Integral Calculus - The Art of Summation and Accumulation

January 10, 2023 — #Mathematics #Calculus

Calculus is a discipline of two fundamental halves: differential calculus, which studies rates of change, and integral calculus, which deals with accumulation and the summation of infinitesimal parts. While differentiation breaks functions down to analyze their instantaneous behavior, integration builds them up, allowing us to find total areas, volumes, and accumulated quantities. This exploration delves into the core concepts of integral calculus, from its definition as the reverse of differentiation to its powerful application as the limit of a sum, covering the essential techniques and theorems that make it a cornerstone of modern mathematics, science, and engineering.

The Indefinite Integral: The Reverse of Differentiation

The first way to understand integration is as the inverse process of differentiation. This is the concept of the indefinite integral or antiderivative.

A function $F(x)$ is called an antiderivative of a function $f(x)$ if $F'(x) = f(x)$ . The indefinite integral of $f(x)$ with respect to $x$ is the family of all its antiderivatives and is denoted by: $\int f(x) dx = F(x) + C$ where $F(x)$ is any particular antiderivative of $f(x)$ , and $C$ is an arbitrary constant of integration. The constant $C$ is necessary because the derivative of any constant is zero, meaning a function can have infinitely many antiderivatives that differ only by a constant.

Elementary Integration: Fundamental Formulas

These formulas are derived directly from the known results of differentiation.

Algebraic:
- $\int x^n dx = \dfrac{x^{n+1}}{n+1} + C \quad (n \neq -1)$
- $\int \dfrac{1}{x} dx = \ln|x| + C$
Exponential & Logarithmix:
- $\int e^x dx = e^x + C$
- $\int e^{ax} dx = \dfrac{1}{a}e^{ax} + C$
- $\int a^x dx = \dfrac{a^x}{\ln a} + C$
- $\int \ln x dx = x \ln x - x + C$
Trigonometric:
- $\int \cos x dx = \sin x + C$
- $\int \sin x dx = -\cos x + C$
- $\int \sec^2 x dx = \tan x + C$
- $\int \csc^2 x dx = -\cot x + C$
- $\int \sec x \tan x dx = \sec x + C$
- $\int \csc x \cot x dx = -\csc x + C$
- $\int \tan x dx = \ln|\sec x| + C$
- $\int \cot x dx = \ln|\sin x| + C$
- $\int \sec x dx = \ln|\sec x + \tan x| + C$
- $\int \csc x dx = \ln|\csc x + \cot x| + C$
Inverse Trigonometric:
- $\int \dfrac{1}{a^2+x^2} dx = \dfrac{1}{a}\arctan\left(\dfrac{x}{a}\right) + C$
- $\int \dfrac{1}{\sqrt{a^2-x^2}} dx = \arcsin\left(\dfrac{x}{a}\right) + C$
- $\int \dfrac{1}{x\sqrt{x^2-a^2}} dx = \dfrac{1}{a}\text{arcsec}\left(\dfrac{|x|}{a}\right) + C$
Hyperbolic:
- $\int \sinh x dx = \cosh x + C$
- $\int \cosh x dx = \sinh x + C$
- $\int \tanh x dx = \ln|\cosh x| + C$
- $\int \coth x dx = \ln|\sinh x| + C$
- $\int \text{sech}^2 x dx = \text{tanh} x + C$
- $\int \text{csch}^2 x dx = -\text{coth} x + C$
- $\int \text{sech} x \text{tanh} x dx = \text{sech} x + C$
- $\int \text{csch} x \text{coth} x dx = -\text{csch} x + C$

Properties of Indefinite Integrals

Constant Multiple Rule: $\int k \cdot f(x) dx = k \int f(x) dx$ , where $k$ is a constant.
Sum/Difference Rule: $\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx$ .
if $\int f(x)dx = F(x) + C$ , then $\int f(ax+b) dx = \dfrac{F(ax+b)}{a} + C$

Special Integration Formulas

$\int \dfrac{f'(x)}{f(x)} dx = \ln|f(x)| + C$ (for $f(x) \neq 0$ ).
$\int (f(x))^n f'(x) dx = \dfrac{(f(x))^{n+1}}{n+1} + C$ (for $n \neq -1$ ).
$\int \dfrac{1}{a \sin x + b \cos x} dx = \ln \left|\tan \left(\dfrac{x}{2} + \dfrac{1}{2} \tan^{-1} \dfrac{b}{a} \right)\right| + C$ .
$\int e^{ax} \sin(bx) dx = \dfrac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) + C$ .
$\int e^{ax} \cos(bx) dx = \dfrac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) + C$ .

