You have spent two months mastering derivatives — the art of finding rates of change. Now we ask the reverse: given the rate, can we recover the original? Welcome to integration.
The Big Shift
Two Months of Derivatives. Now, Everything Runs Backwards.
Think back to the beginning of this course. A car moves along a road and you asked: "Given the position, what is the speed at this instant?" You differentiated. The derivative gave you the rate of change.
Now flip the entire question around. The speedometer is broken, but the odometer works perfectly — you know exactly how fast the car is moving at every moment. The question is now: "Given the speed, what was the total distance traveled?"
This reverse question, recovering a quantity from its rate of change, is the soul of integration. If differentiation is the art of finding the derivative, integration is the art of undoing it. And just as the derivative turned out to be one of the most powerful ideas in mathematics, so too will its inverse.
Where we are in the course: Chapters 1–4 built the machinery of differentiation. Chapter 5 opens a second world entirely — the world of accumulation, area, and antiderivatives. By the end of this chapter, you will see that these two worlds are secretly the same, connected by the most beautiful theorem in calculus.
🚗 A Concrete Example of the Reverse Problem
A delivery rider in Lahore tracks her speed (km/h) every 30 minutes during a 3-hour shift:
Time (hr)
0.0
0.5
1.0
1.5
2.0
2.5
Speed (km/h)
30
45
60
40
55
50
Best estimate of distance: $(30+45+60+40+55+50)(0.5) = \mathbf{140 \text{ km}}$
But speed changes continuously. If speed were given by a formula $v(t)$, could we find the exact distance? Yes — that is precisely what integration will give us.
The Core Question of §5.1: Given a function $f(x)$, can we find a function $F(x)$ whose derivative is $f(x)$? That is, can we find $F$ such that $F'(x) = f(x)$?
5.1a — Antidifferentiation
Antidifferentiation
If differentiation is the art of finding rates of change, antidifferentiation is the art of running backwards — recovering a function from its derivative.
Definition — Antiderivative
A function $F(x)$ is an antiderivative of $f(x)$ if $F'(x) = f(x)$ for all $x$ in the domain.
First Examples
Example 1. Find an antiderivative of $f(x) = 3x^2$. Ask: what function differentiates to $3x^2$? We know $\frac{d}{dx}(x^3) = 3x^2$. So $F(x) = x^3$ is an antiderivative.
Example 2. Find an antiderivative of $f(x) = \cos x$. Since $\frac{d}{dx}(\sin x) = \cos x$, we have $F(x) = \sin x$.
Fundamental Property of Antiderivatives
Theorem
If $F(x)$ is an antiderivative of continuous $f(x)$, then every other antiderivative has the form $G(x) = F(x) + C$ for some constant $C \in \mathbb{R}$.
💡 Why the +C? Quick Proof
Let $G$ be any antiderivative of $f$, and $F$ another. Define $H = G - F$. Then $H'(x) = G'(x) - F'(x) = f(x) - f(x) = 0$ everywhere. A function with zero derivative everywhere must be a constant. So $G - F = C$, giving $G = F + C$. ∎
Geometric meaning: All antiderivatives are vertical shifts of each other. The family $x^3$, $x^3+7$, $x^3 - \pi$ all have derivative $3x^2$.
5.1b — The Indefinite Integral & Its Notation
The Indefinite Integral
The collection of all antiderivatives of f(x) is called the indefinite integral of f, written with the integral sign:
The Indefinite Integral
If $F'(x)=f(x)$, then:
$$\int f(x)\,dx = F(x) + C$$
Symbol
Name
Meaning
$\int$
Integral sign
Stretched "S" — Leibniz invented it in 1675 for Summa (sum)
$f(x)$
Integrand
The function being antidifferentiated
$dx$
Differential
Identifies the variable; connects to the limit $\Delta x \to 0$ later
$C$
Constant of integration
Represents the entire family of antiderivatives
Key difference: The indefinite integral $\int f(x)\,dx$ is a family of functions (with $+C$). The definite integral $\int_a^b f(x)\,dx$ is a number. We study the definite integral in §5.3.
