Learning goals and preparation

This chapter assumes limits, continuity, and basic differentiation, including the product and chain rules. It connects a rate with its accumulated total, introduces the definite integral, develops substitution and integration by parts, and applies integrals to hypothetical continuous probability models. Chapter 1 introduces limits and differentiation; Chapter 2 develops optimization and sensitivity.

Historical optional enrichment. Section 3.6 and its probability exercises are an optional historical extension.

An integral adds many small contributions. When the contributions have signs, the integral is a net total rather than an unsigned area. When a density is integrated, the result is a probability, not the height of the density at a point. Keeping those meanings separate prevents several common errors.

3.1 From small contributions to a definite integral

Divide $[a,b]$ into subintervals with endpoints $a=x_0<x_1<\cdots<x_n=b$. Choose a sample point $\xi_i\in[x_{i-1},x_i]$ and form the Riemann sum

$$\sum_{i=1}^n f(\xi_i)\Delta x_i,\qquad\Delta x_i=x_i-x_{i-1}.$$

The mesh is the largest $\Delta x_i$. A bounded function is Riemann integrable when these sums approach the same finite number as the mesh tends to zero, regardless of the sample points. That number is $\int_a^b f(x)\,dx$. Every continuous function on a closed bounded interval is integrable.

If $f$ is a flow rate in litres per minute and $x$ is time in minutes, the integral has units litres. The sample rectangle's height is a rate and its width is a time; their product is an amount.

Uniform continuity

Definition

One delta for the entire domain

A function $f:A\to\mathbb R$ is uniformly continuous on $A$ if for every $\varepsilon>0$ there exists $\delta>0$ such that for all $x,y\in A$, $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$. The chosen $\delta$ can depend on $\varepsilon$ but not on $x$ or $y$.

Theorem

Heine–Cantor theorem

Every continuous real-valued function on a compact set is uniformly continuous.

Proof

Contradiction using compactness

If uniform continuity failed, some $\varepsilon_0>0$ would permit, for every $n$, points $x_n,y_n\in K$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge\varepsilon_0$. Compactness supplies a subsequence $x_{n_k}\to x\in K$. Since the paired distances tend to zero, also $y_{n_k}\to x$. Continuity gives both $f(x_{n_k})\to f(x)$ and $f(y_{n_k})\to f(x)$, contradicting the output separation.

Worked example

Pointwise but not uniform on an unbounded domain

The function $f(x)=x^2$ is continuous on $\mathbb R$ but not uniformly continuous there. Take $x_n=n$ and $y_n=n+1/n$. Then $|x_n-y_n|=1/n\to0$, but $|f(y_n)-f(x_n)|=2+1/n^2\not\to0$. On any closed bounded interval, Heine–Cantor does give uniform continuity. It is also direct: if $|x|,|y|\le M$, then $|x^2-y^2|\le2M|x-y|$.

Learning goals and definitions

Develop the proof behind integrability statements and distinguish boundedness from integrability. Let $a<b$ and $f$ be bounded on $[a,b]$. For a partition $P:a=x_0<\cdots<x_n=b$, put $m_i=\inf_{[x_{i-1},x_i]}f$, $M_i=\sup_{[x_{i-1},x_i]}f$, and define $L(f,P)=\sum m_i\Delta x_i$, $U(f,P)=\sum M_i\Delta x_i$. The lower integral $\underline I$ is the supremum of all lower sums; the upper integral $\overline I$ is the infimum of all upper sums. Refining a partition makes lower sums nondecreasing and upper sums nonincreasing.

For any two partitions $P,Q$, their common refinement gives $L(f,P)\le U(f,Q)$, so $\underline I\le\overline I$. Boundedness ensures these quantities are finite. The function is Riemann integrable when they agree.

Theorem: Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$.

Proof. If the two integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, using supremum and infimum approximation. Their common refinement has gap less than $\varepsilon$. Conversely the inequalities $0\le\overline I-\underline I\le U(f,P)-L(f,P)$ force the integral gap to be zero when such partitions exist for every positive $\varepsilon$.