Mastering Antidifferentiation: Key Integration Techniques

Many functions cannot be integrated directly using the basic formulas. Several powerful techniques exist to transform complex integrals into simpler ones.

Integration by Substitution (u-substitution)

This technique is the reverse of the chain rule for differentiation. It is used when an integrand contains a function and its derivative.

Method: If an integral is of the form $\int f(g(x))g'(x) dx$ , we can let $u = g(x)$ , which implies $du = g'(x)dx$ . The integral transforms to: $\int f(u) du$ which may be simpler to evaluate.
Standard Substitutions:
- For $a^2 - x^2$ , use $x = a \sin \theta$ or $x = a \cos \theta$ .
- For $a^2 + x^2$ , use $x = a \tan \theta$ or $x = a \sec \theta$ .
- For $x^2 - a^2$ , use $x = a \sec \theta$ or $x = a \csc \theta$ .
- For $\sqrt{\dfrac{a-x}{a+x}}$ or $\sqrt{\dfrac{a+x}{a-x}}$ , use $x = a \cos (2\theta)$ .
- For $\sqrt{\dfrac{x-\alpha}{\beta-x}}$ or $\sqrt{(x-\alpha)(x-\beta)}$ , use $x = \alpha\cos^2\theta + \beta\sin^2\theta$
- Few derived integrals from these substitutions:
  - For $\int \dfrac{dx}{\text{Quadratic}}$ ,
    - $\int \dfrac{dx}{a^2 + x^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{x}{a}\right) + C$
    - $\int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right| + C$
  - For $\int \dfrac{dx}{\sqrt{\text{Quadratic}}}$ ,
    - $\int \dfrac{dx}{\sqrt{a^2 + x^2}} = \ln\left|x + \sqrt{a^2 + x^2}\right| + C = \sinh^{-1}\left(\dfrac{x}{a}\right) + C$
    - $\int \dfrac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C = \cosh^{-1}\left(\dfrac{x}{a}\right) + C$
    - $\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\dfrac{x}{a}\right) + C$
For $\int \dfrac{px + q}{ax^2 + bx + c} dx$ or $\int \dfrac{px + q}{\sqrt{ax^2 + bx + c}} dx$ , express the numerator in terms of the derivative of denominator together with a constant: $px + q = \lambda(2ax + b) + \mu$ Find $\lambda$ and $\mu$ by comparing coefficients, then integrate each term separately: $\int \dfrac{px + q}{ax^2 + bx + c} dx = \lambda \int \dfrac{2ax + b}{ax^2 + bx + c} dx + \mu \int \dfrac{dx}{ax^2 + bx + c}$
For $\int (\sin^m x \cos^n x) dx$ , where $m$ and $n$ are natural numbers,
- If one of them is odd, substitute for the term of even power.
- If both are odd, substitute either of them.
- If both are even, use trigonometric identities only.
- If $m$ and $n$ are rational numbers and $\dfrac{m+n-2}{2}$ is a negative integer, then substitute $\cot x = p$ or $\tan x = p$ .
For $\int \dfrac{dx}{a \cos^2 x + b \sin^2 x}$ , $\int \dfrac{dx}{a + b \sin^2 x}$ , $\int \dfrac{dx}{a + b \cos^2 x}$ , $\int \dfrac{dx}{(a\sin x + b\cos x)^2}$ , $\int \dfrac{dx}{a + b \sin^2 x + \cos^2 x}$ , divide both the numerator and denominator by $\cos^2 x$ , replace $\sec^2 x$ , if any, in the denominator by $(1+\tan^2 x)$ and put $\tan x = t$ So that $\sec^2 x dx = dt$ .
For $\int \dfrac{dx}{a + b \sin x + c \cos x}$ , write $\sin x$ and $\cos x$ in terms of $\tan \dfrac{x}{2}$ using the Weierstrass substitution: $\sin x = \dfrac{2t}{1+t^2}, \quad \cos x = \dfrac{1-t^2}{1+t^2}$ where $t = \tan \dfrac{x}{2}$ .
For $\int \dfrac{p\cos x + q\sin x + r}{a\cos x + b\sin x + c} dx$ , express numerator as $\Lambda(\text{denominator}) + \mu(\text{derivative of denominator}) + \gamma$ , Find $\Lambda$ , $\mu$ , and $\gamma$ by comparing coefficients, and then integrate each term separately. $\Lambda \int dx + \mu \int \dfrac{\text{derivative of denominator}}{\text{denominator}} dx + n \int \dfrac{dx}{a \sin x + b \cos x + c}$

