Integrals reference sheet

This sheet follows the seven units of the course in order. Each unit starts with the big idea in a few sentences, then gives the antiderivatives, methods and theorems you will look up most often, each with the conditions it needs. Most topics end with the integral of the matching chapter worked as an example, and the links under each topic lead to the chapter pages with the full explanation.

Angles are in radians throughout. Definite integrals show their limits, as in $\int_0^1 x\,dx$, and an antiderivative carries $+\,C$ for an arbitrary constant on each interval where it holds. Improper integrals are defined as limits of ordinary integrals. A function is continuously differentiable on an interval when its derivative exists and is continuous there, with one-sided derivatives at the endpoints. Vectors are bold, as in $\mathbf{F}$ and $\mathbf{r}$. A formula holds wherever both sides are defined unless a condition says otherwise.

Accumulation and foundations

A definite integral adds up rate times change in input across an interval. For a positive integrand it is the area under the graph. A limit of Riemann sums defines it, the fundamental theorem of calculus evaluates it with an antiderivative, and linearity splits a sum into separate signed parts.

The definite integral

In $\int_a^b f(x)\,dx$ the function $f$ is the integrand, the numbers $a$ and $b$ are the limits of integration and $x$ is the variable of integration. The integral is the signed area between the graph and the $x$-axis, where area above the axis counts as positive and area below as negative. Its units are the units of $f(x)$ times the units of $x$, so a rate in liters per second integrated over seconds gives liters.

For $f$ integrable on an interval that contains every limit in the row,

A flow of $2$ liters per second for $3$ seconds accumulates

liters, the area of a rectangle of height $2$ and width $3$.

Lessons What an integral measures

Riemann sums

A regular partition splits $[a, b]$ into $n$ subintervals of equal width $\Delta x = \frac{b - a}{n}$, with points $x_k = a + k\,\Delta x$. The mesh width of a partition is the width of its widest subinterval, here $\Delta x$. The right-endpoint and left-endpoint sums are

A function $f$ is integrable on $[a, b]$ when its Riemann sums, with any sample point in each subinterval, approach one limit as the mesh width tends to $0$. That limit is $\int_a^b f(x)\,dx$. Continuous functions are integrable, and so are bounded functions with finitely many discontinuities. For an increasing $f$ the integral lies between $L_n$ and $R_n$, and the gap is $R_n - L_n = \big(f(b) - f(a)\big)\,\Delta x$.

For $f(x) = x$ on $[0, 1]$ the points are $x_k = \frac{k}{n}$, and

so $R_4 = 0.625$, $R_{10} = 0.55$ and $R_{100} = 0.505$. The left sums are $L_n = \frac{1}{2} - \frac{1}{2n}$, both tend to the same limit, and

Lessons The Riemann-sum limit

The fundamental theorem of calculus

An antiderivative of $f$ on an interval is a function $F$ with $F' = f$ there. Any two antiderivatives on the same interval differ by a constant, and the indefinite integral $\int f(x)\,dx = F(x) + C$ names the whole family. Let $f$ be continuous on $[a, b]$.

Part 1. The accumulation function $A(x) = \int_a^x f(t)\,dt$ is continuous on $[a, b]$ and differentiable on $(a, b)$, with $A'(x) = f(x)$.

Part 2. If $F$ is an antiderivative of $f$ on $[a, b]$, then

The constant of integration cancels in the difference. The reverse power rule reads the power rule backward,

for $r \ne -1$ on each interval where $x^r$ is defined. For example $F(x) = \frac{x^3}{3}$ has $F'(x) = x^2$, so

The theorem needs $f$ continuous on all of $[a, b]$. See the improper integrals topic for what goes wrong at a point where $f$ blows up.

Lessons The fundamental theorem

Standard antiderivatives

Each entry holds on every interval inside the stated set, and adding $+\,C$ gives all antiderivatives there. Differentiating an answer checks it.

Lessons The fundamental theorem, Integration by parts, Trigonometric substitution, Completing the square

Linearity and average value

For $f$ and $g$ integrable on $[a, b]$ and constants $\alpha$ and $\beta$,

The sum rule is the case $\alpha = \beta = 1$ and the constant multiple rule the case $\beta = 0$. Each term keeps its own sign, so a term can make a negative signed contribution even when the whole integrand is positive. The integral of a product is not the product of the integrals. For $a < b$,

For example the three terms of $3x^2 - 4x + 2$ contribute $8$, $-8$ and $4$ on $[0, 2]$, so

although the integrand $3\left(x - \frac{2}{3}\right)^2 + \frac{2}{3}$ is positive everywhere. One antiderivative $F(x) = x^3 - 2x^2 + 2x$ confirms $F(2) - F(0) = 4$. The average value is $2$, which the integrand takes at $c = \frac{4}{3}$.

Lessons Linearity and signed contributions

Recognizing symbolic methods

Each method rewrites an integrand until it matches an entry of the table of standard antiderivatives. The work is mostly recognition, choosing the method from the shape of the integrand, and every answer can be checked by differentiating it.

Substitution

Substitution rule. If $g$ is continuously differentiable on $[a, b]$ and $f$ is continuous on the set of values $g([a, b])$, then

The indefinite form is $\int f\big(g(x)\big)\,g'(x)\,dx = F\big(g(x)\big) + C$ for an antiderivative $F$ of $f$, the chain rule in reverse. Look for an inner function $u = g(x)$ whose derivative also appears as a factor, write the differential $du = g'(x)\,dx$, and either change the bounds to $g(a)$ and $g(b)$ or substitute back before using the original bounds. When leftover factors of $x$ must be written in terms of $u$, the substitution must be one-to-one on the interval. For example $u = x^2$ is one-to-one for $x \ge 0$, where $x = \sqrt{u}$.

With $u = x^2$ and $du = 2x\,dx$, the bounds $0$ and $\sqrt{\pi/2}$ become $0$ and $\frac{\pi}{2}$, and

The check is $\frac{d}{dx}\sin(x^2) = 2x\cos(x^2)$.

Lessons Substitution

Integration by parts

Integration by parts. If $u$ and $v$ are continuously differentiable on $[a, b]$, then

The indefinite form is $\int u\,dv = uv - \int v\,du$. It is the product rule $(uv)' = u'v + uv'$ integrated and rearranged, and the boundary term $[uv]_a^b$ must be kept. Choose $u$ as the factor that gets simpler when differentiated, such as a polynomial or $\ln x$, and $dv$ as a factor you can integrate, such as $e^x\,dx$ or $\sin x\,dx$.

With $u = x$ and $v = e^x$,

The same number is a complementary area. The rectangle $[0, 1] \times [0, e]$ has area $e$, the region under $y = e^x$ has area $e - 1$, and the region between the curve and the $y$-axis has area $\int_1^e \ln y\,dy = 1$. In general, for $f$ continuously differentiable and strictly increasing on $[a, b]$,

Lessons Integration by parts

Powers of sine and cosine

The cosine of a doubled angle, $\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1$, gives the half-angle or power-reducing identities

and the antiderivatives that follow from them,

Here $\sin^2 x$ means $(\sin x)^2$, not $\sin(x^2)$. Over a full period the oscillation cancels. For a nonzero integer $k$,

For $\int \sin^m x \cos^n x\,dx$ with whole numbers $m$ and $n$,

For example $\int \sin^3 x\,dx = \int (1 - \cos^2 x)\sin x\,dx = -\cos x + \frac{1}{3}\cos^3 x + C$. As $x$ runs from $0$ to $\pi$ the doubled angle $2x$ runs through one full period of cosine, so

Lessons The half-angle identity

Trigonometric substitution

For $a > 0$, a square root of $a^2 - x^2$, $a^2 + x^2$ or $x^2 - a^2$ becomes a single trigonometric function after the substitution below. The angle range makes the substitution one-to-one and makes the trigonometric function in the last column nonnegative, so the square root needs no absolute value.

The differentials are $dx = a\cos\theta\,d\theta$, $a\sec^2\theta\,d\theta$ and $a\sec\theta\tan\theta\,d\theta$. Change the bounds with the angle, or return to $x$ with a right triangle, as in $\theta = \arcsin\frac{x}{a}$ and $\cos\theta = \frac{\sqrt{a^2 - x^2}}{a}$. The same substitution gives the quarter circle $\int_0^a \sqrt{a^2 - x^2}\,dx = \frac{\pi a^2}{4}$.

With $x = 2\sin\theta$ and $0 \le \theta \le \frac{\pi}{6}$, which covers $0 \le x \le 1$, the cosine is positive, so $\sqrt{4 - x^2} = 2\cos\theta$ and $dx = 2\cos\theta\,d\theta$. Then

which agrees with $\arcsin\frac{1}{2} = \frac{\pi}{6}$.

Lessons Trigonometric substitution

Partial fractions

A rational function $\frac{P}{Q}$, a ratio of polynomials, is proper when the degree of $P$ is less than the degree of $Q$. If it is not proper, divide first to write it as a polynomial plus a proper fraction. Factor $Q$ over the real numbers into linear factors and irreducible quadratics, those with a negative discriminant. A proper fraction is then a sum of terms, with one group for each distinct factor of $Q$, where $m$ is the power to which that factor appears.

Find the constants by clearing denominators and comparing coefficients, or by putting the roots of $Q$ into the cleared identity. That identity is between polynomials, so it holds at every $x$, even where the original fraction is undefined. The pieces integrate to logarithms, powers and, after completing the square, arctangents,

the second for $k \ge 2$.

The poles of $\frac{1}{x^2 - 1}$ are $x = \pm 1$, and neither lies in $[2, 3]$, where both $x - 1$ and $x + 1$ are positive. From $\frac{1}{x^2 - 1} = \frac{1}{2(x - 1)} - \frac{1}{2(x + 1)}$,

With a pole inside the interval the integral is improper. The integral of $\frac{1}{x^2 - 1}$ over $[0, 2]$ diverges, because near $x = 1$ the integrand behaves like $\frac{1}{2(x - 1)}$.

Lessons Partial fractions

Completing the square

For a quadratic $x^2 + bx + c$,

When the discriminant $b^2 - 4c$ is negative, the quadratic is irreducible over the real numbers and $c - \frac{b^2}{4} = k^2$ with $k > 0$. With $h = -\frac{b}{2}$, the shift and scale substitution $u = \frac{x - h}{k}$ leads to $\int \frac{du}{1 + u^2}$, and

the second for $\lvert x - h \rvert < k$. The inverse tangent takes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. A linear numerator is split into a multiple of the derivative $2x + b$ plus a constant, and $\int \frac{2x + b}{x^2 + bx + c}\,dx = \ln(x^2 + bx + c) + C$ when the quadratic is positive.

The quadratic $x^2 + 4x + 8 = (x + 2)^2 + 4$ has discriminant $16 - 32 = -16$. With $u = \frac{x + 2}{2}$ and $dx = 2\,du$ the bounds become $1$ and $2$, and

Lessons Completing the square

Choosing a method

Look at the integrand and try the first row that fits.

Some integrands, such as $e^{-x^2}$, $\frac{\sin x}{x}$ and $\frac{1}{\ln x}$, have no elementary antiderivative. Their definite integrals come from symmetry, numerical rules, series or the methods of the last unit.

Lessons Substitution, Integration by parts, The half-angle identity, Trigonometric substitution, Partial fractions, Completing the square

Structure, limits and approximation

Some integrals are settled by structure rather than by an antiderivative. Symmetry cancels or doubles whole terms, an improper integral is a limit of ordinary integrals, Simpson’s rule estimates a value with a guaranteed error, and a power series can be integrated one term at a time.

Symmetry

A function is even when $f(-x) = f(x)$ and odd when $f(-x) = -f(x)$, for every $x$ in a domain symmetric about $0$. For $f$ integrable on a symmetric interval $[-a, a]$,

the first when $f$ is odd and the second when $f$ is even. A product of two even or two odd functions is even, and a product of an even and an odd function is odd. Any $f$ on $[-a, a]$ splits into an even part $\frac{f(x) + f(-x)}{2}$ and an odd part $\frac{f(x) - f(-x)}{2}$, so read the parity of each term before calculating. Since $x^3$ is odd and $\cos x$ is even,

For an improper integral the rule for odd functions applies only after the integral is known to converge. The integral of $x$ over $(-\infty, \infty)$ diverges, even though $\int_{-R}^{R} x\,dx = 0$ for every $R$, and the integral of $\frac{1}{x}$ over $[-1, 1]$ diverges too.

Lessons Symmetry

Improper integrals

An integral over an unbounded interval, or of a function that is unbounded near a point, is defined as a limit. For $f$ continuous on $[a, \infty)$, and for $f$ continuous on $(a, b]$ and unbounded near $a$,

and similarly at the other end. The integral converges when the limit exists and is finite, and diverges otherwise. Over $(-\infty, \infty)$, or with a problem at both ends or at an interior point, split the integral so that each piece has one problem, and it converges only when every piece converges.

