Integrals

If the derivative answers "how fast is this changing right now?", the integral answers the mirror-image question: "given the rate of change, how much has accumulated in total?" It is the mathematics of adding up infinitely many infinitely small pieces — the distance covered by a moving car, the area trapped under a curve, the total charge that flows through a wire. That sounds impossible, and for two thousand years it was done only in clever special cases. The breakthrough was the discovery that this "adding up" problem is secretly the reverse of finding slopes. That single insight — that area and slope are two sides of one coin — is one of the most powerful ideas humans have ever had.

This page builds the integral from the ground up: first as an area, then as a limit of sums, and finally as the antiderivative that the Fundamental Theorem of Calculus hands you for free.

Learning Objectives

By the end of this page, you should be able to:

• Explain the integral as both accumulated area under a curve and as the reverse of differentiation.
• Approximate a definite integral using Riemann sums and understand how the limit gives an exact value.
• Distinguish between an indefinite integral (a family of antiderivatives) and a definite integral (a number).
• State and apply the Fundamental Theorem of Calculus in both its forms.
• Evaluate integrals using the power rule, constant-multiple rule, and sum rule, and check answers by differentiating.

Quick Answer

An integral measures accumulation. The definite integral $\int_a^b f(x)\,dx$ is the (signed) area between the curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$, defined precisely as the limit of Riemann sums — rectangles whose widths shrink to zero. The indefinite integral $\int f(x)\,dx = F(x) + C$ is the family of all antiderivatives of $f$, meaning functions whose derivative is $f$. The Fundamental Theorem of Calculus ties them together: to find the area, you don't add up infinitely many rectangles — you just find an antiderivative $F$ and compute $F(b) - F(a)$. This turns an impossible-looking infinite sum into a two-step evaluation.

Where It Came From

The integral was born from a stubborn, ancient problem: how do you find the area of a shape with curved sides? For straight-sided figures the Greeks were comfortable, but a circle, a parabola, or the region under any curve resisted every simple formula.

Archimedes (c. 250 BCE) attacked this with the method of exhaustion. To find the area of a circle or the region under a parabola, he inscribed and circumscribed shapes made of straight pieces — triangles and polygons — with more and more sides. The true area was squeezed between the two approximations, and as the pieces multiplied, the gap "exhausted" toward zero. He correctly found the area of a parabolic segment to be $\tfrac{4}{3}$ the area of an inscribed triangle. This is exactly the spirit of a Riemann sum, done by hand, brilliantly, two millennia early. But it was ad hoc: every new shape needed a new stroke of genius.

In the 1600s Bonaventura Cavalieri pushed further with his "method of indivisibles," imagining an area as built from infinitely many parallel line segments (and volumes from stacked cross-sections). His principle — two regions with equal cross-sections at every height have equal area — let him compute areas that would have stumped Archimedes, though the reasoning about "infinitely thin" pieces was logically shaky.

The revolution came when Isaac Newton and Gottfried Wilhelm Leibniz, independently in the late 1600s, saw what no one before them had: the area-accumulation problem is the inverse of the tangent-slope problem. If you know the rate at which area accumulates, you know the height of the curve, and vice versa. This is the Fundamental Theorem of Calculus. It transformed integration from a bag of one-off tricks into a systematic machine: to find an area, undo a derivative. Leibniz even gave us the elongated-S symbol $\int$ (for "summa," a sum) and the $dx$ notation we still use. Rigorous definitions of the limit and the sum came later, from Cauchy and Bernhard Riemann in the 1800s, giving the definite integral the airtight footing it has today.

The Integral as Area: Riemann Sums

Start with the concrete question. You have a function $f(x)$, and you want the area under its graph between $x = a$ and $x = b$. If the top were flat you'd just multiply width by height. It isn't flat — so approximate it with rectangles.