Integration by Parts

This technique is the reverse of the product rule for differentiation and is used to integrate products of functions.

Formula: $\int uv \, dx = u \left(\int v \, dx \right) - \int \left(\dfrac{du}{dx} \int v \, dx\right)dx$
Derivation: Start with the product rule for derivatives: $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$ . Integrating both sides with respect to $x$ : $\int \dfrac{d}{dx}(uv) dx = \int u\dfrac{dv}{dx} dx + \int v\dfrac{du}{dx} dx$ . Rearranging gives the formula. A common mnemonic for choosing which function to set as $u$ is the ILATE rule (Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential).

Integration by Cancellation

This technique is useful when the integrand can be simplified by canceling terms.

Method: If the integrand is of the form $\dfrac{f(x)}{g(x)}$ where $f(x)$ and $g(x)$ are polynomials, and if $f(x)$ is a factor of $g(x)$ , then: $\int \dfrac{f(x)}{g(x)} dx = \int \dfrac{1}{\text{remaining terms}} dx$ This can often lead to simpler integrals that can be solved using basic techniques.
Few Formulas:
- $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$
- $\int e^{ax} \sin(bx) dx = \dfrac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) + C$
- $\int e^{ax} \cos(bx) dx = \dfrac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)) + C$

Integration by Partial Fractions

This technique is used to integrate rational functions ( $P(x)/Q(x)$ where $P, Q$ are polynomials).

Method: If the degree of the numerator is less than the degree of the denominator, the fraction can be decomposed into a sum of simpler "partial" fractions based on the factors of the denominator $Q(x)$ .
- Case 1 (Distinct Linear Factors): $\dfrac{P(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$ .
- Case 2 (Repeated Linear Factors): $\dfrac{P(x)}{(x-a)^k} = \dfrac{A_1}{x-a} + \dfrac{A_2}{(x-a)^2} + \dots + \dfrac{A_k}{(x-a)^k}$ .
- Case 3 (Irreducible Quadratic Factors): $\dfrac{P(x)}{(x-a)(x^2+bx+c)} = \dfrac{A}{x-a} + \dfrac{Bx+C}{x^2+bx+c}$ .
Once decomposed, each term can be integrated using standard rules.

Integration of Irrational Functions

This involves a variety of techniques, often aiming to rationalize the integrand through substitution.

