IB Mathematics AI AHL Definite and indefinite integration MAI Study Notes - New Syllabus

IB Mathematics AI AHL Definite and indefinite integration MAI Study Notes

LEARNING OBJECTIVE

Definite and indefinite integration

Key Concepts:

Definite and indefinite integration
Integration by inspection, or substitution

Definition:

Integration is the reverse process of differentiation. It allows us to find a function when its derivative is known and calculate the area under curves.

Types of Integration:

Indefinite Integration:

It represents a general form of the antiderivative of a function and includes a constant of integration \( C \). It does not have limits.

Definite Integration:

It calculates the net area under the curve between specific limits \( a \) and \( b \). The result is a number, no constant \( C \).

Key Differences:

Aspect	Indefinite Integration	Definite Integration
Limits	No limits	Specific limits \( a \) to \( b \)
Result	Function + \( C \)	Numerical value
Represents	Family of functions	Area under the curve

Standard integrals include:

Power rule: \( \displaystyle\int x^n \, dx = \dfrac{x^{n+1}}{n+1} + C \) for \( n \neq -1 \)
\( \displaystyle\int \dfrac{1}{x} \, dx = \ln|x| + C \)
\( \displaystyle\int \sin x \, dx = -\cos x + C \)
\(\displaystyle \int \cos x \, dx = \sin x + C \)
\(\displaystyle \int \dfrac{1}{cos^2(x)} \, dx =\ tan x+ C \)
\( \displaystyle\int e^x \, dx = e^x + C \)

Example: Indefinite Integral

Evaluate the integral: \( \int \left( 2x^3 – 3x + 4 \right) dx \)

▶️Answer/Explanation

Use the power rule on each term:

\( \int 2x^3 \, dx = \frac{2x^4}{4} = \frac{x^4}{2} \)
\( \int -3x \, dx = \frac{-3x^2}{2} \)
\( \int 4 \, dx = 4x \)

Final Answer: \( \frac{x^4}{2} – \frac{3x^2}{2} + 4x + C \)

Example: Definite Integral

Evaluate the integral: \( \int_{1}^{3} (x^2 + 2) dx \)

▶️Answer/Explanation

\( \int x^2 \, dx = \frac{x^3}{3} \)
\( \int 2 \, dx = 2x \)

Combined integral: \( \frac{x^3}{3} + 2x \)

Apply limits:

\( \left( \frac{3^3}{3} + 2 \times 3 \right) – \left( \frac{1^3}{3} + 2 \times 1 \right) = \left( \frac{27}{3} + 6 \right) – \left( \frac{1}{3} + 2 \right) \)

\( = (9 + 6) – \left( \frac{1}{3} + 2 \right) = 15 – \left( \frac{1}{3} + 2 \right) = 15 – \frac{7}{3} = \frac{38}{3} \)

Final Answer: \( \frac{38}{3} \)

Definition:

This method involves recognizing the derivative of a known function within the integrand, allowing for a direct “guess and check” anti-derivative.

It is often used when the integrand matches a known derivative form, especially with composite expressions.

Example

Evaluate: \( \int \frac{1}{3x + 2} \, dx \)

▶️Answer/Explanation

Recognize the integral as matching \( \int \frac{1}{ax + b} \, dx = \frac{1}{a} \ln|ax + b| + C \)

Here: \( a = 3, b = 2 \)

Answer: \( \frac{1}{3} \ln|3x + 2| + C \)

Example

Evaluate: \( \int \frac{\sin x}{\cos x} \, dx \)

▶️Answer/Explanation

Let \( u = \cos x \Rightarrow \frac{du}{dx} = -\sin x \Rightarrow dx = -\frac{du}{\sin x} \)
Rewrite integral: \( \int \frac{\sin x}{\cos x} \, dx = \int \frac{1}{u} \cdot (-du) = -\int \frac{1}{u} \, du \)
Integral: \( -\ln |u| + C = -\ln |\cos x| + C \)

Final Answer: \( -\ln |\cos x| + C \)

Definition:

This method simplifies an integral by substituting a part of the expression with a new variable. It is effective when the integrand is the product of a function and its derivative.

General form:

\( \int f(g(x))g'(x) \, dx = \int f(u) \, du \), where \( u = g(x) \)

General Steps of Substitution

Identify the inner function \( g(x) \) such that its derivative \( g'(x) \) appears in the integrand.
Substitute \( u = g(x) \).
Find \( du = g'(x) \, dx \) and rewrite the integral in terms of \( u \).
Integrate with respect to \( u \).
Substitute back \( x \) to get the final answer.

Example

Evaluate: \( \int \sin(2x + 5) \, dx \)

▶️Answer/Explanation

Let \( u = 2x + 5 \Rightarrow \frac{du}{dx} = 2 \Rightarrow dx = \frac{du}{2} \)
Substitute: \( \int \sin(u) \cdot \frac{1}{2} du = \frac{1}{2} \int \sin(u) \, du \)
Integral: \( \frac{1}{2}(-\cos u) + C = -\frac{1}{2} \cos(2x + 5) + C \)

Final Answer: \( -\frac{1}{2} \cos(2x + 5) + C \)

Example

Evaluate: \( \int 4x \sin x^2 \, dx \)

▶️Answer/Explanation

Let \( u = x^2 \Rightarrow \frac{du}{dx} = 2x \Rightarrow dx = \frac{du}{2x} \)
Rewrite integral: \( \int 4x \sin(x^2) \, dx = 4 \int x \sin(x^2) \, dx \)
Substitute: \( 4 \int \sin(u) \cdot \frac{du}{2} = 2 \int \sin(u) \, du \)
Integral: \( 2(-\cos u) + C = -2\cos(x^2) + C \)

Final Answer: \( -2\cos(x^2) + C \)

The Moscow Mathematical Papyrus (circa 1850 BCE) contains the oldest known correct formula for the volume of a frustum of a square pyramid, found in Problem 14.

The formula given by the Egyptians is:

\( V = \frac{h}{3} \left( a^2 + ab + b^2 \right) \)

This formula shows the advanced practical geometry skills of ancient Egyptian scribes, used for architecture, construction, and land surveying.

It is exactly equivalent to the modern formula used today for calculating the volume of a frustum of a square pyramid, showing their remarkable mathematical understanding.

Example

Calculate the volume of a pyramidal frustum with: