The Definite Integral

The derivative answers how fast something changes at an instant. The integral answers the complementary question: given the rate, how much accumulates over an interval.

When the rate is constant the answer is a multiplication, travelling at 60 km/h for two hours covers 120 km. When the rate varies, no single multiplication applies, and the resolution mirrors the derivative's: chop the interval into pieces short enough that the rate is nearly constant on each, multiply and add, then take more and shorter pieces.

For ${\int }_0^1{x}^{2}dx$ the approximations from below run $0.219$, $0.285$, $0.328$, $0.3328$ as the number of pieces goes $4$, $10$, $100$, $1000$, while those from above come down through $0.469$, $0.385$, $0.338$, $0.3338$. The true value is squeezed between them, and it is $\frac{1}{3}$.

What makes the subject workable is that this limit is almost never computed. The fundamental theorem shows that accumulating and differentiating are inverse operations, so the integral of $x^2$ is found by asking what differentiates to $x^2$. Nothing in the two definitions hints at that connection, one concerns tangent slopes, the other rectangle areas, which is why the theorem earns its name.

This unit builds the limit, states and uses both parts of the theorem, derives the substitution rule as the chain rule reversed, and marks the two places the machinery misleads: when the integrand dips below the axis, and when it is discontinuous inside the interval.

The canonical intuition gives the construction and the theorem. What follows is the numerical detail that makes the limit credible rather than merely asserted.

The squeeze. For ${\int }_0^1{x}^{2}dx$, take $n$ equal subintervals of width $1/n$. Because $x^2$ increases on $[0,1]$, sampling at each subinterval's left endpoint underestimates and at the right endpoint overestimates:

The gap is exactly $1/n$. It is $f(1)-f(0)$ times the subinterval width, since the two sums differ only in their first and last rectangles. So the true value is pinned to an interval shrinking to nothing, and no appeal to intuition is needed to say the limit exists.

Exactly, without a limit process. The right sum has a closed form. Sampling at $x_i=i/n$,

At $n=4$ this is $\frac{5\cdot 9}{96}=\frac{15}{32}=0.46875$; at $n=10$, $\frac{11\cdot 21}{600}=\frac{77}{200}=0.385$; at $n=100$, $\frac{6767}{20000}=0.33835$. Expanding, the expression is $\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}$, whose limit is visibly $\frac{1}{3}$.

Why the midpoint rule does better. At $n=100$ the midpoint sum is already correct to four decimal places while the endpoint sums are correct to two. Sampling in the middle makes the rectangle's overshoot on one half cancel its undershoot on the other, leaving an error of order $1/n^2$ rather than $1/n$. This is the same first-order-versus-second-order distinction the derivative's tangent-line error showed.

That the limit equals $\frac{1}{3}$ is visible here, but that it can be obtained without summing anything is not. The fundamental theorem supplies $x^3/3$ as an antiderivative and the answer follows from $\frac{1^3}{3}-\frac{0^3}{3}$. One subtraction against a thousand-term sum.

The canonical definition gives the Riemann sum, the theorem and the properties. What follows is why each is stated with the conditions it carries.

Why the limit must be independent of the sample points. The definition requires the same limit whatever $x_i^{\ast }$ is chosen in each subinterval. This is not pedantry: for a badly behaved function the choice changes the answer. On $[0,1]$, the function that is 1 at rationals and 0 at irrationals gives a sum of 1 when every sample is rational and 0 when every sample is irrational, so no limit exists and the function is not Riemann integrable. Requiring independence is what excludes it.

Why continuity is sufficient but not necessary. A continuous function on a closed bounded interval is uniformly continuous, so the oscillation within each subinterval can be forced below any threshold by taking $n$ large. The upper and lower sums converge together. But a step function is discontinuous and plainly integrable: finitely many jumps contribute finitely many rectangles of vanishing width. What must be avoided is discontinuity on a set large enough to matter.