The p-test. For real $p$,

Both follow from $\int_1^R x^{-p}\,dx = \frac{R^{1-p} - 1}{1 - p}$ for $p \ne 1$. In the boundary case $p = 1$ both diverge, since $\int_1^R \frac{dx}{x} = \ln R \to \infty$ and $\int_t^1 \frac{dx}{x} = -\ln t \to \infty$, so $\int_0^\infty x^{-p}\,dx$ diverges for every $p$. For $p > 1$ the remaining tail is

For example $p = 2$ gives $\int_1^R x^{-2}\,dx = 1 - \frac{1}{R} \to 1$, and $p = 1.1$ gives $10$, approached slowly.

• Comparison test. If $f$ and $g$ are continuous with $0 \le g \le f$ on $[a, \infty)$, convergence of $\int_a^\infty f\,dx$ implies convergence of $\int_a^\infty g\,dx$, and divergence of the second implies divergence of the first. Since $x^2 \ge x$ for $x \ge 1$, the integral $\int_1^\infty e^{-x^2}\,dx$ converges and is at most $\int_1^\infty e^{-x}\,dx = \frac{1}{e}$.
• Absolute convergence. For $f$ continuous on $[a, \infty)$, if $\int_a^\infty \lvert f \rvert\,dx$ converges, so does $\int_a^\infty f\,dx$.
• Singularities inside. The fundamental theorem needs $f$ continuous on $[a, b]$. Applied blindly to $\frac{1}{x^2}$ on $[-1, 1]$ it gives $\big[-\frac{1}{x}\big]_{-1}^{1} = -2$, a negative value for a positive integrand, while in fact $\int_0^1 \frac{dx}{x^2}$ diverges and so does the whole integral. Any integral of $\frac{1}{x}$ over an interval containing $0$ diverges in the same way.

Lessons When an infinite region has finite area

Simpson’s rule

Composite Simpson’s rule. Take an even number $n$ of subintervals of $[a, b]$ with width $h = \frac{b - a}{n}$, points $x_j = a + jh$ and values $f_j = f(x_j)$. Then

On each pair of subintervals the rule integrates the quadratic interpolant through the three points, which gives $\frac{h}{3}(f_0 + 4f_1 + f_2)$, and it is exact for every polynomial of degree at most $3$.

Simpson’s error bound. If $f^{(4)}$ is continuous on $[a, b]$ and $\lvert f^{(4)}(x) \rvert \le M$ there, then

Any valid bound $M$ can be used, and a smaller one needs fewer subintervals. The function $e^{-x^2}$ has no elementary antiderivative. On $[0, 1]$ its fourth derivative is $(16x^4 - 48x^2 + 12)\,e^{-x^2}$, and since $e^{-x^2} \le 1$ the crude bound $M = 16 + 48 + 12 = 76$ holds. Two subintervals give $S_2 = \frac{1}{6}\big(1 + 4e^{-1/4} + e^{-1}\big) \approx 0.747180$. With $n = 100$ the error is at most $\frac{76}{180 \cdot 100^4} \approx 4.22 \times 10^{-9}$ and $S_{100} \approx 0.7468241329$, so

to six decimal places. The largest value of $\lvert f^{(4)} \rvert$ on $[0, 1]$ is in fact $12$, at $x = 0$.

Lessons Approximate an integral with a controlled error

Power series and term-by-term integration

A power series $\sum_{n=0}^{\infty} c_n (x - a)^n$ has a radius of convergence $R$ with $0 \le R \le \infty$. It converges absolutely for $\lvert x - a \rvert < R$ and diverges for $\lvert x - a \rvert > R$.

• Term-by-term integration. If $R > 0$, then for $\lvert x - a \rvert < R$ the integral from $a$ to $x$ of the series is $\sum_{n=0}^{\infty} \frac{c_n}{n + 1}(x - a)^{n+1}$, a power series with the same radius $R$. The series converges uniformly on every closed interval inside $(a - R, a + R)$, which justifies exchanging the sum and the integral there.
• Weierstrass M-test. If each $f_n$ is continuous on $[\alpha, \beta]$ with $\lvert f_n(x) \rvert \le M_n$ there and $\sum M_n$ converges, then $\sum f_n$ converges uniformly on $[\alpha, \beta]$ and its integral is the sum of the integrals of the $f_n$.
• Geometric tail bound. If $a_n > 0$ and $a_{n+1} \le r\,a_n$ for all $n \ge N + 1$ with $r < 1$, then $\sum_{n > N} a_n \le \frac{a_{N+1}}{1 - r}$.