Slice $[a, b]$ into $n$ strips each of width $\Delta x = \dfrac{b - a}{n}$. On each strip, build a rectangle whose height is the function value at some sample point. The total rectangle area is a Riemann sum:

$$S_n = \sum_{i=1}^{n} f(x_i)\,\Delta x$$

As $n \to \infty$, the rectangles hug the curve ever more tightly and the sum converges to the exact area. That limit is the definite integral:

$$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,\Delta x$$

The notation is a frozen picture of the process: $\int$ is a stretched "S" for sum, $f(x)$ is the height, and $dx$ is the infinitely thin width that $\Delta x$ becomes.

Worked Example: approximating and then finding an exact area

Estimate the area under $f(x) = x^2$ from $x = 0$ to $x = 2$ using $n = 4$ right-endpoint rectangles.

Width: $\Delta x = \dfrac{2 - 0}{4} = 0.5$. Right endpoints: $x = 0.5, 1.0, 1.5, 2.0$.

Heights: $f(0.5) = 0.25$, $f(1.0) = 1$, $f(1.5) = 2.25$, $f(2.0) = 4$.

$$S_4 = 0.5\,(0.25 + 1 + 2.25 + 4) = 0.5 \times 7.5 = 3.75$$

That overestimates, because $x^2$ is increasing and right endpoints sit above the curve. The true value, which we'll confirm below, is $\int_0^2 x^2\,dx = \tfrac{8}{3} \approx 2.667$. With $n = 4$ we're already in the neighborhood; with $n = 100$ the right-sum gives about $2.707$, and the limit lands exactly on $\tfrac{8}{3}$. The rectangles never lie — they just need to get thin enough.

Note "signed area": where $f(x)$ dips below the axis, its contribution counts as negative. So $\int_0^{2\pi} \sin x\,dx = 0$, because the area above the axis from $0$ to $\pi$ exactly cancels the area below from $\pi$ to $2\pi$.

Antiderivatives and the Indefinite Integral

Now the other thread. An antiderivative of $f$ is any function $F$ whose derivative is $f$: that is, $F'(x) = f(x)$. Since the derivative of a constant is zero, antiderivatives come in families — if $F$ works, so does $F + C$ for any constant $C$. We write this whole family as the indefinite integral:

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

The $+C$ is not optional bookkeeping; it says "we recovered the shape of the function but lost its vertical position when we differentiated, and we can't get that back without more information."

The single most useful rule is the power rule for integration (the power rule for derivatives run backward):

$$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$$

To check any integral, differentiate your answer — you should get back the integrand. That habit will save you constantly.

Worked Example: an indefinite integral, verified

Find $\int (3x^2 + 4x - 5)\,dx$.

Integrate term by term:

$$\int 3x^2\,dx = 3\cdot\frac{x^3}{3} = x^3, \quad \int 4x\,dx = 4\cdot\frac{x^2}{2} = 2x^2, \quad \int -5\,dx = -5x$$

So the answer is $x^3 + 2x^2 - 5x + C$.

Check by differentiating: $\dfrac{d}{dx}(x^3 + 2x^2 - 5x + C) = 3x^2 + 4x - 5$. It matches the integrand exactly, so we're right.

The Fundamental Theorem of Calculus

Here is where the two ideas collide and something magical happens. Why should the area problem (Riemann sums) have anything to do with antiderivatives (undoing slopes)? The Fundamental Theorem of Calculus (FTC) says it does — completely.

FTC Part 1 (the integral undoes the derivative): if $\displaystyle F(x) = \int_a^x f(t)\,dt$ is the accumulated area up to $x$, then $F'(x) = f(x)$. The rate at which area piles up equals the current height of the curve — which is intuitively obvious once you say it: a taller curve adds area faster.

FTC Part 2 (the evaluation shortcut): if $F$ is any antiderivative of $f$, then

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

This is the payoff. Instead of building infinitely many rectangles, find an antiderivative and subtract two numbers. We write $F(b) - F(a)$ compactly as $\big[F(x)\big]_a^b$.

Worked Example: closing the loop on our area

Compute $\int_0^2 x^2\,dx$ exactly.

