IB Mathematics AI – Topic 5

Definition:

If \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\)

\[ \int f(x)\,dx = F(x) + C \]

where C is the constant of integration

Basic Integration Rules:

Power Rule:

\[ \int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1 \]

Special case (n = -1):

\[ \int \frac{1}{x}\,dx = \ln|x| + C \]

Standard Integrals (HL):

• \(\int e^x\,dx = e^x + C\)
• \(\int \sin x\,dx = -\cos x + C\)
• \(\int \cos x\,dx = \sin x + C\)
• \(\int \sec^2 x\,dx = \tan x + C\)

Properties:

\[ \int cf(x)\,dx = c\int f(x)\,dx \]

\[ \int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx \]

⚠️ Common Pitfalls & Tips:

• Don't forget +C for indefinite integrals
• Power rule: add 1 to exponent, then divide by new exponent
• Check answer by differentiating
• All standard integrals in formula booklet

Definite Integral:

\[ \int_a^b f(x)\,dx = [F(x)]_a^b = F(b) - F(a) \]

Evaluates to a NUMBER (no +C needed)

Fundamental Theorem of Calculus:

If \(F'(x) = f(x)\), then:

\[ \int_a^b f(x)\,dx = F(b) - F(a) \]

Steps:

1. Find antiderivative F(x)
2. Evaluate F(b) - F(a)
3. No constant C needed

Properties:

• \(\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx\) (swap limits → change sign)
• \(\int_a^a f(x)\,dx = 0\) (same limits)
• \(\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx\) (additive)

⚠️ Common Pitfalls & Tips:

• Definite integral = number, indefinite = function + C
• Always subtract lower limit from upper limit
• Use GDC to verify answers
• Swapping limits changes the sign

Area Under Curve (above x-axis):

\[ \text{Area} = \int_a^b f(x)\,dx \]

when \(f(x) \geq 0\) on [a, b]

Area When Function is Below x-axis:

If \(f(x) < 0\), the integral is negative

For area (always positive):

\[ \text{Area} = \left|\int_a^b f(x)\,dx\right| \]

Area Between Curve and x-axis (mixed):

If curve crosses x-axis, split at zeros:

\[ \text{Total Area} = \left|\int_a^c f(x)\,dx\right| + \left|\int_c^b f(x)\,dx\right| \]

where c is where f(x) = 0

⚠️ Common Pitfalls & Tips:

• Area is always positive - use absolute value if needed
• If curve crosses x-axis, split into regions
• Use GDC to visualize and calculate
• Units: area is in square units (cm², m²)

Area Between Two Curves:

For curves \(y = f(x)\) and \(y = g(x)\) where \(f(x) \geq g(x)\) on [a, b]:

\[ \text{Area} = \int_a^b [f(x) - g(x)]\,dx \]

Integration of (upper function - lower function)

Steps:

1. Find intersection points (set f(x) = g(x))
2. Determine which function is upper/lower
3. Integrate (upper - lower) between intersection points
4. Take absolute value for area

Alternative Method:

\[ \text{Area} = \left|\int_a^b [f(x) - g(x)]\,dx\right| \]

⚠️ Common Pitfalls & Tips:

• Always subtract lower from upper function
• Find intersection points first
• Sketch curves to identify which is upper/lower
• Use GDC to find intersections and area

Solution:

Step 1: Find intersection points

Set \(x^2 = 2x + 3\)

\[ x^2 - 2x - 3 = 0 \]

\[ (x-3)(x+1) = 0 \]

x = -1 or x = 3

Step 2: Identify upper and lower functions

Between x = -1 and x = 3:

Line \(y = 2x + 3\) is above parabola \(y = x^2\)

Step 3: Set up integral

\[ \text{Area} = \int_{-1}^3 [(2x+3) - x^2]\,dx \]

\[ = \int_{-1}^3 (-x^2 + 2x + 3)\,dx \]

Step 4: Integrate

\[ = \left[-\frac{x^3}{3} + x^2 + 3x\right]_{-1}^3 \]

Step 5: Evaluate

At x = 3: \(-\frac{27}{3} + 9 + 9 = -9 + 18 = 9\)

At x = -1: \(-\frac{-1}{3} + 1 - 3 = \frac{1}{3} - 2 = -\frac{5}{3}\)

\[ \text{Area} = 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{32}{3} \]

Answer: \(\frac{32}{3}\) or 10.67 square units

Definition:

Method to approximate area under curve using trapezoids

Trapezoidal Rule Formula:

\[ \int_a^b f(x)\,dx \approx \frac{h}{2}[y_0 + 2(y_1 + y_2 + ... + y_{n-1}) + y_n] \]

where:

• \(h = \frac{b-a}{n}\) is the width of each trapezoid
• n is the number of intervals (trapezoids)
• \(y_i = f(x_i)\) are the function values at each point

Pattern:

First and last y-values appear once

All middle y-values appear with coefficient 2

Alternative Form:

\[ \text{Area} \approx \frac{h}{2}[(y_0 + y_n) + 2(y_1 + y_2 + ... + y_{n-1})] \]

⚠️ Common Pitfalls & Tips:

• More trapezoids (larger n) → better approximation
• First and last terms have coefficient 1, middle terms have 2
• Don't forget h/2 factor outside brackets
• Use GDC to calculate or verify

Definition:

Technique for integrating composite functions

Reverse of chain rule in differentiation

Method:

For integrals of form \(\int f(g(x))g'(x)\,dx\):

1. Choose substitution: \(u = g(x)\)
2. Find \(\frac{du}{dx} = g'(x)\), so \(dx = \frac{du}{g'(x)}\)
3. Rewrite integral in terms of u
4. Integrate with respect to u
5. Substitute back to get answer in terms of x

For Definite Integrals:

Two options:

• Change limits: use \(u(a)\) and \(u(b)\)
• Substitute back and use original limits

Common Examples:

1. \(\int 2x(x^2+1)^5\,dx\)

Let \(u = x^2 + 1\), then \(du = 2x\,dx\)

\[ = \int u^5\,du = \frac{u^6}{6} + C = \frac{(x^2+1)^6}{6} + C \]

2. \(\int \cos x \cdot e^{\sin x}\,dx\)

Let \(u = \sin x\), then \(du = \cos x\,dx\)

\[ = \int e^u\,du = e^u + C = e^{\sin x} + C \]

⚠️ Common Pitfalls & Tips:

• Look for function and its derivative in integrand
• For definite integrals, either change limits or substitute back
• Check answer by differentiating
• Practice recognizing substitution patterns

Solution:

Step 1: Choose substitution

Let \(u = x^3 + 1\)

Step 2: Find du

\[ \frac{du}{dx} = 3x^2 \]

\[ du = 3x^2\,dx \]

Step 3: Change limits

When x = 0: \(u = 0^3 + 1 = 1\)

When x = 2: \(u = 2^3 + 1 = 9\)

Step 4: Rewrite integral

\[ \int_0^2 (x^3+1)^4 (3x^2\,dx) = \int_1^9 u^4\,du \]

Step 5: Integrate

\[ = \left[\frac{u^5}{5}\right]_1^9 \]

Step 6: Evaluate

\[ = \frac{9^5}{5} - \frac{1^5}{5} = \frac{59049 - 1}{5} = \frac{59048}{5} = 11809.6 \]

Answer: 11810 (or \(\frac{59048}{5}\))

Definition:

Volume of solid formed when region is rotated about an axis

Rotation About x-axis:

\[ V = \pi\int_a^b [f(x)]^2\,dx \]

For region between \(y = f(x)\) and x-axis, from x = a to x = b

Rotation About y-axis:

\[ V = \pi\int_c^d [g(y)]^2\,dy \]

For region between \(x = g(y)\) and y-axis, from y = c to y = d

Key Points:

• Square the function: \([f(x)]^2\)
• Don't forget π factor
• Volume is in cubic units (cm³, m³)
• Visualize the solid using GDC

Method:

1. Identify axis of rotation
2. Write function appropriately
3. Square the function
4. Set up integral with π factor
5. Evaluate using GDC

⚠️ Common Pitfalls & Tips:

• Must square the function inside integral
• Don't forget π factor
• Check which axis rotation is about
• Use GDC for calculation

Basic Rules

• Power: \(\frac{x^{n+1}}{n+1}\) + C
• Indefinite: Always +C
• Definite: Gives number

Areas

• Under curve: \(\int_a^b f(x)\,dx\)
• Between: \(\int_a^b [f-g]\,dx\)
• Always positive

Trapezoidal Rule

• h/2[first + 2(middles) + last]
• Approximation method
• More trapezoids = better

Volume (HL)

• \(\pi\int [f(x)]^2\,dx\)
• Square function
• Don't forget π