Why the sign convention is forced rather than chosen. Each term is $f(x_i^{\ast })\mathrm{\Delta }x$ with , so the sign of the term is the sign of $f$. Nothing in the construction takes an absolute value, and inserting one would break the additivity property ${\int }_a^c={\int }_a^b+{\int }_b^c$. The pieces would stop summing correctly whenever $f$ changed sign in between.

Why ${\int }_a^b=-{\int }_b^a$. Reversing the limits reverses the orientation, making $\mathrm{\Delta }x$ negative. The convention is not an extra rule but the same formula read with , and it is what keeps additivity true for every arrangement of $a$, $b$ and $c$ rather than only for .

Why Part 2 needs an antiderivative on the whole interval. The proof applies the mean value theorem on each subinterval of a partition of $[a,b]$, which requires $F$ to be differentiable throughout. If $f$ blows up inside the interval, no such $F$ exists there, and the formula $F(b)-F(a)$, though still computable as arithmetic, is answering a question that was never posed.

Why the constant of integration is real. If $F^{\prime }=G^{\prime }=f$ on an interval then $(F-G{)}^{\prime }=0$, and a function with vanishing derivative on an interval is constant, so $F-G=C$. The word interval is load-bearing: on a disconnected domain such as $\{x\ne 0\}$ the difference can be one constant on each piece, which is why $\int \frac{dx}{x}=\ln |x|+C$ hides a subtlety about the two branches.

Part 1. Let $f$ be continuous on $[a,b]$ and define the accumulation function

Then $G$ is differentiable on $(a,b)$ and $G^{\prime }(x)=f(x)$.

Proof. By additivity, $G(x+h)-G(x)={\int }_x^{x+h}f(t)dt$. Since $f$ is continuous on the closed interval between $x$ and $x+h$, it attains a minimum $m_h$ and a maximum $M_h$ there, and the integral of a function between two constants lies between their integrals:

Dividing by gives $m_h\le \frac{G(x+h)-G(x)}{h}\le M_h$. As $h\to 0$ both $m_h$ and $M_h$ tend to $f(x)$ by continuity, so the difference quotient is squeezed to $f(x)$. The same argument with inequalities reversed covers . $◼$

Check. With $f(t)=t^2$ and $a=0$, $G(x)=x^3/3$. Differentiating numerically at $x=2$ and $x=3$ gives $4.000000$ and $9.000000$, matching $x^2$.

Part 2. If $F$ is any antiderivative of $f$ on $[a,b]$, that is, $F^{\prime }=f$ throughout, then

Proof. Part 1 gives one antiderivative, $G(x)={\int }_a^{x}f$. Any other differs from it by a constant, so $F=G+C$. Then

since $G(a)={\int }_a^{a}f=0$. The constant cancels, which is why any antiderivative serves. $◼$

What the two parts say together. Part 1: differentiating an accumulation returns the integrand. Part 2: accumulating a derivative returns the net change. They are the two directions of a single inverse relationship, and Part 2 follows from Part 1 rather than standing independently.

Verification on a case with a negative integrand. For $f(x)=x^2-4x$ on $[1,3]$, take $F(x)=\frac{x^3}{3}-2x^2$. Then $F(3)=9-18=-9$ and $F(1)=\frac{1}{3}-2=-\frac{5}{3}$, so

A midpoint Riemann sum with $\mathrm{200,000}$ subintervals returns $-7.33333333$. The value is negative because throughout $[1,3]$, indeed $f(1)=-3$, $f(2)=-4$, $f(3)=-3$.

The substitution rule as a corollary. If $F^{\prime }=f$ then $\frac{d}{dx}F(g(x))=f(g(x))g^{\prime }(x)$ by the chain rule, so $F(g(x))$ is an antiderivative of $f(g(x))g^{\prime }(x)$. Applying Part 2,

The limits transform because the antiderivative is evaluated at $g(a)$ and $g(b)$, not at $a$ and $b$, which is exactly the step that is forgotten when a substitution is done in a definite integral without converting.

Computing one integral from the definition, with no appeal to antiderivatives, shows what the fundamental theorem saves.