An antiderivative of $x^2$ is $F(x) = \dfrac{x^3}{3}$. By FTC Part 2:

$$\int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \approx 2.667$$

This is the exact value our Riemann sums were creeping toward — obtained in one line, no limits required. That is the whole point of the theorem.

Worked Example: total distance from velocity

A particle moves with velocity $v(t) = 6t$ metres per second. How far does it travel between $t = 1$ and $t = 3$ seconds?

Distance is accumulated velocity, so it's the integral of $v$:

$$\int_1^3 6t\,dt = \big[3t^2\big]_1^3 = 3(9) - 3(1) = 27 - 3 = 24 \text{ metres}$$

Notice the antiderivative $3t^2$ is exactly the position function whose derivative is the velocity $6t$ — the integral rebuilt position from speed.

Real-World Applications

• Physics — distance, work, charge. Integrating velocity gives displacement; integrating force over distance gives work done; integrating current over time gives total electric charge. Anywhere a rate is known and a total is wanted, you integrate.
• Engineering. Areas and volumes of irregular shapes, the centre of mass of a beam, and the total heat flow through a material are all integrals. Signal processing uses integrals to measure energy and to build the Fourier transforms behind audio and image compression.
• Economics. The area under a marginal-cost curve gives total cost; consumer and producer surplus are literally areas between curves. The Gini coefficient measuring inequality is an integral comparing an income-distribution curve to perfect equality.
• Medicine and biology. "Area under the curve" (AUC) of a drug-concentration-versus-time graph tells pharmacologists the total exposure a patient receives to a medication — a routine, life-affecting integral.
• Probability and statistics. The probability that a continuous random variable falls in a range is the integral of its density function over that range; the total area under any density curve is exactly 1.

Common Mistakes

Mistake 1: Forgetting the $+C$ on indefinite integrals. Why it's wrong: $\int f(x)\,dx$ is not a single function but an entire family of antiderivatives differing by a constant. Dropping $C$ discards infinitely many valid answers and will bite you when solving differential equations or applying initial conditions. Correction: always write $+C$ for an indefinite integral. (A definite integral, by contrast, is a number and has no $+C$ — the constant cancels in $F(b) - F(a)$.)

Mistake 2: Treating "area under the curve" as always positive. Why it's wrong: the definite integral gives signed area. Regions below the $x$-axis count negatively, so $\int_a^b f\,dx$ can be zero or negative even when the graph clearly encloses space. Correction: if you want the total geometric area, split the integral where $f$ crosses the axis and integrate the absolute value of each piece.

Mistake 3: Misapplying the power rule at $n = -1$. Why it's wrong: the formula $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$ divides by $n + 1$, which is zero when $n = -1$ — undefined. Students still write $\int x^{-1}\,dx = \dfrac{x^0}{0}$. Correction: $\int x^{-1}\,dx = \int \dfrac{1}{x}\,dx = \ln|x| + C$. This single exception is worth memorising.

Mistake 4: Swapping the limits without a sign change. Why it's wrong: the order of the limits matters. $\int_a^b f\,dx = -\int_b^a f\,dx$. Correction: keep the lower limit at the bottom, and if you deliberately flip them, flip the sign.

Comparison and Connections

Derivatives and integrals are inverse operations, and keeping their roles straight prevents most confusion.

Feature	Derivative	Integral (definite)	Integral (indefinite)
Core question	How fast is it changing?	How much accumulated?	What has this rate as its derivative?
Geometric meaning	Slope of the tangent	Signed area under the curve	Family of curves
Output	A function (or a number at a point)	A number	A family of functions $F(x)+C$
Notation	$f'(x)$, $\dfrac{dy}{dx}$	$\int_a^b f(x)\,dx$	$\int f(x)\,dx$
Linked by	the Fundamental Theorem of Calculus	the Fundamental Theorem of Calculus	supplies the antiderivative used

The FTC is the bridge: differentiation and integration are inverse processes, exactly as multiplication and division are. Riemann sums are the rigorous definition of the definite integral; the FTC is the practical shortcut for computing it. And the indefinite integral is the tool the FTC needs — you must find an antiderivative before you can evaluate $F(b) - F(a)$.

