Learning goals and preparation

We assume limits, continuity, derivatives, the product and chain rules, and elementary trigonometric identities. This chapter develops definite integrals and the main hand-calculation techniques: substitution, integration by parts, trigonometric identities and substitutions, and rational partial fractions. It also explains how an infinite interval changes an integration problem into a limit problem.

The aim is not to force every integrand into one formula. A useful method exploits structure, preserves domains, and leaves an answer that can be verified by differentiation or by a limiting calculation.

3.1 Definite integrals and antiderivatives

Why the real numbers need more than arithmetic

The rational numbers support the familiar ordered arithmetic, yet the set $\{q\in\mathbb Q:q>0,\ q^2<2\}$ has no rational least upper bound. The real number system adds a completeness property: bounded nonempty sets have least upper bounds. This is the source of many existence results later in analysis, not simply another rule for manipulating inequalities.

Definition

Upper bound and supremum

For a nonempty $A\subset\mathbb R$, $u$ is an upper bound if $a\le u$ for every $a\in A$. A number $s$ is $\sup A$ if it is an upper bound and $s\le u$ for every upper bound $u$. A maximum is an element of $A$ that is also an upper bound. The maximum, when it exists, is the supremum; a supremum need not belong to $A$. Lower bounds, infimum and minimum are defined by reversing inequalities.

Theorem

Completeness axiom

Every nonempty subset of $\mathbb R$ that is bounded above has a real supremum. Equivalently, every nonempty subset bounded below has a real infimum, via $\inf A=-\sup(-A)$.

For example, $\sup(0,1)=1$ although $1$ is not a maximum of that set. Conversely, $\sup[0,1]=1$ and it is a maximum. The empty set and an unbounded-above set do not have a real supremum under this definition.

Theorem

Supremum approximation criterion

An upper bound $s$ of a nonempty set $A$ is $\sup A$ if and only if, for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a\le s$.

Proof

Proving both directions

If $s=\sup A$ and no element exceeds $s-\varepsilon$, then $s-\varepsilon$ is a smaller upper bound, contradiction. Conversely, suppose $s$ is an upper bound and the approximation property holds. If another upper bound $u$ were smaller than $s$, choose $\varepsilon=s-u>0$. The property gives $a>u$, contradicting that $u$ is an upper bound. Thus $s$ is the least upper bound.

Worked example

Use both halves of a supremum proof

Let $A=\{n/(n+1):n\in\mathbb N, n\ge1\}$. Every element is less than $1$, so $1$ is an upper bound. Given $\varepsilon>0$, choose $n>1/\varepsilon$ by the Archimedean property below. Then $1-n/(n+1)=1/(n+1)<\varepsilon$. Thus $\sup A=1$. No element equals $1$, so $A$ has no maximum. An upper-bound calculation alone would not prove leastness: $2$ is also an upper bound.

Archimedean property and density

Theorem

Archimedean property

For every real $x$ there exists a positive integer $n>x$. Consequently, for every $\varepsilon>0$ there is $n\in\mathbb N$ such that $1/n<\varepsilon$.

Proof

Completeness forces integers to be unbounded

If the positive integers were bounded above, completeness would give $s=\sup\mathbb N$. Since $s-1$ is not an upper bound, some $n\in\mathbb N$ satisfies $n>s-1$. But $n+1>s$ and $n+1\in\mathbb N$, contradicting that $s$ bounds the set. Apply this to $1/\varepsilon$ to obtain the second statement.

Between any real numbers $x<y$ there is a rational number. Indeed choose $n$ with $n(y-x)>1$; the Archimedean property and the integer-part property yield an integer $m$ with $nx<m<ny$, so $x<m/n<y$. The integer-part property follows from the unboundedness of integers in both directions and the well-ordering of the natural numbers. An irrational also lies between $x$ and $y$: choose a rational $q$ with $x-\sqrt2<q<y-\sqrt2$; then $q+\sqrt2$ is irrational and lies in the interval. Thus density does not say rationals and irrationals are equal sets.

Worked example

An infimum via reflection

For $B=\{2+1/n:n\ge1\}$, every element exceeds $2$. Given $\varepsilon>0$, choose $n>1/\varepsilon$ to obtain $2<2+1/n<2+\varepsilon$. Therefore $2$ is the infimum, but not a minimum. The maximum is $3$, attained at $n=1$.

Quantifiers in convergence

Definition

Convergence of a real sequence

A sequence $(x_n)$ converges to $L\in\mathbb R$ if for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that for every $n\ge N$, $|x_n-L|<\varepsilon$. Write $x_n\to L$. The index $N$ may depend on $\varepsilon$ but must work for all later indices.

To negate convergence to $L$, there must be some $\varepsilon_0>0$ such that for every $N$ there is an $n\ge N$ with $|x_n-L|\ge\varepsilon_0$. A few early terms do not affect convergence; a recurring failure arbitrarily late does.

Worked example

Construct an epsilon–N proof

For $x_n=(3n+1)/(n+2)$, guess $L=3$. Since

$$\left|\frac{3n+1}{n+2}-3\right|=\frac5{n+2}<\frac5n,$$

choose $N>5/\varepsilon$. Then every $n\ge N$ gives $|x_n-3|<\varepsilon$. The guess is motivation; the inequality and quantified choice establish the proof.

Theorem

Uniqueness, boundedness and limit laws

A convergent real sequence has a unique limit and is bounded. If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$, $x_ny_n\to xy$, and $x_n/y_n\to x/y$ when $y\ne0$ and the quotient is defined eventually. If $x_n\le z_n\le y_n$ eventually and both outer sequences converge to $L$, then $z_n\to L$.

Proof

Uniqueness and boundedness

If $x_n\to L$ and $x_n\to M$, then for any $\varepsilon>0$, sufficiently late terms satisfy both $|x_n-L|<\varepsilon/2$ and $|x_n-M|<\varepsilon/2$. The triangle inequality yields $|L-M|<\varepsilon$. Since this holds for every positive $\varepsilon$, $L=M$. For boundedness choose $N$ so that $|x_n-L|<1$ for $n\ge N$. The finite initial terms and the bound $|x_n|\le|L|+1$ for the tail together give a finite bound for the whole sequence.

The converse of boundedness is false: $(-1)^n$ is bounded but does not converge. If it converged to $L$, its even subsequence would converge to $1$ and its odd subsequence to $-1$, violating uniqueness.

Proof

Product law mechanism

A convergent sequence $(y_n)$ is bounded, say $|y_n|\le M$. Write

$$|x_ny_n-xy|\le|y_n||x_n-x|+|x||y_n-y|.$$

Both terms tend to zero. For the quotient, $y_n\to y\ne0$ implies $|y_n|\ge|y|/2$ eventually; this keeps the denominator away from zero and gives $1/y_n\to1/y$.

Subsequences

Definition

Subsequence

A subsequence of $(x_n)$ is $(x_{n_k})$ where $n_1<n_2<\cdots$ are positive integers. If $x_n\to L$, then every subsequence converges to $L$ because $n_k\ge k$ and therefore eventually lies beyond any prescribed index.

Theorem

Bolzano–Weierstrass

Every bounded real sequence has a convergent subsequence.

Proof

Nested-interval proof

Place every term in a closed bounded interval $I_1$. Bisect it. At least one half contains infinitely many terms; choose that half as $I_2$. Continue, obtaining nested closed intervals $I_k$ of lengths tending to zero, each containing infinitely many terms. Choose indices $n_k$ strictly increasing with $x_{n_k}\in I_k$; this is possible because each interval contains infinitely many terms. Completeness gives a common point $L$ in all intervals: the supremum of their left endpoints lies in each interval because every left endpoint is at most every right endpoint. Since $x_{n_k},L\in I_k$, their distance is at most the interval length and tends to zero.

The conclusion is about a subsequence, not necessarily the whole sequence. For example, $(-1)^n$ has constant subsequences converging to $1$ and $-1$, but does not converge.

Real closed sets: the sequence criterion and the open complement

Work on the real line in the usual classical set-theoretic setting, with countable witness selections available. A set $U$ is open when every $y$ in $U$ has some $r>0$ with $(y-r,y+r)$ contained in $U$. We use the Archimedean consequence that $1/n$ tends to zero. For every $K$ subset of the real line, the following conditions are equivalent:

1. Every sequence of points of $K$ that converges in the real line has its limit in $K$.
2. The complement of $K$ is open.

Proof of 2 implies 1. Suppose $x_n$ belongs to $K$ and $x_n$ tends to $x$. If $x$ were outside $K$, openness would give $r>0$ with $(x-r,x+r)$ disjoint from $K$. Convergence gives a threshold after which $|x_n-x|<r$, contradicting $x_n$ in $K$. Therefore $x$ belongs to $K$.

Proof of 1 implies 2. Fix $y$ outside $K$. If no positive-radius interval around $y$ avoided $K$, then for every positive integer $n$ there would be a point $x_n$ in $K$ with $|x_n-y|<1/n$. Choose these witnesses as a sequence. Given $\varepsilon>0$, choose $N$ with $1/N<\varepsilon$. For $n\ge N$, $|x_n-y|<1/n\le1/N<\varepsilon$, so $x_n$ tends to $y$. Condition 1 would force $y$ into $K$, a contradiction. Hence some interval around $y$ avoids $K$. This holds for every point of the complement, proving it open. The witness-sequence step uses the stated classical setting; this proof is not offered as a theorem about choice-free foundations. If $K$ is empty there is no sequence in $K$, and its complement is the whole open line. If $K$ is the whole line, its empty complement is open vacuously. Both edge cases also satisfy the equivalence.

Learning goals and definitions

A set $K\subseteq\mathbb R$ is compact if every family of open sets whose union contains $K$ has a finite subfamily still covering $K$. A set is closed if it contains all limits of convergent sequences of its points, equivalently if its complement is open. Sequential compactness means every sequence in $K$ has a subsequence converging to a member of $K$.

The empty set is compact, closed and bounded, and sequentially compact vacuously. In proofs using a selected point or an infinite sequence, treat that case separately. We may assume the real completeness property and its nested-interval and Bolzano–Weierstrass consequences from the earlier sequence units.

Theorem: Heine–Borel in the real line

For $K\subseteq\mathbb R$, the following are equivalent: compactness by open covers; closedness and boundedness; sequential compactness.

Compact implies closed and bounded. The intervals $(-n,n)$, $n\ge1$, cover $\mathbb R$, hence $K$. A finite subcover gives a bound. To show closedness, take $y\notin K$. For each $x\in K$, the interval $B(x,|x-y|/2)$ avoids $y$. A finite subfamily covers $K$. Let $\delta$ be the minimum of their positive half-distances to $y$. Every point in the chosen interval centered at $x$ has distance from $y$ strictly greater than $|x-y|/2\ge\delta$. Thus $B(y,\delta)$ misses $K$, showing its complement open. If $K$ is empty this is immediate.

Closed and bounded implies compact. Enclose nonempty $K$ in a closed bounded interval $I_0$. Suppose an open cover of $K$ has no finite subcover. Bisect $I_0$. At least one closed half meets $K$ in a set not finitely coverable; otherwise two finite collections would cover $K$. Choose such a half and repeat. We obtain nested closed intervals $I_n$ with lengths tending to zero, and $K\cap I_n$ not finitely coverable. Each intersection is nonempty; choose $x_n\in K\cap I_n$. The nested-interval property gives a unique common point $p$, and $x_n\to p$ because both lie in $I_n$. Closedness gives $p\in K$. Some member $O$ of the cover contains $p$, and openness gives a ball around $p$ inside $O$. For large $n$, the whole interval $I_n$ lies in that ball, so $O$ alone covers $K\cap I_n$, a contradiction.

Closed and bounded implies sequentially compact. Every sequence in $K$ is bounded, so Bolzano–Weierstrass supplies a convergent subsequence. Closedness retains its limit in $K$.

Sequentially compact implies closed and bounded. If unbounded, choose $x_n\in K$ with $|x_n|>n$; no subsequence can converge to a finite real number, a contradiction. If $p$ is a limit of points in $K$, choose $x_n\in K$ with $|x_n-p|<1/n$. A subsequence converges to a point of $K$ by sequential compactness, but it also converges to $p$; uniqueness of real limits gives $p\in K$. Thus $K$ is closed. This completes all implications.

Compact subsets of the real line

A set $K\subset\mathbb R$ is compact if every open cover has a finite subcover. In $\mathbb R$, Heine–Borel says this is equivalent to being closed and bounded; it is also equivalent to every sequence in $K$ having a subsequence converging to a point of $K$. The closedness requirement keeps the subsequential limit in the set; boundedness permits extraction. We use the sequential form in the proofs below.

Theorem

Extreme value theorem

If $K\subset\mathbb R$ is nonempty and compact and $f:K\to\mathbb R$ is continuous, then $f$ is bounded and attains both a maximum and a minimum on $K$.