Theorem: finitely many discontinuities are permitted

If $f$ is bounded and continuous except possibly at finitely many points, it is integrable.

Proof. Let $|f|\le M$, taking $M>0$. Surround the exceptional points by finitely many small intervals, with endpoints included in the partition, having total length less than $\varepsilon/(4M)$. At endpoints of $[a,b]$ use one-sided intervals. Their contribution to the upper-lower gap is at most $2M$ times their total length, hence less than $\varepsilon/2$. On the remaining finitely many closed intervals, $f$ is continuous and therefore uniformly continuous. Subdivide each sufficiently finely that its oscillation on every resulting subinterval is less than $\varepsilon/[2(b-a)]$. Their total gap is less than $\varepsilon/2$. Combining all partition points gives gap below $\varepsilon$. If an exceptional interval endpoint would itself be exceptional, move it slightly so all actual exceptional points lie inside the selected intervals or at the domain endpoints. This keeps the remaining closed intervals continuous.

Continuous functions and all tagged meshes

Let $f:[a,b]\to\mathbb R$ be continuous, where $a<b$. The extreme value theorem makes $f$ bounded, and Heine–Cantor makes it uniformly continuous. For a partition $P$, write $L(f,P)$ and $U(f,P)$ for its lower and upper Darboux sums. The proved Darboux integrability theorem for continuous functions supplies the common integral $I$.

Fix $\varepsilon>0$. Uniform continuity supplies $\delta>0$ such that $|f(x)-f(y)|<\varepsilon/[2(b-a)]$ whenever $|x-y|<\delta$. If the mesh of $P$ is less than $\delta$, every subinterval has oscillation at most $\varepsilon/[2(b-a)]$. Therefore

$$U(f,P)-L(f,P)\le\frac{\varepsilon}{2}<\varepsilon.$$

For every selection of tags, its tagged sum $S(P)$ lies between $L(f,P)$ and $U(f,P)$; so does $I$. Thus $|S(P)-I|\le U(f,P)-L(f,P)<\varepsilon$. The same $\delta$ works for every partition of sufficiently small mesh and every choice of tags. This proves the tagged-mesh definition for continuous functions on the stated closed interval.

For $a<b$, take a partition $a=x_0<x_1<\cdots<x_n=b$, with $\Delta x_i=x_i-x_{i-1}$, tags $\xi_i\in[x_{i-1},x_i]$, and mesh $\|P\|=\max_i\Delta x_i$. A finite number $I$ is the definite integral $\int_a^bf(x)dx$ if, for every $\varepsilon>0$, there is $\delta>0$ such that every partition with $\|P\|<\delta$ and every choice of tags satisfy $|\sum_i f(\xi_i)\Delta x_i-I|<\varepsilon$. In that case $I$ is the common limit. Continuous functions on a closed interval are integrable.

Let $f$ be continuous on $[a,b]$ and $A(x)=\int_a^xf(t)dt$. For $x\in(a,b)$ and nonzero $h$ with $x+h\in[a,b]$, the integral estimate gives

$$\left|\frac{A(x+h)-A(x)}h-f(x)\right|\le\sup_{t\in[\min(x,x+h),\max(x,x+h)]}|f(t)-f(x)|\longrightarrow0\quad(h\to0).$$

Thus $A'(x)=f(x)$ in the interior; at the endpoints the same argument gives one-sided derivatives. If $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $F'=f$ there, the mean value theorem makes $F-A$ constant on $[a,b]$. Therefore $\int_a^bf=F(b)-F(a)$.

Worked example 1: a sum with an exact limit. For $f(x)=x$ on $[0,1]$, use $n$ equal widths and right endpoints $i/n$. The sum is

$$\sum_{i=1}^n\frac in\frac1n=\frac{n(n+1)}{2n^2}=\frac12+\frac1{2n}.$$

It tends to $1/2$. Integrability, supplied here by continuity, ensures the same limit for all sufficiently fine tagged partitions; computing one convenient family alone would not prove that fact for an arbitrary function.

