands
Find the equation of a curve given its gradient function and a boundary condition
Calculate area under a curve and area between two curves
Indefinite Integration
Power rule, +C, term-by-term, simplify first
Definite Integration
F(b) − F(a), signed vs unsigned area
Trig & Exponential
∫sin x, ∫cos x, ∫sec²x, ∫eˣ and reverse chain rule
Substitution
u = g(x), change dx, back-substitute or change limits
Finding Curves
Integrate dy/dx, apply boundary condition to find C
Area Between Curves
∫[a,b] (top − bottom) dx
Reverse Chain Rule
∫(ax+b)ⁿ and ∫e^(ax+b) short-cuts
Absolute Area
Split at roots, take absolute values of each section
Learn 1 — Indefinite Integration
Integration as Reverse Differentiation
Integration is the reverse (anti-derivative) of differentiation. If d/dx[F(x)] = f(x), then ∫f(x) dx = F(x) + C. The symbol ∫ is the integral sign, dx indicates the variable of integration, and C is the constant of integration.
The Power Rule for Integration
∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1)
In words: increase the power by 1, then divide by the new power. This works for any rational power n except n = −1 (which gives division by zero).
The Special Case: n = −1
∫(1/x) dx = ∫x⁻¹ dx = ln|x| + C
The modulus signs are needed because ln is only defined for positive numbers — but 1/x exists for negative x too. For exam purposes, writing ln x + C is accepted when the context implies x > 0.
Why the Constant of Integration C is Essential
When we differentiate, any constant term disappears. So when reversing the process, we cannot know what constant was there. The family of functions F(x) + C all have the same derivative f(x):
Integrate f(x) = 3x² + 2/x³ = 3x² + 2x⁻³
= x³ + 2x⁻²/(−2) + C = x³ − x⁻² + C = x³ − 1/x² + C
Simplifying Before Integrating
Always expand brackets or split fractions before applying the power rule. You cannot integrate products or quotients directly using these rules (unless one is a constant).
Integrate (x³ + 4x) / x² = x + 4/x = x + 4x⁻¹ first, then integrate:
∫(x + 4x⁻¹) dx = x²/2 + 4 ln|x| + C
Integrate x(2x − 1) = 2x² − x first:
∫(2x² − x) dx = 2x³/3 − x²/2 + C
The most important preparation step: always rewrite expressions as a sum of terms of the form axⁿ before integrating. Expand all brackets. Split all fractions where the denominator is a monomial (single power of x). Never try to integrate a product or quotient directly with the power rule.
Learn 2 — Definite Integration
The Fundamental Theorem of Calculus
A definite integral has limits of integration (numbers a and b) and gives a numerical answer — no + C needed. The result equals the area under the curve from x = a to x = b (with sign).
∫[a,b] f(x) dx = [F(x)]ₐᵇ = F(b) − F(a)
where F(x) is any antiderivative of f(x).
Step-by-Step Method
Step 1: Find the indefinite integral F(x) (no + C needed for definite integrals). Step 2: Write [F(x)] with limits as square bracket notation [F(x)]ₐᵇ. Step 3: Substitute the upper limit: F(b). Step 4: Substitute the lower limit: F(a). Step 5: Calculate F(b) − F(a).
When the curve y = f(x) lies below the x-axis on [a, b], the definite integral gives a negative value. This is the signed area. The unsigned (actual geometric) area is the absolute value.
Example: ∫[0,2] (x² − 4) dx
Note: x² − 4 < 0 for x ∈ [0, 2) since 0 − 4 = −4 < 0.
[x³/3 − 4x]₀² = (8/3 − 8) − 0 = 8/3 − 24/3 = −16/3
The signed area is −16/3. The geometric (absolute) area is 16/3.
Absolute Area — Splitting at Roots
When a curve crosses the x-axis between a and b, split the integral at each root. Add the absolute values of each part.
Find the total area enclosed by y = x² − 1, x = −1 and x = 2.
Root at x = 1 (where x² − 1 = 0 in [−1, 2]).
∫[−1,1] (x² − 1) dx = [x³/3 − x]₋₁¹ = (1/3 − 1) − (−1/3 + 1) = −2/3 − 2/3 = −4/3 → area = 4/3
∫[1,2] (x² − 1) dx = [x³/3 − x]₁² = (8/3 − 2) − (1/3 − 1) = 2/3 − (−2/3) = 4/3 → area = 4/3 Total absolute area = 4/3 + 4/3 = 8/3
Geometric Interpretation
The definite integral ∫[a,b] f(x) dx equals the net signed area between the curve y = f(x) and the x-axis from x = a to x = b.