$\int \sqrt{\text{Quadratic}} \, dx$
- $\int \sqrt{a^2 + x^2} dx = \dfrac{x}{2} \sqrt{a^2 + x^2} + \dfrac{a^2}{2} \ln\left|x + \sqrt{a^2 + x^2}\right| + C$
- $\int \sqrt{x^2 - a^2} dx = \dfrac{x}{2} \sqrt{x^2 - a^2} - \dfrac{a^2}{2} \ln\left|x + \sqrt{x^2 - a^2}\right| + C$
- $\int \sqrt{a^2 - x^2} dx = \dfrac{x}{2} \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1}\left(\dfrac{x}{a}\right) + C$
For $\int \text{Linear} \sqrt{\text{Quadratic}} \, dx$ , substitute $\text{Linear} = m(\text{Quadratic})' + n$ , find $m$ and $n$ by comparing coefficients, then integrate each term separately.
For $\int \dfrac{dx}{\text{Linear}_1\sqrt{\text{Linear}_2}}$ , $\int \dfrac{\text{Linear}_1}{\sqrt{\text{Linear}_2}} \, dx$ , $\int \dfrac{\sqrt{\text{Linear}_2}}{\text{Linear}_1} \, dx$ , substitute $\text{Linear}_2 = t^2$ ,
For $\int \dfrac{dx}{\text{Linear}\sqrt{\text{Quadratic}}}$ , substitute $\text{Linear} = \dfrac{1}{t}$ ,
For $\int \dfrac{dx}{(ax^2+b)\sqrt{cx^2+d}}$ , substitute $x = \dfrac{1}{t}$ , which reduces the integral to $\int \dfrac{tdt}{(pt^2 + q)\sqrt{rt^2 + s}}$ , and then substitute $u^2 = rt^2 + s$ .

The Definite Integral: The Limit of a Sum

The second, and perhaps more powerful, concept of integration is the definite integral, which is defined as the limit of a sum.

Geometric Interpretation: Geometrically, the definite integral $\int_a^b f(x)dx$ represents the signed area of the region bounded by the curve $y=f(x)$ , the x-axis, and the vertical lines $x=a$ and $x=b$ . Areas above the x-axis are positive, and areas below are negative.
Definition as the Limit of a Riemann Sum:
1. Take an interval $[a,b]$ .
2. Partition it into $n$ subintervals of width $\Delta x_k$ .
3. In each subinterval, choose a sample point $c_k$ .
4. Form the sum $\sum_{k=1}^n f(c_k)\Delta x_k$ . This is a Riemann Sum.
5. The definite integral is the limit of this sum as the width of the largest subinterval approaches zero (which corresponds to $n \to \infty$ for uniform partitions). $\int_a^b f(x)dx = \lim_{n \to \infty} \sum_{k=1}^n f(x_k)\Delta x \quad (\text{for uniform partition } \Delta x = \dfrac{b-a}{n})$
Some Important Series
- $\sum_{k=1}^n k = \dfrac{n(n+1)}{2}$ (Triangular numbers)
- $\sum_{k=1}^n k^2 = \dfrac{n(n+1)(2n+1)}{6}$ (Square numbers)
- $\sum_{k=1}^n k^3 = \left(\dfrac{n(n+1)}{2}\right)^2$ (Cubic numbers)
- $\sum_{k=1}^n a^k = \dfrac{a(a^n-1)}{a-1}$ (Geometric series, for $a \neq 1$ )
- $\sum_{k=1}^n \sin(\alpha + (k-1)\beta) = \dfrac{\sin\left(\dfrac{n\beta}{2}\right)\sin\left(\alpha + \dfrac{(n-1)}{2} \beta\right)}{\sin\left(\dfrac{\beta}{2}\right)}$ (Sum of sine series)
- $\sum_{k=1}^n \cos(\alpha + (k-1)\beta) = \dfrac{\sin\left(\dfrac{n\beta}{2}\right)\cos\left(\alpha + \dfrac{(n-1)}{2} \beta\right)}{\sin\left(\dfrac{\beta}{2}\right)}$ (Sum of cosine series)
- $\sum_{k=1}^{\inf} \dfrac{1}{k^2} = \dfrac{\pi^2}{6}$ (Basel problem)
- $\sum_{k=1}^{\inf} \dfrac{(-1)^{k-1}}{k} = \ln(2)$ (Alternating harmonic series)
- $\sum_{k=1}^{\inf} \dfrac{(-1)^{k-1}}{k^2} = \dfrac{\pi^2}{12}$ (Alternating Basel problem)
- $\sum_{k=1}^{\inf} \dfrac{1}{(2k-1)^2} = \dfrac{\pi^2}{8}$ (Sum of reciprocals of odd squares)
- $\sum_{k=1}^{\inf} \dfrac{1}{(2k)^2} = \dfrac{\pi^2}{24}$ (Sum of reciprocals of even squares)

The Fundamental Theorems of Calculus

These theorems establish the crucial link between differentiation and integration, showing they are inverse operations.