The integral is linear and additive across intervals:

$$\int_a^b(\alpha f+\beta g)=\alpha\int_a^bf+\beta\int_a^bg,\qquad \int_a^cf=\int_a^bf+\int_b^cf.$$

If $f\le g$ on an interval, their integrals obey the same order. Reversing endpoints changes the sign.

3.2 The fundamental theorem of calculus

If $f$ is continuous on $[a,b]$ and $A(x)=\int_a^x f(t)\,dt$, then $A'(x)=f(x)$ for $a<x<b$. To understand why,

$$\frac{A(x+h)-A(x)}h=\frac1h\int_x^{x+h}f(t)\,dt.$$

This is an average of $f$ over a short interval. Its difference from $f(x)$ is bounded in magnitude by the largest $|f(t)-f(x)|$ on that interval, which tends to zero by continuity. The same conclusion holds for positive and negative $h$.

A second form states that if $f$ is continuous on $[a,b]$, $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and $F'=f$ on $(a,b)$, then

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

Indeed the accumulated-integral function and $F$ have the same derivative; their difference is constant by the mean value theorem. This converts an accumulation problem into an antiderivative calculation.

An indefinite integral $\int f(x)\,dx=F(x)+C$ denotes the family of antiderivatives on an interval. The constant is needed because differentiation erases constants. It is not appended after evaluating a definite integral.

Worked example 2: rate to total. Suppose a tank's net inflow is $r(t)=6-2t$ L/min for $0\le t\le4$. The change in volume is

$$\int_0^4(6-2t)\,dt=[6t-t^2]_0^4=8\text{ L}.$$

The inflow changes sign at $t=3$. The total amount moved, counting flow in either direction positively, is $\int_0^4|6-2t|\,dt=9+1=10$ L. A net change and a total movement need not agree.

3.3 Substitution reverses the chain rule

The chain rule says $dF(g(x))/dx=F'(g(x))g'(x)$. Thus, under the usual continuity and differentiability assumptions,

$$\int F'(g(x))g'(x)\,dx=F(g(x))+C.$$

For definite integrals, transform the bounds as well:

$$\int_a^b f(g(x))g'(x)\,dx=\int_{g(a)}^{g(b)}f(u)\,du.$$

This formula follows by differentiating an antiderivative composition; it does not require $g$ to be globally one-to-one when used in this one-variable form.

2. Definite substitution and parts

Let $a<b$, $g\in C^1([a,b])$, and let $f$ be continuous on an interval $I$ containing $g([a,b])$. Define oriented integrals in the usual way if the bounds are reversed. Then

$$\int_a^b f(g(x))g'(x)\,dx=\int_{g(a)}^{g(b)} f(u)\,du.$$

Proof. Extend $f$ continuously to an open interval containing $g([a,b])$ by constant continuation at any finite endpoint of $I$ that lies in the image. Pick $u_0$ in that open interval and define $F(u)=\int_{u_0}^{u}f(t)\,dt$. By the fundamental theorem $F'=f$ throughout that open interval. The chain rule gives $(F\circ g)'=f(g)g'$ on $(a,b)$. The fundamental theorem applied to the continuous integrand $f(g)g'$ gives $F(g(b))-F(g(a))$, which is the right-hand side. No monotonicity or injectivity of $g$ is required under these hypotheses.

If $u,v\in C^1([a,b])$, then

$$\int_a^b u(x)v'(x)\,dx=[u(x)v(x)]_a^b-\int_a^b u'(x)v(x)\,dx.$$

Proof. Integrate the product rule $(uv)'=u'v+uv'$ and rearrange using linearity and the fundamental theorem. Continuity makes all integrals proper Riemann integrals.

Example. With $g(x)=x^2$ on $[-1,1]$ and $f(u)=e^u$, substitution gives $\int_{-1}^{1}2xe^{x^2}dx=\int_1^1 e^u du=0$ even though $g$ is not one-to-one. For parts, $\int_0^1 xe^x dx=[xe^x]_0^1-\int_0^1e^x dx=1$.

