INTEGRATION TECHNIQUES — CORE RESULTS & APPLICATIONS

Concept	Key Idea
Power Rule	Integrates xn for all rational n except −1
Logarithmic Case	Special rule for ∫1/x dx
Trig Integrals	Reverse differentiation of trig functions
Exponential Integrals	ex remains unchanged under integration
Substitution	Undoing the chain rule

📌 1. Power Rule for Integration (n ≠ −1)

• Core rule: ∫xⁿ dx = xⁿ⁺¹ / (n+1) + C
• Applicability: Works for positive, negative, and fractional rational exponents
• Why n ≠ −1: Division by zero would occur if n+1 = 0
• Conceptual meaning: Measures accumulated area under polynomial curves
• Exam habit: Always rewrite roots and fractions as powers before integrating

📌 2. Logarithmic Special Case

• Unique exception: ∫1/x dx = ln|x| + C
• Reason: Power rule fails because exponent −1 causes division by zero
• Absolute value: Required because ln(x) is undefined for x ≤ 0
• Key appearance: In inverse proportionality models
• Common trap: Never write ln(x) without absolute value in IB exams

📌 3. Trigonometric Integrals

• Reverse differentiation: Integration undoes derivatives
• Key results: ∫cos x dx = sin x + C
• Negative signs: ∫sin x dx = −cos x + C
• Tan result: ∫sec²x dx = tan x + C
• Use case: Oscillatory and wave motion

📌 4. Exponential Integrals

• Unique property: e^x is its own derivative and integral
• Basic rule: ∫e^x dx = e^x + C
• Scaled exponent: ∫e^kx dx = (1/k)e^kx + C
• Growth meaning: Continuous accumulation processes
• IB relevance: Central to modelling questions

📌 5. Integration by Substitution (u-sub)

• Purpose: Simplifies composite integrals
• Core idea: Replace inner function with u
• Derivative link: du must appear in the integral
• Back-substitution: Required for indefinite integrals
• IB marking: Full method marks need full substitution steps

📌 Practice Questions (IB Style)

Multiple Choice Questions

Select the one correct option for each question.

MCQ 1. Evaluate the integral
∫ x³ dx.

A. x⁴ + C
B. 3x² + C
C. x⁴⁄4 + C
D. 4x³ + C

Answer & Explanation

Correct answer: C
Using the power rule, ∫xⁿ dx = xⁿ⁺¹/(n+1) + C.
Here n = 3, so the integral becomes x⁴/4 + C.

MCQ 2. Which of the following integrals evaluates to ln|x| + C?

A. ∫ x⁻² dx
B. ∫ 1/x dx
C. ∫ eˣ dx
D. ∫ x dx

Answer & Explanation

Correct answer: B
The logarithmic result occurs only when integrating x⁻¹.
All other powers are handled using the power rule.

MCQ 3. Evaluate the integral
∫ sec²x dx.

A. sin x + C
B. cos x + C
C. tan x + C
D. −cos x + C

Answer & Explanation

Correct answer: C
Since d/dx(tan x) = sec²x, the antiderivative of sec²x is tan x + C.

Short Answer Questions

Answer clearly, showing all necessary reasoning.

Question 1.
Explain why the power rule cannot be applied to evaluate ∫ x⁻¹ dx.

Model Answer

The power rule requires division by n + 1.
When n = −1, this results in division by zero, which is undefined.
Therefore, a different function is required, and the integral of x⁻¹ is defined as ln|x| + C instead.

Question 2.
Find ∫ e⁴ˣ dx and explain the method used.

Model Answer

The derivative of e⁴ˣ is 4e⁴ˣ, so the integral must compensate for this factor.
Dividing by 4 gives ∫ e⁴ˣ dx = (1/4)e⁴ˣ + C.
This follows directly from reversing the chain rule.

Long Answer Questions

Question 1.
Let f(x) = (2x² + 1)³.

(a) Show that ∫ (2x² + 1)³ · 4x dx can be evaluated using substitution.
(b) Find the value of the integral.
(c) Verify your result by differentiation.

Full Solution

(a) Let u = 2x² + 1.
Then du = 4x dx, which exactly matches the remaining factor in the integrand.

(b) The integral becomes ∫ u³ du = u⁴/4 + C.
Substituting back gives (2x² + 1)⁴/4 + C.

(c) Differentiating (2x² + 1)⁴/4 using the chain rule returns
(2x² + 1)³ · 4x, confirming the result.

Question 2.
A particle moves along a straight line with velocity
v(t) = 3t² − 6t.

(a) Find the displacement of the particle for 0 ≤ t ≤ 3.
(b) Find the total distance travelled in this interval.
(c) Explain the difference between displacement and total distance in this context.

Full Solution

(a) Displacement is given by ∫ v(t) dt.
∫ (3t² − 6t) dt = t³ − 3t².
Evaluating from 0 to 3 gives 0.

(b) To find total distance, solve v(t) = 0 → t = 0 and t = 2.
The velocity is negative on (0,2) and positive on (2,3).
Distance = |∫₀² v(t) dt| + ∫₂³ v(t) dt = 5.

(c) Displacement measures net change in position, which is zero here.
Distance measures total motion regardless of direction, which is non-zero.