• Area above x-axis contributes positively.
• Area below x-axis contributes negatively.
• Absolute (total shaded) area requires splitting at x-intercepts.
Always sketch the curve or check where it crosses the x-axis before evaluating a definite integral representing an area. If the question says "find the area", it means the geometric (positive) area — you must handle any regions below the x-axis separately and take their absolute values.
Learn 3 — Trig & Exponential Integration
Standard Trigonometric Integrals
∫sin x dx = −cos x + C ∫cos x dx = sin x + C ∫sec²x dx = tan x + C
These three must be memorised. Notice that integrating sin gives −cos (the negative sign is critical and frequently lost in exams).
∫sin x dx = −cos x + C ✓ note the minus
∫3 cos x dx = 3 sin x + C
∫sec²x dx = tan x + C (used whenever you see sec²x or 1/cos²x)
Standard Exponential Integral
∫eˣ dx = eˣ + C
The exponential function eˣ integrates to itself — just as it differentiates to itself.
Reverse Chain Rule — Linear Composite Functions
When the argument is a linear function (ax + b) rather than just x, divide by the inner derivative a:
∫(3x + 2)⁴ dx
a = 3, n = 4: Result = (3x+2)⁵/(3×5) + C = (3x+2)⁵/15 + C
∫e^(5x−1) dx
a = 5: Result = e^(5x−1)/5 + C
∫sin(2x) dx
a = 2: Result = −cos(2x)/2 + C
∫cos(4x + π/3) dx
a = 4: Result = sin(4x + π/3)/4 + C
∫e^(−3x) dx
a = −3: Result = e^(−3x)/(−3) + C = −e^(−3x)/3 + C
∫1/(2x + 7) dx
a = 2: Result = (1/2) ln|2x + 7| + C
∫(2 − x)³ dx
a = −1: Result = (2−x)⁴/(−1×4) + C = −(2−x)⁴/4 + C
Why This Works
To verify ∫cos(ax+b) dx = sin(ax+b)/a + C, differentiate the result:
d/dx[sin(ax+b)/a] = a·cos(ax+b)/a = cos(ax+b) ✓
The division by a compensates for the chain rule factor that would appear on differentiation. This is why this technique is called the reverse chain rule.
The reverse chain rule only works when the inner function is linear (ax + b). If the inner function is quadratic (e.g. x² + 1), you cannot use this shortcut — you need integration by substitution instead. Always check: is the argument of the function a simple linear expression in x?
Learn 4 — Integration by Substitution
What Is Substitution?
Integration by substitution is the integration equivalent of the chain rule. When an integrand contains a composite function f(g(x)) and g'(x) appears as a factor (or can be arranged to appear), we substitute u = g(x) to simplify the integral.
The Step-by-Step Method
Step 1: Choose a substitution u = g(x) — typically the "inside" function of the composite. Step 2: Differentiate: du/dx = g'(x), so du = g'(x) dx. Rearrange to express dx in terms of du. Step 3: Replace all occurrences of x (and dx) with expressions in u. Step 4: Integrate with respect to u. Step 5: Back-substitute u = g(x) to write the answer in terms of x. For definite integrals: Either change the limits to u-values using u = g(a) and u = g(b), OR back-substitute and apply the original limits.
Example 1: ∫2x(x² + 1)⁴ dx
Let u = x² + 1 → du/dx = 2x → du = 2x dx
Replace 2x dx with du and (x²+1)⁴ with u⁴:
∫u⁴ du = u⁵/5 + C
Back-substitute: (x² + 1)⁵/5 + C
Example 2: ∫x·e^(x²) dx
Let u = x² → du = 2x dx → x dx = du/2
∫e^u · (du/2) = (1/2)eᵘ + C
Back-substitute: e^(x²)/2 + C
Example 3: ∫cos x · sin³x dx
Let u = sin x → du = cos x dx
∫u³ du = u⁴/4 + C
Back-substitute: sin⁴x/4 + C
Example 4: ∫x√(x² + 3) dx (indefinite)
Let u = x² + 3 → du = 2x dx → x dx = du/2
∫√u · (du/2) = (1/2)∫u^(1/2) du = (1/2) · u^(3/2)/(3/2) + C = u^(3/2)/3 + C
Back-substitute: (x² + 3)^(3/2)/3 + C
Definite Integral with Substitution — Changing Limits
∫[0,2] 2x(x² + 1)³ dx
Let u = x² + 1, du = 2x dx
When x = 0: u = 1 When x = 2: u = 5
∫[1,5] u³ du = [u⁴/4]₁⁵ = 625/4 − 1/4 = 624/4 = 156
When to Use Substitution
Spot a composite function f(g(x)) where the derivative g'(x) appears as a factor (possibly up to a constant multiple). Common patterns:
• f(x²+1) · x → let u = x²+1
• f(sin x) · cos x → let u = sin x
• f(eˣ) · eˣ → let u = eˣ
• f(ln x) · (1/x) → let u = ln x
The key skill is recognising when substitution is needed. If the inner function's derivative is a constant multiple of what appears in the integrand, substitution will work cleanly. If not, check whether the reverse chain rule shortcut applies. After substituting, confirm that no x terms remain before integrating — if x is still present, the substitution was not complete.