Supplemental calculus practice: CR problems

Full solutions to the CR problems

Worked example 3: an inner expression and its derivative. Calculate

$$\int_0^2\frac{2x}{1+x^2}\,dx.$$

Let $u=1+x^2$, so $du=2x\,dx$; the bounds become $u=1$ and $u=5$. The result is $\int_1^5du/u=\ln5$. Alternatively evaluate $\ln(1+x^2)$ between the original bounds. Do not mix transformed integrands with untransformed bounds.

For $\int x\sqrt{1+x^2}\,dx$, the same substitution gives $\frac12\int u^{1/2}du=\frac13(1+x^2)^{3/2}+C$. A derivative check confirms the missing-factor risk has been handled correctly.

3.4 Integration by parts reverses the product rule

Integrating $(uv)'=u'v+uv'$ gives

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

For definite integrals, include the boundary term $[uv]_a^b$. The method helps when differentiating one factor simplifies it and the other factor can be integrated.

Worked example 4: a time-weighted exponential. For $k>0$,

$$\int_0^T te^{-kt}\,dt =\left[-\frac tk e^{-kt}\right]_0^T+\frac1k\int_0^T e^{-kt}\,dt =\frac1{k^2}\bigl(1-e^{-kT}(1+kT)\bigr).$$

The negative sign in the antiderivative of $e^{-kt}$ is essential. At $T=0$ the result is zero, as it must be. Differentiating the final expression with respect to $T$ returns $Te^{-kT}$, a second check.

3.5 Area and admissible splitting

If a continuous curve crosses the axis, its signed integral may cancel positive and negative contributions. The unsigned area between $y=f(x)$ and the axis is $\int|f(x)|dx$, calculated by splitting at zeros. The area between two curves is $\int|f-g|dx$; the “top curve” may change across the interval.

Worked example 5: crossing curves. On $[0,2]$, compare $y=x$ and $y=x^2$. They meet at $0$ and $1$. The area is

$$\int_0^1(x-x^2)\,dx+\int_1^2(x^2-x)\,dx =\frac16+\frac56=1.$$

Using $\int_0^2(x^2-x)dx$ would give $2/3$, the signed difference, not the area.

3.6 Density, distribution, and expectation

A continuous probability model can be specified by a nonnegative density $p$ with total integral $1$. For an interval,

$$\Pr(a\le X\le b)=\int_a^b p(x)\,dx.$$

A density is not a point probability: an individual point has probability zero in this model, and density heights can exceed $1$. The cumulative distribution function is $F(x)=\Pr(X\le x)=\int_{-\infty}^xp(t)dt$. Where $p$ is continuous, $F'(x)=p(x)$.

When the relevant integrals converge absolutely, the expectation is $E[X]=\int xp(x)dx$ and the variance is $E[(X-E[X])^2]=E[X^2]-E[X]^2$. These definitions require a chosen probability model; an arbitrary nonnegative graph is not a density until it is normalized.

Worked example 6: normalize first. Let $p(x)=cx$ for $0\le x\le2$, and zero otherwise. Normalization requires $c\int_0^2x\,dx=2c=1$, so $c=1/2$. Then

$$\Pr(X\le1)=\int_0^1\frac x2dx=\frac14,\quad E[X]=\int_0^2\frac{x^2}2dx=\frac43.$$

Its CDF is $0$ for $x<0$, $x^2/4$ for $0\le x\le2$, and $1$ for $x>2$. The mean lies in the support, but not at its midpoint because the density weights larger values more heavily.

An improper integral on $[0,\infty)$ is the limit $\lim_{T\to\infty}\int_0^T$. For the exponential density $p(t)=\lambda e^{-\lambda t}$, $t\ge0$, $\lambda>0$, this limit is $1$. Its tail probability is $e^{-\lambda s}$ for $s\ge0$, and its mean is $1/\lambda$, using Worked example 4 and the limit $Te^{-\lambda T}\to0$. An improper integral is not obtained by treating infinity as an ordinary endpoint value.