First Fundamental Theorem of Calculus (FTC1): If $f$ is a continuous function on $[a,b]$ , then the function $F$ defined by: $F(x) = \int_a^x f(t)dt$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and its derivative is $f(x)$ : $F'(x) = \dfrac{d}{dx} \int_a^x f(t)dt = f(x)$ This theorem states that the process of integrating a function and then differentiating it returns the original function.
Second Fundamental Theorem of Calculus (FTC2 - Evaluation Theorem): If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ on $[a,b]$ (i.e., $F'(x) = f(x)$ ), then: $\int_a^b f(x)dx = F(b) - F(a)$ Proof Sketch: Let $G(x) = \int_a^x f(t)dt$ . By FTC1, $G'(x) = f(x)$ . Since $F(x)$ is also an antiderivative, $F(x) = G(x) + C$ for some constant $C$ . Then $F(b) - F(a) = (G(b)+C) - (G(a)+C) = G(b) - G(a)$ . $G(b) = \int_a^b f(t)dt$ and $G(a) = \int_a^a f(t)dt = 0$ . Thus, $F(b) - F(a) = \int_a^b f(t)dt$ . This theorem provides a practical method for calculating definite integrals without using the limit of sums.

Properties and Advanced Techniques for Definite Integrals

Properties of Definite Integrals

Let $f$ and $g$ be continuous functions.

$\int_a^a f(x)dx = 0$
$\int_a^b f(x)dx = \int_a^b f(t)dt$
$\int_a^b f(x)dx = -\int_b^a f(x)dx$
$\int_a^b c \, dx = c(b-a)$ , where $c$ is a constant.
$\int_a^b [f(x) \pm g(x)]dx = \int_a^b f(x)dx \pm \int_a^b g(x)dx$
$\int_a^c f(x)dx + \int_c^b f(x)dx = \int_a^b f(x)dx$
$\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ .
$\int_0^{2a} f(x)dx = \int_0^a f(x) dx + \int_0^a f(2a-x)dx$ .
$\int_0^{2a} f(x)dx = \int_0^a (f(a-x) + f(a+x))dx$ .
$\int_a^b f(x)dx = (b-a)\int_0^1 f((b-a)x+a)dx$ .
Odd/Even Functions:
- If $f$ is an even function ( $f(-x)=f(x)$ ), then $\int_{-a}^a f(x)dx = 2\int_0^a f(x)dx$ .
- If $f$ is an odd function ( $f(-x)=-f(x)$ ), then $\int_{-a}^a f(x)dx = 0$ .
- If $f(t)$ is an odd function, then $\phi(x) = \int_a^x f(t)dt$ is an even function.
Periodic Functions:

* If $f$ has period $T$, then $\int_0^{nT} f(x)dx = n\int_0^T f(x)dx$.
* $\int_a^{a+nT} f(x)dx = n\int_0^T f(x)dx$.
* $\int_{mT}^{nT} f(x)dx = (n-m)\int_0^T f(x)dx$.
* $\int_{a+nT}^{b+nT} f(x)dx = \int_a^b f(x)dx$.

Important Results:

$\int_0^{\dfrac{\pi}{2}} \ln \sin x \, dx = \int_0^{\dfrac{\pi}{2}} \ln \cos x \, dx = -\dfrac{\pi}{2} \ln 2$ (known as the Dirichlet integral).

