Unit 5 - Notes

MTH005

Unit 5: Integral Calculus

5.1 Introduction to Integration (Indefinite Integrals)

Calculus consists of two main branches: Differential Calculus (rates of change) and Integral Calculus (accumulation). Integration is essentially the inverse process of differentiation.

5.1.1 The Indefinite Integral

An indefinite integral represents a family of functions. If $F(x)$ is a function such that its derivative is $f(x)$ (i.e., $F'(x) = f(x)$ ), then $F(x)$ is called an antiderivative of $f(x)$ .

Notation:

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

$\int$ : The integral symbol (an elongated 'S' for summation).
$f(x)$ : The integrand (the function being integrated).
$dx$ : The variable of integration (indicates we are integrating with respect to $x$ ).
$F(x)$ : The antiderivative.
$C$ : The Constant of Integration. Because the derivative of a constant is zero, there are infinite antiderivatives for a function, differing only by a constant vertical shift.

5.1.2 Basic Rules and Formulas

1. The Power Rule
For any real number $n \neq -1$ :

\int x^n \, dx = \frac{x^{n+1}}{n+1} + C

Example: $\int x^4 \, dx = \frac{x^5}{5} + C$
Example: $\int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$

2. The Constant Multiple Rule

\int k \cdot f(x) \, dx = k \int f(x) \, dx

3. The Sum and Difference Rule

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

4. Special Functions

Exponential: $\int e^x \, dx = e^x + C$
Logarithmic (Reciprocal): $\int \frac{1}{x} \, dx = \ln|x| + C$
Trigonometric:
- $\int \sin x \, dx = -\cos x + C$
- $\int \cos x \, dx = \sin x + C$
- $\int \sec^2 x \, dx = \tan x + C$

5.2 Integration by Substitution

Integration by substitution is the counterpart to the Chain Rule in differentiation. It is used when the integrand is a composite function, or when the integrand contains a function and its derivative.

5.2.1 The Method (u-Substitution)

If the integral is of the form $\int f(g(x))g'(x) \, dx$ :

Choose $u$ : Let $u = g(x)$ (usually the "inner" part of a composite function).
Differentiate: Find $du = g'(x) \, dx$ .
Solve for $dx$ : Isolate $dx$ or match the terms in the integrand to $du$ .
Substitute: Rewrite the entire integral in terms of $u$ . The variable $x$ must disappear completely.
Integrate: Perform the integration with respect to $u$ .
Back-substitute: Replace $u$ with the original expression $g(x)$ .

5.2.2 Example

Solve: $\int 2x(x^2 + 1)^3 \, dx$

Let $u = x^2 + 1$ .
Differentiate: $\frac{du}{dx} = 2x \implies du = 2x \, dx$ .
Substitute: The term $2x \, dx$ in the original integral becomes $du$ .
$\int (x^2 + 1)^3 \cdot (2x \, dx) = \int u^3 \, du$
Integrate:
$\frac{u^4}{4} + C$
Back-substitute:
$\frac{(x^2 + 1)^4}{4} + C$

5.3 Integration by Parts

Integration by parts is the counterpart to the Product Rule in differentiation. It is used to integrate the product of two different types of functions (e.g., algebraic multiplied by exponential).

5.3.1 The Formula

\int u \, dv = uv - \int v \, du

5.3.2 Selection Strategy (LIATE Rule)

To use the formula successfully, you must choose one part of the integrand to be $u$ (which you differentiate) and the rest to be $dv$ (which you integrate). Choose $u$ based on the LIATE priority list:

L - Logarithmic functions ( $\ln x$ )
I - Inverse trigonometric functions ( $\arctan x$ )
A - Algebraic functions ( $x^2, 3x$ )
T - Trigonometric functions ( $\sin x, \cos x$ )
E - Exponential functions ( $e^x$ )

Whichever function comes first in this list is your $u$ .