Setting up. Partition $[0,1]$ into $n$ equal subintervals, each of width $\mathrm{\Delta }x=1/n$. Sample at right endpoints, $x_i=i/n$ for $i=1,\dots ,n$. The Riemann sum is

The sum of squares. The identity $\sum _{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}$ is proved by induction: it holds at $n=1$, since $1=\frac{1\cdot 2\cdot 3}{6}$, and assuming it at $n$,

which is the formula at $n+1$.

Substituting.

Checking before the limit. At $n=4$: $\frac{1}{3}+\frac{1}{8}+\frac{1}{96}=\frac{15}{32}=0.46875$. At $n=10$: $\frac{77}{200}=0.385$. At $n=100$: $\frac{6767}{20000}=0.33835$. These match the right-sum column of the table exactly.

Taking the limit. The two correction terms vanish, so

The same answer in one line. $F(x)=x^3/3$ satisfies $F^{\prime }=x^2$, so Part 2 gives $F(1)-F(0)=\frac{1}{3}-0=\frac{1}{3}$.

The direct route needed a closed form for $\sum i^2$, which exists for squares and cubes but becomes unwieldy quickly and is unavailable for most integrands. There is no comparable identity waiting for $\sum \sin (i/n)$. The antiderivative route needed only the power rule read backwards. This is why the fundamental theorem is the organising result of the subject rather than one theorem among many: it converts an intractable limit into an algebra problem.

1. A polynomial by the fundamental theorem: ${\displaystyle {\int }_1^3(x^2-4x)dx}$.

An antiderivative is $F(x)={\displaystyle \frac{x^3}{3}}-2x^2$, since $F^{\prime }(x)=x^2-4x$.

Check: a midpoint sum with $\mathrm{200,000}$ subintervals gives $-7.33333333$.

The answer is negative, and correctly so: $f(1)=-3$, $f(2)=-4$, $f(3)=-3$, so the curve is below the axis across the whole interval. This is signed area, not total area.

Order matters. Computing $F(1)-F(3)$ instead gives $+\frac{22}{3}$, the right magnitude with the wrong sign. The most common slip in applying Part 2.

2. Signed area with a sign change: ${\displaystyle {\int }_0^3(x^2-4x)dx}$.

Same antiderivative: $F(3)-F(0)=-9-0=-9$.

The integrand is negative on all of $(0,4)$, so no cancellation occurs here. The value is simply more negative than the previous one because the interval is longer. To obtain the total area between curve and axis on $[0,3]$ one would compute ${\int }_0^3|x^2-4x|dx=9$, which is a different number from the integral whenever the integrand takes negative values.

3. Substitution in a definite integral: ${\displaystyle {\int }_0^{2}2x(x^2+1{)}^{3}dx}$.

Let $u=x^2+1$, so $du=2xdx$. The factor $2x$ is present, which is what makes this substitution available.

Convert the limits: $x=0\Rightarrow u=1$, and $x=2\Rightarrow u=5$.

Check: a midpoint sum with $\mathrm{400,000}$ subintervals.

The error to avoid: evaluating $\left[\frac{u^4}{4}\right]$ at $x=0$ and $x=2$ rather than at $u=1$ and $u=5$ gives $\frac{16-0}{4}=4$, which is wrong by a factor of 39 and carries no visible warning.

4. Part 1 in use: ${\displaystyle \frac{d}{dx}}{\displaystyle {\int }_0^x{t}^{2}dt}$.

Part 1 answers immediately: the derivative is $x^2$, with no integration performed.

Confirming the long way: ${\int }_0^x{t}^{2}dt=\frac{x^3}{3}$, whose derivative is $x^2$. Numerically at $x=2$ and $x=3$, symmetric difference quotients of $x^3/3$ give $4.000000$ and $9.000000$, matching $x^2=4$ and $9$.

The point of Part 1 is that the first answer required no antiderivative at all, accumulating then differentiating returns the integrand, whatever it was.