Learn 5 — Finding Curves & Constants
Finding a Curve from Its Gradient Function
If you are given dy/dx = f(x) and a point (a, b) that lies on the curve, integrate to find y = F(x) + C, then substitute the known point to determine C.
y = ∫(dy/dx) dx then sub in (a, b) to find C
Example 1 — Polynomial Gradient
Given dy/dx = 3x² − 4x + 1 and the curve passes through (2, 3). Find y.
Integrate: y = x³ − 2x² + x + C
Substitute (2, 3): 3 = 8 − 8 + 2 + C → 3 = 2 + C → C = 1
Equation of curve: y = x³ − 2x² + x + 1
Example 2 — Rational Gradient
Given dy/dx = 2x + 3/x² and the curve passes through (1, 5). Find y.
Rewrite: dy/dx = 2x + 3x⁻²
Integrate: y = x² + 3x⁻¹/(−1) + C = x² − 3/x + C
Substitute (1, 5): 5 = 1 − 3 + C → C = 7
Equation of curve: y = x² − 3/x + 7
Area Under a Curve
The area between the curve y = f(x), the x-axis, and the lines x = a and x = b is given by the definite integral (taking the absolute value if the curve dips below the axis):
Area = ∫[a,b] f(x) dx (provided f(x) ≥ 0 on [a,b])
Find the area under y = x² + 1 from x = 0 to x = 3.
∫[0,3] (x² + 1) dx = [x³/3 + x]₀³ = (9 + 3) − 0 = 12
Area Between Two Curves
When two curves y = f(x) and y = g(x) intersect, the area between them from x = a to x = b (where f is the top curve) is:
Area = ∫[a,b] (f(x) − g(x)) dx
Find the area enclosed between y = 4 − x² (top) and y = x + 2 (bottom).
Intersection: 4 − x² = x + 2 → x² + x − 2 = 0 → (x+2)(x−1) = 0 → x = −2, x = 1
Area = ∫[−2,1] (4 − x² − x − 2) dx = ∫[−2,1] (2 − x² − x) dx
= [2x − x³/3 − x²/2]₋₂¹
= (2 − 1/3 − 1/2) − (−4 + 8/3 − 2)
= (12/6 − 2/6 − 3/6) − (−6 + 8/3)
= 7/6 − (−10/3) = 7/6 + 20/6 = 27/6 = 9/2
Area Between Curve and y-axis
To find the area between a curve and the y-axis, integrate x as a function of y:
Area between y = x², y-axis, y = 0 to y = 4:
x = √y = y^(1/2)
Area = ∫[0,4] y^(1/2) dy = [2y^(3/2)/3]₀⁴ = 2·8/3 = 16/3
When finding the area between two curves, always identify which curve is on top (greater y-value) in the region of interest. If the two curves swap which is on top at some point, you must split the integral there. Also remember: the area found by integration is always expressed as a positive number of square units.
Worked Examples
Eight fully worked examples covering all key integration techniques for Cambridge A-Level 9709.
Example 1 — Polynomial Indefinite Integration
Q: Find ∫(6x² − 4x + 3) dx.
M1: Apply power rule term by term. Increase each power by 1, divide by new power.
M1: At x = 3: 9/2 + 18 − 9 = 9/2 + 9 = 27/2. At x = −2: 4/2 − 12 + 8/3 = 2 − 12 + 8/3 = −28/3.
A1: Area = 27/2 − (−28/3) = 81/6 + 56/6 = 125/6 square units. 4 marks
Common Mistakes in Integration
These are the errors examiners see most often in Cambridge 9709 scripts. Knowing why each is wrong is the best way to avoid them.
Mistake 1 — Forgetting the constant of integration +C
✗ ∫(3x² + 2) dx = x³ + 2x
✓ ∫(3x² + 2) dx = x³ + 2x + C. The constant C is lost every time you differentiate a constant — so it must be added back every time you integrate. In Cambridge exams, a missing +C in an indefinite integral typically costs 1 mark.