Proof

Extract a maximizing subsequence

If $f$ were unbounded above, choose $x_n\in K$ with $f(x_n)>n$. A subsequence $x_{n_k}\to x\in K$ exists; continuity would give $f(x_{n_k})\to f(x)$, contradicting divergence to infinity. The same reasoning applied to $-f$ gives boundedness below. Let $s=\sup f(K)$. Choose $x_n\in K$ with $s-1/n<f(x_n)\le s$. A convergent subsequence tends to some $x\in K$, and continuity implies $f(x)=s$. Apply the argument to $-f$ for a minimum.

Worked example

Which hypothesis fails?

The continuous function $f(x)=x$ on $(0,1)$ is bounded but has no maximum or minimum: $0$ and $1$ are not in the domain. The set is bounded but not closed, hence not compact. On $[0,1]$, both endpoints are present and the extrema are attained. Compactness is a sufficient global hypothesis, not a condition required for every individual function to attain extrema.

Fermat's necessary condition. At a differentiable interior local maximum or minimum, $f'(a)=0$.

Proof. At a local maximum, sufficiently small positive increments give nonpositive difference quotients and negative increments give nonnegative ones. A two-sided derivative can satisfy both sign limits only by being zero. Negating the function treats a minimum. $\square$

The interior hypothesis matters: $f(x)=x$ on $[0,1]$ has extrema at endpoints despite derivative 1. The converse fails: $x^3$ has zero derivative at zero but no extremum there.

Rolle's theorem. If $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$.

Proof. The extreme-value theorem supplies maximum and minimum. If the function is constant, every interior point works. Otherwise a value differs from the equal endpoint values, so either a maximum or a minimum occurs in the interior. Fermat's condition applies there. $\square$

Mean value theorem. Under the same continuity and differentiability hypotheses but without equal endpoint values, some $c\in(a,b)$ satisfies $f(b)-f(a)=f'(c)(b-a)$.

Subtract the secant line $f(a)+[f(b)-f(a)](x-a)/(b-a)$ from $f$ and apply Rolle. This proves the theorem without assuming $f'$ is continuous. Consequences include: derivative zero on an interval makes the function constant; nonnegative derivative makes it nondecreasing; and $|f'|\le M$ on an interval gives $|f(y)-f(x)|\le M|y-x|$ for points in it. An interval-wide derivative bound cannot be inferred from one derivative value.

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 $\int_a^bf=F(b)-F(a)$.

The integral is signed; negative values of $f$ contribute negatively.

The notation $\int f(x)dx=F(x)+C$ describes antiderivatives on an interval. Different constants may be required on disconnected domain components. For example, $\int dx/x=\ln|x|+C$ separately on each of $(-\infty,0)$ and $(0,\infty)$. It is not an integral across the singularity at zero.

A working table. For $n\ne-1$ on a suitable real interval,

$$\int x^n dx=\frac{x^{n+1}}{n+1}+C,\quad \int\frac{dx}{x}=\ln|x|+C,\quad \int e^{kx}dx=\frac1k e^{kx}+C\ (k\ne0).$$

Also $\int\cos xdx=\sin x+C$, $\int\sin xdx=-\cos x+C$, $\int\sec^2xdx=\tan x+C$, and $\int dx/(1+x^2)=\arctan x+C$. Differentiate these formulas to check coefficients, signs and domains. A definite integral is a number and does not acquire a final $+C$.

3.2 Substitution and correct bounds

The chain rule implies

$$\int f(g(x))g'(x)dx=F(g(x))+C\quad\text{when }F'=f.$$

For definite integrals,

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

A one-variable substitution of this form does not require global injectivity of $g$ when justified by an antiderivative composition. More general change-of-variable interpretations do require attention to orientation and multiplicity; do not treat every transformation as an unsigned replacement of intervals.

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, $\int_0^1 xe^x dx=[xe^x]_0^1-\int_0^1e^x dx=1$.

Worked example 1: a missing coefficient. For $\int x(4+x^2)^3dx$, set $u=4+x^2$, $du=2x\,dx$. The integral is $\frac12\int u^3du=(4+x^2)^4/8+C$. Differentiation gives $x(4+x^2)^3$, confirming the factor $1/8$.

Worked example 2: a nonmonotone inner function. Evaluate $\int_{-1}^1 2x\cos(x^2)dx$. The antiderivative is $\sin(x^2)$, so the answer is $\sin1-\sin1=0$. Transforming both endpoints by $u=x^2$ gives the same oriented integral from $1$ to $1$. This is valid because the factor $2x$ retains orientation. It does not mean the different integral $\int_{-1}^1\cos(x^2)dx$ is zero.