Practice Questions

Recall

State the power rule for integration and give the one value of $n$ for which it fails. Answer: $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ for $n \neq -1$; it fails at $n = -1$, where instead $\int x^{-1}\,dx = \ln|x| + C$.

Understanding

Explain in words why $\int_0^{2\pi}\sin x\,dx = 0$ even though the sine graph clearly encloses area. Answer: The integral measures signed area. From $0$ to $\pi$ the sine is positive (area above the axis); from $\pi$ to $2\pi$ it is negative and equal in size. The two cancel, giving a net of zero, even though the total geometric area is $4$.

Application

Evaluate $\int_1^4 (2x + 1)\,dx$. Answer: An antiderivative is $x^2 + x$. Then $\big[x^2 + x\big]_1^4 = (16 + 4) - (1 + 1) = 20 - 2 = 18$.

Analysis

A car's velocity is $v(t) = 3t^2$ m/s. Find the exact distance travelled in the first 2 seconds, and explain how a Riemann sum with 2 rectangles would compare. Answer: $\int_0^2 3t^2\,dt = \big[t^3\big]_0^2 = 8 - 0 = 8$ metres. A right-endpoint sum with $\Delta t = 1$ gives $v(1)\cdot1 + v(2)\cdot1 = 3 + 12 = 15$ — a large overestimate, because $3t^2$ is increasing and steep; the rectangles overshoot until they become thin enough to converge to $8$.

FAQ

Is the integral just the opposite of the derivative? For computation, essentially yes — that is the content of the FTC. But conceptually the definite integral is defined as a limit of sums (accumulated area), and it's a genuine theorem, not a definition, that this equals an antiderivative evaluated at the endpoints. Both viewpoints matter.

What does the $dx$ actually mean? Think of it as the infinitely thin width of each rectangle in the Riemann sum — the limit of $\Delta x$. It also tells you which variable you're integrating with respect to, which becomes essential once several variables are floating around.

Why does the indefinite integral need $+C$ but the definite one doesn't? Differentiating erases any constant, so undoing it can't recover that constant — hence $+C$ for the family. In a definite integral the constant appears in both $F(b)$ and $F(a)$ and cancels when you subtract, so it never affects the answer.

Can I always find an antiderivative with a formula? No. Some functions, like $e^{-x^2}$ (the bell curve), have no antiderivative expressible with elementary functions. The definite integral still exists as an area; we just compute it numerically or with special functions.

How is a definite integral different from just the area? The definite integral is signed area: parts below the axis subtract. If you want plain geometric area, integrate $|f(x)|$, splitting at every axis crossing.

Quick Revision

• Definite integral $= \displaystyle\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\,\Delta x$ = signed area under $f$ from $a$ to $b$.
• Indefinite integral $\int f(x)\,dx = F(x) + C$, the family of antiderivatives ($F' = f$).
• Power rule: $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ for $n\neq-1$; and $\int x^{-1}\,dx = \ln|x| + C$.
• Linearity: $\int (af + bg)\,dx = a\int f\,dx + b\int g\,dx$.
• FTC Part 2: $\int_a^b f(x)\,dx = F(b) - F(a)$ for any antiderivative $F$.
• FTC Part 1: $\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)$.
• Sign rules: $\int_a^b f = -\int_b^a f$, and $\int_a^a f = 0$.
• Always check an integral by differentiating the answer.

Related Topics

Prerequisites

• Derivatives — integration is the reverse process, so you must be fluent with slopes first.
• Limits — the definite integral is defined as a limit of Riemann sums.
• Functions — you integrate functions, so knowing their behaviour is essential.

Related Topics

• Applications of Calculus — areas, volumes, work, and accumulation problems in depth.
• Distributions — probability densities are integrated to get probabilities.

Next Topics

• Applications of Calculus — put integrals to work on real geometric and physical problems.
• The Calculus overview: Calculus for how limits, derivatives, and integrals fit together.