Mistake 2 — The n = −1 edge case
✗ ∫x⁻¹ dx = x⁰/0 + C (undefined)
✓ ∫x⁻¹ dx = ∫(1/x) dx = ln|x| + C. The power rule fails for n = −1 because dividing by (n+1) = 0 is undefined. This case must be treated separately using the logarithm integral.
Mistake 3 — Wrong sign when integrating cos x
✗ ∫cos x dx = −sin x + C
✓ ∫cos x dx = sin x + C (positive). It is d/dx(cos x) = −sin x (negative derivative). But the integral reverses: ∫cos x dx = sin x. The minus sign belongs to ∫sin x dx = −cos x, not to ∫cos x dx.
Mistake 4 — Not changing limits in definite substitution
✗ ∫[0,2] 2x(x²+1)³ dx with u = x²+1 → ∫[0,2] u³ du (still using x-limits)
✓ When u = x²+1: at x=0, u=1; at x=2, u=5. The limits must become ∫[1,5] u³ du. Alternatively, back-substitute to x before applying the original limits — but never use x-limits with a u-integrand.
Mistake 5 — Absolute vs signed area
✗ Area between y = x²−4 and x-axis from x=0 to x=3 = ∫[0,3](x²−4)dx (gives −7/3, not the area)
✓ When the curve goes below the x-axis, split at the root (x=2). Take |∫[0,2](x²−4)dx| + ∫[2,3](x²−4)dx = 16/3 + 7/3 = 23/3. Always check whether the curve crosses the axis in the interval.
Mistake 6 — Incorrect reverse chain rule — not dividing by inner derivative
✗ ∫(3x+2)⁵ dx = (3x+2)⁶/6 + C
✓ ∫(3x+2)⁵ dx = (3x+2)⁶/(6×3) + C = (3x+2)⁶/18 + C. You must divide by both (n+1) and the coefficient a of x. Forgetting to divide by a is the most common reverse chain rule error.
Mistake 7 — Splitting a fraction with a variable denominator incorrectly
✗ ∫(x² + 3)/(x²) dx treated as ∫1 dx + ∫3/x² dx = x − 3/x + C... then getting confused about whether this is correct
✓ Split correctly: (x² + 3)/x² = 1 + 3x⁻² → ∫(1 + 3x⁻²) dx = x + 3x⁻¹/(−1) + C = x − 3/x + C. The method is correct but you must rewrite each term as axⁿ first.
Mistake 8 — Trying to integrate a product without substitution
✗ ∫x·(x²+1)³ dx = (x²/2)·(x²+1)⁴/4 + C
✓ You cannot integrate products by multiplying the separate integrals. Use substitution u = x²+1, du = 2x dx → (1/2)∫u³ du = u⁴/8 + C = (x²+1)⁴/8 + C. Integration has no product rule analogue.
Mistake 9 — Forgetting to find C when a boundary condition is given
✗ Given dy/dx = 2x + 1 passing through (1, 4): writes y = x² + x (no C, no substitution)
✓ Integrate to get y = x² + x + C. Substitute (1, 4): 4 = 1 + 1 + C → C = 2. Final answer: y = x² + x + 2. The specific curve requires a specific C.
Key Formulas — Integration Techniques
Function f(x)
Integral ∫f(x) dx
Notes
xⁿ (n ≠ −1)
xⁿ⁺¹/(n+1) + C
power rule; n can be negative or fractional
x⁻¹ = 1/x
ln|x| + C
special case n = −1; note modulus
k (constant)
kx + C
= kx¹ by power rule
eˣ
eˣ + C
self-integral
e^(ax+b)
e^(ax+b)/a + C
reverse chain rule, divide by a
sin x
−cos x + C
note the minus sign
cos x
sin x + C
positive
sec²x
tan x + C
1/cos²x → tan x
sin(ax+b)
−cos(ax+b)/a + C
reverse chain rule
cos(ax+b)
sin(ax+b)/a + C
reverse chain rule
sec²(ax+b)
tan(ax+b)/a + C
reverse chain rule
(ax+b)ⁿ (n≠−1)
(ax+b)ⁿ⁺¹/[a(n+1)] + C
reverse chain rule; divide by a AND (n+1)
1/(ax+b)
(1/a)ln|ax+b| + C
reverse chain rule for n=−1
f(g(x))·g'(x)
F(g(x)) + C
substitution u = g(x)
Definite ∫[a,b]f(x)dx
F(b) − F(a)
no + C; upper minus lower
Area under curve
∫[a,b] f(x) dx
f(x) ≥ 0; absolute value if below x-axis
Area between curves
∫[a,b] (top−bottom) dx
find intersections first for limits
Find curve from dy/dx
y = ∫f(x) dx + C, then find C
use given point (a,b) to solve for C
Proof Bank
Three proofs that underpin the theory of integration at A-Level. Understanding these deepens your grasp of why the rules work.
Proof 1 — Integration as Anti-Differentiation (Reverse Power Rule)
We verify that d/dx[xⁿ⁺¹/(n+1)] = xⁿ, confirming the power rule for integration.
Let F(x) = xⁿ⁺¹/(n+1) for n ≠ −1.
Differentiate using the power rule for differentiation:
F'(x) = d/dx[xⁿ⁺¹/(n+1)] = (n+1)xⁿ/(n+1) = xⁿ
Since F'(x) = xⁿ = f(x), we have confirmed: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C Q.E.D.
The constant C arises because any constant has derivative zero, so we cannot determine it from differentiation alone.
Proof 2 — The Fundamental Theorem of Calculus (Statement & Intuition)
Statement: If F is an antiderivative of f (i.e. F'(x) = f(x)) and f is continuous on [a,b], then:
∫[a,b] f(x) dx = F(b) − F(a)
Intuition via Riemann sums:
Define A(x) = ∫[a,x] f(t) dt as the area function from a to x.
A small increase in x by h adds a thin strip of width h and height ≈ f(x):
A(x + h) − A(x) ≈ f(x) · h
Dividing by h and taking the limit: A'(x) = f(x)
So A is an antiderivative of f. Since F is also an antiderivative, F(x) = A(x) + constant.
At x = a: F(a) = A(a) + constant = 0 + constant.
Therefore: ∫[a,b] f(x) dx = A(b) = F(b) − F(a). Q.E.D.
Proof 3 — Riemann Sum Concept
The definite integral is defined as the limit of a Riemann sum. Divide [a,b] into n equal subintervals each of width Δx = (b−a)/n.
Choose a sample point xᵢ* in each subinterval [xᵢ₋₁, xᵢ].
The area of the i-th rectangle is f(xᵢ*) · Δx.
Sum all rectangles: S_n = Σᵢ₌₁ⁿ f(xᵢ*) · Δx
As n → ∞ (and Δx → 0), if f is continuous, this sum converges: ∫[a,b] f(x) dx = lim[n→∞] Σᵢ₌₁ⁿ f(xᵢ*) · (b−a)/n
This definition shows that integration literally sums infinitely many infinitely thin rectangles to compute exact area. The Fundamental Theorem then provides an efficient computational method avoiding this limit.
Area Under Curve Visualiser
Drag the upper limit slider to see the shaded area under y = x² from 0 to b. Area above x-axis is shown in pink; area below in blue. The exact value is computed and displayed.
The curve C has equation y = 2x^(3/2) − 3x + 1. (i) Find the x-coordinates where C crosses the x-axis in the interval x ≥ 0. (ii) Find the area bounded by C and the x-axis between these two crossings.
(i) Set 2x^(3/2) − 3x + 1 = 0. Let t = x^(1/2): 2t³ − 3t² + 1 = 0 [M1]
Factor: (t−1)(2t²−t−1) = (t−1)(2t+1)(t−1) = (t−1)²(2t+1) = 0 [M1]
t = 1 (double root) → x = 1; t = −1/2 rejected (t ≥ 0). Also x = 0: y = 1 ≠ 0. Check: x = 0 gives y = 1.
Wait — recheck at x=0: y=1. Root only at x=1 with multiplicity 2 (tangency). Also check x=1: y = 2−3+1 = 0 ✓ [A1 x=1]
For the region: curve from x=0 to x=1 (y ≥ 0 since y(0)=1, y(1)=0) [B1]
(ii) Area = ∫[0,1] (2x^(3/2) − 3x + 1) dx = [4x^(5/2)/5 − 3x²/2 + x]₀¹ [M1]
= 4/5 − 3/2 + 1 = 4/5 − 1/2 = 8/10 − 5/10 = 3/10 square units [A1]
Past Paper Questions
5 questions drawn from Cambridge A-Level 9709 Pure 1 past papers on Integration. Attempt each before revealing the solution.
The gradient of a curve is given by dy/dx = (2x − 1)/(x²(x − 1)) which can be written as A/x + B/x² + C/(x−1). Given the curve passes through (2, 3), find the equation of the curve.
The curve y = f(x) has a gradient function f'(x) = 3x^(1/2) − 6x^(−1/2). Find the area enclosed between the curve, the x-axis and the lines x = 1 and x = 9, given that f(4) = 0.