5.1c — Rules for Integrating Common Functions
Integration Rules
Each rule below is simply a differentiation rule read in reverse. Always verify by differentiating your answer.
Rule 1 — The Constant Rule
$$\int k\,dx = kx + C$$
Since $\frac{d}{dx}(kx)=k$, the antiderivative of any constant $k$ is $kx$.
A LUMS student cycles to campus with velocity $v(t) = 6t^2 - 4t + 3$ km/h, where $t$ is hours. If the student starts at position $s=0$, find the position function $s(t)$.
A differential equation is an equation involving an unknown function and its derivatives — the language in which physics, biology, and economics write their deepest laws.
Definition
The order of a DE is the order of the highest derivative appearing. In Calc I we study separable first-order equations where variables can be separated onto opposite sides.
Type
Form
Note
Separable
$\dfrac{dy}{dx}=f(x)g(y)$
Our focus in Calc I
Linear (1st order)
$y'+P(x)y=Q(x)$
Integrating factor method
Autonomous
$dy/dt=f(y)$
Right side depends only on y
Higher order
$y''+4y=\sin x$
Studied in Diff. Equations course
Method — Separation of Variables
Given $\dfrac{dy}{dx}=f(x)\cdot g(y)$:
Separate variables: $\dfrac{dy}{g(y)}=f(x)\,dx$
Integrate both sides independently
Solve algebraically for $y$
Apply initial condition to find the specific constant $C$
Why It Works
Treating $\dfrac{dy}{dx}$ as a fraction and cross-multiplying is justified by the Chain Rule. When we integrate $\int\dfrac{dy}{g(y)}$, we integrate the left side with respect to $y$, and the right side with respect to $x$ — legitimate because $dy = \dfrac{dy}{dx}dx$.
💼 Revenue IVP
The marginal revenue for a LUMS canteen is $\dfrac{dR}{dq}=50-2q$, with $R(0)=0$. Find $R(q)$ and the revenue-maximising quantity.
$R(q)=\displaystyle\int(50-2q)\,dq = 50q-q^2+C$
Apply $R(0)=0$: $C=0$
$$R(q)=50q-q^2 \qquad \text{(max at } q=25\text{, giving }R=625\text{)}$$
🌿 Exponential Growth/Decay
Solve $\dfrac{dy}{dx}=ky$, $y(0)=y_0$. This models population growth, radioactive decay, and compound interest simultaneously.
Separate: $\dfrac{dy}{y}=k\,dx$
Integrate: $\ln|y|=kx+C_1$
Exponentiate: $y=Ce^{kx}$ where $C=e^{C_1}$
Apply $y(0)=y_0$: $C=y_0$
$$\boxed{y=y_0 e^{kx}}$$
If $k>0$: exponential growth. If $k<0$: exponential decay.
5.1f — Continuous Compounding
Continuous Compounding via Differential Equations
🏠 The Model
$\dfrac{dB}{dt}=rB$ — the balance grows at a rate proportional to itself. This is the DE of continuous compounding: the more you have, the faster it grows.
Theorem — Continuous Compounding Formula
If $P$ rupees are invested at annual rate $r$ compounded continuously for $t$ years:
$$B(t)=Pe^{rt}$$
✏️ Full Derivation from the DE
IVP: $\dfrac{dB}{dt}=rB$, $B(0)=P$.
Separate: $\dfrac{dB}{B}=r\,dt$
Integrate: $\ln|B|=rt+C_1$
Exponentiate: $B=Ce^{rt}$
$B(0)=C=P$, therefore $B(t)=Pe^{rt}$ ✓
💵 Example
PKR 5,000 invested at $6\%$ annual rate compounded continuously. Find the balance after 10 years.
📈 Continuous Compounding Explorer — $B(t)=Pe^{rt}$
$B(t)$ = PKR —
Interest = PKR —
Doubles in ~ — yrs
Rule of 70: At rate $r\%$, money doubles in approximately $70/r$ years. Derived from $Pe^{rt}=2P \Rightarrow t=\dfrac{\ln 2}{r} \approx \dfrac{0.693}{r}$.