Leibnitz Rule for Differentiation Under the Integral Sign

This rule allows us to differentiate an integral whose limits are also functions of the variable of differentiation. $\dfrac{d}{dx} \int_{a(x)}^{b(x)} f(t)dt = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$ A more general form exists when the integrand $f$ also depends on $x$ : $\dfrac{d}{dx} \int_{a(x)}^{b(x)} f(x,t)dt = f(x,b(x))\cdot b'(x) - f(x,a(x))\cdot a'(x) + \int_{a(x)}^{b(x)} \dfrac{\partial f}{\partial x} dt$

Inequalities Involving Integrals

If $f(x) \ge g(x)$ for all $x$ in $[a,b]$ , then $\int_a^b f(x)dx \ge \int_a^b g(x)dx$ .
Bounding Property: If $m \le f(x) \le M$ for all $x$ in $[a,b]$ , then: $m(b-a) \le \int_a^b f(x)dx \le M(b-a)$
Triangle Inequality for Integrals: $\left| \int_a^b f(x)dx \right| \le \int_a^b |f(x)|dx$
Cauchy-Schwarz Inequality: If $f$ and $g$ are square-integrable functions on $[a,b]$ , then: $\left( \int_a^b f(x)g(x)dx \right)^2 \le \int_a^b f(x)^2dx \cdot \int_a^b g(x)^2dx$
Minkowski's Inequality: If $f$ and $g$ are integrable functions on $[a,b]$ , then: $\left( \int_a^b |f(x) + g(x)|dx \right) \le \left( \int_a^b |f(x)|dx + \int_a^b |g(x)|dx \right)$

Visualizing Integration: Curve Tracing and Area Under Curves

The most direct application of the definite integral is finding the area of a region bounded by curves.

Area Under a Curve: If $f(x) \ge 0$ on $[a,b]$ , the area $A$ of the region bounded by the curve $y=f(x)$ , the x-axis, and the lines $x=a$ and $x=b$ is given by: $A = \int_a^b f(x)dx$ If the curve is below the x-axis, the integral will be negative, so the area is $|\int_a^b f(x)dx|$ .
Area Between Two Curves: If $f(x) \ge g(x)$ on $[a,b]$ , the area of the region between the curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is: $A = \int_a^b [f(x) - g(x)]dx$
Curve Tracing: Before setting up an integral for area (or volume), it is often essential to trace the curves to understand the region of integration. This involves finding intercepts, checking for symmetry, finding asymptotes, and analyzing the function's derivative to find intervals of increase/decrease and local extrema.
Some Standard Areas:
- Area of a Circle ( $x^2+y^2=r^2$ ): This can be calculated as $A = \int_{-r}^r 2\sqrt{r^2-x^2} dx$ . This integral evaluates to $\pi r^2$ using trigonometric substitution.
- Area of an Ellipse ( $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ): This can be calculated as $A = 4 \int_0^a b\sqrt{1-x^2/a^2} dx$ , which evaluates to $\pi ab$ .

Key Takeaways: The Dual Nature and Power of Integration

Integral calculus offers a profound set of tools for understanding accumulation and the continuous sum, acting as the counterpart to differential calculus.

Integration as Antidifferentiation: The indefinite integral, $\int f(x)dx = F(x)+C$ , is fundamentally the process of finding the family of functions whose derivative is $f(x)$ . Mastery of this requires various techniques like substitution, integration by parts, and partial fractions.
The Definite Integral as a Limit of a Sum: The definite integral, $\int_a^b f(x)dx$ , is formally defined as the limit of a Riemann sum, representing the accumulated sum of infinitesimal quantities. Geometrically, this corresponds to the signed area under a curve.
The Fundamental Theorem of Calculus: This monumental theorem provides the bridge between the two concepts. It states that differentiation and integration are inverse operations and gives us the practical method for evaluating definite integrals using antiderivatives: $\int_a^b f(x)dx = F(b)-F(a)$ .
Properties and Rules: Definite integrals have a rich set of algebraic and analytical properties (linearity, additivity of intervals, properties of odd/even/periodic functions) that aid in their evaluation.
Broad Applications: Beyond calculating areas, integration is essential for finding volumes, arc lengths, centers of mass, work done by a variable force, probabilities for continuous random variables, and solving differential equations that model countless phenomena in science and engineering.

Together, differentiation and integration form the two pillars of calculus, providing a complete language for describing and analyzing functions and the dynamic systems they represent.