3.3 Integration by parts and repeated use

The product rule yields

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

For definite integrals replace $uv$ with $[uv]_a^b$ and keep both bounds on the remaining integral. Choose $u$ so its derivative simplifies, if possible.

Worked example 3: polynomial times exponential. For $\int x^2e^xdx$, choose $u=x^2$, $dv=e^xdx$. Then

$$\int x^2e^xdx=x^2e^x-2\int xe^xdx=e^x(x^2-2x+2)+C.$$

The second integration by parts reduces the polynomial degree again. A derivative check cancels the unwanted lower-degree terms.

Worked example 4: a logarithm without a visible product. Write $\ln x=(\ln x)\cdot1$ for $x>0$. Let $u=\ln x$, $dv=dx$. Then $\int\ln xdx=x\ln x-x+C$. This method uses the logarithm's domain; it is not an antiderivative formula for a real $\ln x$ at negative $x$.

Some products lead back to the original integral. For $I=\int e^x\cos xdx$, integration by parts twice gives $I=e^x\cos x+e^x\sin x-I$. Hence $I=e^x(\cos x+\sin x)/2+C$. Treating the repeated integral as an unknown and solving the resulting linear equation is legitimate, with the arbitrary constant absorbed at the end.

3.4 Trigonometric powers and substitutions

For an odd power of sine or cosine, save one factor for a differential and convert the remaining even power using $\sin^2x+\cos^2x=1$.

Worked example 5: an odd power. For $\int\sin^3x\cos^2xdx$, use $u=\cos x$, $du=-\sin xdx$, and $\sin^2x=1-u^2$. The integral becomes

$$-\int(1-u^2)u^2du=-\frac{u^3}{3}+\frac{u^5}{5}+C.$$

For even powers, half-angle identities often help: $\sin^2x=(1-\cos2x)/2$ and $\cos^2x=(1+\cos2x)/2$. Thus $\int_0^{\pi/2}\sin^2xdx=\pi/4$.

A radical can suggest a trigonometric substitution. For $\sqrt{a^2-x^2}$ with $a>0$, set $x=a\sin\theta$ on $\theta\in[-\pi/2,\pi/2]$. Then the radical equals $a\cos\theta$, not $\pm a\cos\theta$, because cosine is nonnegative on the selected branch. Also $dx=a\cos\theta\,d\theta$.

Worked example 6: a semicircle integral with branch control. Calculate $\int_0^a\sqrt{a^2-x^2}dx$. The substitution gives

$$a^2\int_0^{\pi/2}\cos^2\theta d\theta=\frac{\pi a^2}{4}.$$

The geometric quarter-circle area independently checks the result. Without a branch restriction one would have $\sqrt{a^2\cos^2\theta}=a|\cos\theta|$, not automatically $a\cos\theta$.

The related choices $x=a\tan\theta$ for $\sqrt{a^2+x^2}$ and $x=a\sec\theta$ for $\sqrt{x^2-a^2}$ can simplify other radicals, but their intervals and signs must likewise be stated. Choose a substitution because it simplifies the integrand, not just because a square root appears.

3.5 Rational functions and partial fractions

For a rational function $P(x)/Q(x)$, first divide polynomials if the numerator degree is at least the denominator degree. Then factor the denominator over the reals. Distinct linear factors produce terms $A/(x-a)$; repeated factors need all powers up to their multiplicity; an irreducible quadratic needs a linear numerator.

Worked example 7: distinct factors. Decompose

$$\frac{3x+5}{(x+1)(x+2)}=\frac A{x+1}+\frac B{x+2}.$$

Equating numerators gives $A+B=3$ and $2A+B=5$, so $A=2,B=1$. Therefore the antiderivative is $2\ln|x+1|+\ln|x+2|+C$ on intervals avoiding $-1,-2$.

For a repeated factor, $1/[x(x+1)^2]$ requires $A/x+B/(x+1)+C/(x+1)^2$. Equating coefficients gives $A=1,B=-1,C=-1$. An antiderivative is $\ln|x|-\ln|x+1|+1/(x+1)+C$. Omitting the first power or repeated power would prevent a general decomposition.

For an irreducible denominator, split a numerator into a multiple of the derivative and a remainder. For example,

$$\int\frac{2x+3}{x^2+1}dx=\ln(x^2+1)+3\arctan x+C.$$

The logarithm needs no absolute value because $x^2+1>0$.