3.7 Graded exercises

Foundation

1. Evaluate $\int_1^3(2x+1)\,dx$ and give a derivative check of the antiderivative.
2. Find $d/dx\,[\int_0^{x^2}\cos t\,dt]$.
3. A rate is $r(t)=4-t$ for $0\le t\le6$. Find its net integral and its total absolute accumulation.

Core

4. Evaluate $\int_0^1 3x^2e^{x^3}\,dx$ by substitution, including transformed bounds.
5. Evaluate $\int_0^1 xe^x\,dx$ by integration by parts.
6. Find the area between $y=x$ and $y=x^3$ on $[0,2]$.
7. Let $p(x)=k x(2-x)$ on $[0,2]$ and zero elsewhere. Find $k$, $\Pr(X\le1)$, and $E[X]$.

Synthesis

8. For the density in Exercise 7, compute $E[X^2]$ and the variance. Check that the variance is nonnegative.
9. For an exponential density with parameter $\lambda>0$, derive $\Pr(X>s+t\mid X>s)=\Pr(X>t)$ for $s,t\ge0$, using conditional probability and integrals. State why the conditional probability is defined.
10. A student integrates a nonnegative function on $[0,1]$ and obtains $1.4$, calling it a probability. Explain the missing condition. If the function is $q(x)=3x^2+1$, normalize it to a density on $[0,1]$ and compute the probability of $[0,1/2]$.

3.8 Full solutions

1. An antiderivative is $x^2+x$, whose derivative is $2x+1$. The definite integral is $(9+3)-(1+1)=10$.
2. The fundamental theorem followed by the chain rule gives $2x\cos(x^2)$.
3. The net integral is $[4t-t^2/2]_0^6=6$. Split the absolute integral at $4$: the first part is $8$, and the second is $2$, for total $10$.
4. Let $u=x^3$; then $du=3x^2dx$ and the bounds remain $0$ and $1$. The integral is $\int_0^1e^udu=e-1$.
5. Choose $u=x$, $dv=e^xdx$. Then $[xe^x]_0^1-\int_0^1e^xdx=e-(e-1)=1$.
6. The difference $x-x^3$ is positive on $(0,1)$ and negative on $(1,2)$. The area is $\int_0^1(x-x^3)dx+\int_1^2(x^3-x)dx=1/4+9/4=5/2$.
7. $\int_0^2x(2-x)dx=[x^2-x^3/3]_0^2=4/3$, so $k=3/4$. The probability to $1$ is $(3/4)(1-1/3)=1/2$. The mean is $(3/4)\int_0^2(2x^2-x^3)dx=(3/4)(16/3-4)=1$.
8. $E[X^2]=(3/4)\int_0^2(2x^3-x^4)dx=(3/4)(8-32/5)=6/5$. The variance is $6/5-1^2=1/5>0$.
9. The tail integral is $\Pr(X>u)=e^{-\lambda u}$. Since $\{X>s+t\}\subseteq\{X>s\}$, the conditional probability is $e^{-\lambda(s+t)}/e^{-\lambda s}=e^{-\lambda t}=\Pr(X>t)$. The denominator is positive for every finite $s\ge0$.
10. Nonnegativity alone is insufficient; a density's total integral over its support must be $1$. Here $\int_0^1q(x)dx=[x^3+x]_0^1=2$. Thus $p(x)=(3x^2+1)/2$ on $[0,1]$, zero elsewhere. The probability to $1/2$ is $\frac12[(1/2)^3+1/2]=5/16$.

Summary and onward map

An integral is a limit of sums, interpreted through its units and signs. The fundamental theorem connects accumulation with derivatives. Substitution reverses the chain rule; integration by parts reverses the product rule. Unsigned area requires absolute values or a correct sign split. A density must be nonnegative and normalized, and its probabilities come from integrals, not heights.

Prerequisite map. Continuity supports integrability and the fundamental theorem. Product and chain rules justify the two main techniques. Logarithms and exponential limits support improper integrals. Ordinary differential equations and broader probability topics require further chapters; they are not covered by this slice.