5.3.3 Example

Solve: $\int x \cos x \, dx$

Assign variables:
- Algebraic ( $x$ ) comes before Trigonometric ( $\cos x$ ) in LIATE.
- Let $u = x$
- Let $dv = \cos x \, dx$
Differentiate $u$ and Integrate $dv$ :
- $du = 1 \, dx$
- $v = \int \cos x \, dx = \sin x$ (ignore $+C$ for intermediate steps)
Apply Formula:
$\int x \cos x \, dx = x \sin x - \int \sin x \, dx$
Solve remaining integral:
$= x \sin x - (-\cos x) + C$
$= x \sin x + \cos x + C$

5.4 Definite Integrals

While an indefinite integral yields a function, a definite integral yields a numerical value representing the net accumulation of the quantity.

5.4.1 Notation and Definition

\int_{a}^{b} f(x) \, dx

$a$ : Lower limit of integration.
$b$ : Upper limit of integration.

5.4.2 The Fundamental Theorem of Calculus (Part II)

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ , then:

\int_{a}^{b} f(x) \, dx = F(b) - F(a)

Notation usually written as:

[F(x)]_a^b

5.4.3 Properties of Definite Integrals

Zero Width: $\int_{a}^{a} f(x) \, dx = 0$
Reversing Limits: $\int_{a}^{b} f(x) \, dx = - \int_{b}^{a} f(x) \, dx$
Splitting Intervals: $\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx$ (where $c$ is between $a$ and $b$ )

5.4.4 Example

Evaluate $\int_{1}^{3} x^2 \, dx$ .

Find antiderivative: $F(x) = \frac{x^3}{3}$ .
Apply limits:
$F(3) - F(1) = \left(\frac{3^3}{3}\right) - \left(\frac{1^3}{3}\right)$
$= \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{26}{3}$

5.5 Area Under Simple Well-Known Curves

One of the primary geometric applications of the definite integral is calculating the area between a curve and the x-axis.

5.5.1 The Geometric Interpretation

The definite integral $\int_{a}^{b} f(x) \, dx$ represents the net signed area:

Positive Area: If $f(x) > 0$ (graph is above the x-axis), the integral is positive.
Negative Area: If $f(x) < 0$ (graph is below the x-axis), the integral is negative.

To find the total geometric area, one must take the absolute value of regions below the axis.

\text{Total Area} = \int_{a}^{b} |f(x)| \, dx

5.5.2 Steps to Find Area

Sketch the curve: Identify where it sits relative to the axes.
Identify boundaries: Determine limits $x=a$ and $x=b$ .
Set up the integral: $\text{Area} = \int_{a}^{b} y \, dx$ .
Evaluate.

5.5.3 Example: Area under a Parabola

Find the area enclosed by $y = x^2$ , the x-axis, $x=0$ , and $x=2$ .

Check graph: $y=x^2$ is non-negative for $x \in [0,2]$ .
Set up Integral:
$A = \int_{0}^{2} x^2 \, dx$
Evaluate:
$A = \left[ \frac{x^3}{3} \right]_0^2$
$A = \left( \frac{2^3}{3} \right) - \left( \frac{0^3}{3} \right)$
$A = \frac{8}{3} \text{ square units.}$

5.5.4 Example: Area with Negative Regions

Find the area between $y = x$ , the x-axis, $x=-1$ , and $x=1$ .

From $x = -1$ to $0$, the graph is below the axis.
From $x = 0$ to $1$, the graph is above the axis.

\text{Area} = \left| \int_{-1}^{0} x \, dx \right| + \int_{0}^{1} x \, dx

Region 1 (Negative):

$\int_{-1}^{0} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^0 = 0 - \frac{(-1)^2}{2} = -\frac{1}{2}$
Absolute value $= \frac{1}{2}$ .
Region 2 (Positive):

$\int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^1 = \frac{1}{2} - 0 = \frac{1}{2}$
Total Area:

$\frac{1}{2} + \frac{1}{2} = 1 \text{ square unit.}$

(Note: If we simply integrated $\int_{-1}^{1} x \, dx$ , the result would be 0, which is the net displacement, not the physical area.)