3.6 Improper integrals are limits

Define $\int_a^\infty f(x)dx=\lim_{T\to\infty}\int_a^Tf(x)dx$ when that limit exists and is finite. Likewise a singular endpoint requires a one-sided limit. An interior singularity must be split into two separately convergent integrals; cancellation across it is not ordinary convergence.

Worked example 8: the power threshold. For $p\ne1$,

$$\int_1^T x^{-p}dx=\frac{T^{1-p}-1}{1-p}.$$

As $T\to\infty$, this converges to $1/(p-1)$ exactly when $p>1$. For $p=1$, the integral is $\ln T$ and diverges. Thus decay to zero of an integrand alone does not imply finite total accumulation.

The integral of $1/x$ across $[-1,1]$ is not zero as an ordinary improper integral: each side diverges. A symmetric cutoff may cancel, but that is a different concept, not justification for evaluating an antiderivative across an undefined point.

3.7 Graded exercises

Foundation

1. Evaluate $\int_0^2(3x^2-2x+1)dx$.
2. Find $\int 4x(1+2x^2)^5dx$ and check it by differentiation.
3. Evaluate $\int_1^e\ln xdx$ by parts.

Core

4. Find $\int x\cos(3x)dx$.
5. Evaluate $\int_0^{\pi/2}\sin^3x\cos^2xdx$.
6. Evaluate $\int_0^2\sqrt{4-x^2}dx$ by a trigonometric substitution with a stated branch.
7. Decompose and integrate $(2x+3)/[(x+1)(x+2)]$.
8. Find $\int(3x+2)/(x^2+4)dx$.

Synthesis

9. Determine whether $\int_1^\infty x^{-3/2}dx$ and $\int_0^1x^{-3/2}dx$ converge, calculating the appropriate limits.
10. Explain precisely why $\int_{-1}^1x^{-2}dx$ cannot be evaluated as $[-1/x]_{-1}^1=-2$. Evaluate the one-sided pieces as improper limits and identify the failed fundamental-theorem hypothesis.

3.8 Full solutions

1. An antiderivative is $x^3-x^2+x$. The result is $8-4+2=6$.
2. Set $u=1+2x^2$, $du=4xdx$. The answer is $u^6/6+C=(1+2x^2)^6/6+C$. Its derivative is $(1+2x^2)^5(4x)$.
3. The antiderivative is $x\ln x-x$. Evaluating gives $(e-e)-(0-1)=1$.
4. Let $u=x$, $dv=\cos3x\,dx$, so $v=\sin3x/3$. The result is $x\sin3x/3+\cos3x/9+C$. Differentiating cancels the extra sine terms and leaves $x\cos3x$.
5. With $u=\cos x$, the bounds change from $1$ to $0$, and the integral is $\int_0^1(u^2-u^4)du=1/3-1/5=2/15$.
6. Set $x=2\sin\theta$, $0\le\theta\le\pi/2$. The radical is $2\cos\theta$ and $dx=2\cos\theta d\theta$, giving $4\int_0^{\pi/2}\cos^2\theta d\theta=\pi$.
7. Equating $A+B=2$, $2A+B=3$ gives $A=B=1$. The antiderivative is $\ln|x+1|+\ln|x+2|+C$, on each interval avoiding the two poles.
8. Split the numerator: the $3x$ part is $(3/2)(2x)$, giving $(3/2)\ln(x^2+4)$. The remaining integral $\int2dx/(x^2+4)=\arctan(x/2)$. Thus the answer is $(3/2)\ln(x^2+4)+\arctan(x/2)+C$.
9. On $[1,T]$, the antiderivative is $-2x^{-1/2}$, giving $2-2/\sqrt T\to2$, so the first converges. On $[\varepsilon,1]$, it gives $-2+2/\sqrt\varepsilon\to\infty$ as $\varepsilon\downarrow0$, so the second diverges.
10. The integrand is undefined and unbounded at zero, so it is not continuous on the whole interval and the ordinary fundamental theorem cannot be used across that point. On $[\varepsilon,1]$ the integral is $-1+1/\varepsilon\to\infty$. On $[-1,-\varepsilon]$ it is $1/\varepsilon-1\to\infty$. Both improper pieces diverge positively, consistent with the nonnegative integrand; the claimed negative value is impossible.

Summary and onward map

The fundamental theorem evaluates continuous definite integrals through antiderivatives. Substitution follows the chain rule; parts follows the product rule. Identities simplify trigonometric powers, branch-controlled substitutions simplify radicals, and partial fractions separate rational terms. Every technique must preserve its domain. Improper integrals require limits, and internal singularities must be treated on each side separately.