Unit 6

Integration and Accumulation of Change: the one-page sheet

6.1 Accumulating Change

When you know a rate, the change in the amount is the area between the rate graph and the horizontal axis. Its units are the rate's units times the input's units. Area above the axis adds and area below subtracts, giving the net change; add it to the starting amount to get the amount at the end.

The area between a rate graph and the horizontal axis is the accumulated change in the amount.
Units of the area = units of the rate times units of the input: (liters per minute) times (minutes) = liters.
Area above the axis adds to the amount; area below the axis subtracts. The result is the net change.
Amount at the end = starting amount + net change.

the change in the amount is rate times time, the area of a rectangle under the rate graph the change is still the area under the rate graph, now a triangle, a trapezoid or a curved region that region lies below the axis and its area is subtracted: the amount decreases there the amount at the end is the starting amount plus the net change

accumulated change: The total change in an amount over an interval of time, built up from its rate of change. On a graph of the rate, it is the area between the graph and the horizontal axis: area above the axis adds, area below subtracts. Its units are the rate's units times the input's units.

6.2 Estimating Area with Riemann Sums

A Riemann sum estimates accumulated change by adding width times height over subintervals; a trapezoidal sum averages the two end heights. Take every width from the table. Left and right sums under- or overestimate according to whether the function increases or decreases; trapezoids and midpoints according to its concavity.

A Riemann sum adds (width) times (height) over the subintervals: left, right or midpoint heights.
A trapezoidal sum uses width times the average of the two end heights; it equals (left + right) / 2.
From a table, each width is the difference of two times. Never assume equal widths.
Increasing function: left sum under, right sum over. Decreasing: the reverse. Concave up: trapezoid over, midpoint under.

treat it as roughly constant on each subinterval and add up width times height the widths are 1, 2, 1 and 3; each product uses its own width left ends give the smallest heights, so a left sum underestimates and a right sum overestimates chords lie above the graph, so a trapezoidal sum overestimates

Riemann sum: An estimate of the accumulated change (the area under a graph) built from rectangles: split the interval into subintervals and add up (width) times (a height of the function on that subinterval). A left sum uses each subinterval's left end, a right sum its right end, a midpoint sum its middle.
trapezoidal sum: An estimate of the area under a graph built from trapezoids: on each subinterval, the width times the average of the function's values at the two ends. It equals the average of the left and right Riemann sums that use the same subintervals.
subinterval: One of the pieces an interval is split into for a Riemann or trapezoidal sum. Subintervals may have equal widths (from a formula) or unequal widths (from the times in a table); each width is the difference of its two ends.

6.3 Riemann Sums, Sigma Notation and the Definite Integral

The definite integral ∫ from a to b of f(x) dx is the limit of Riemann sums ∑ f(x_i) Δx as n → ∞ , with Δx = (b − a)/n and x_i = a + i Δx . Translate both ways by matching the width, the points and the height. For a rate, the integral is the accumulated change.

$\int_a^b f(x)\,dx$ $= \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\,\Delta x$ with $\Delta x = \frac{b - a}{n}$ and $x_i = a + i\,\Delta x$ .
In $\int_a^b f(x)\,dx$ , $f(x)$ is the integrand, $a$ and $b$ are the limits of integration, and $dx$ names the variable.
For a positive function, the definite integral is the area under the graph; for a rate, the accumulated change, in the accumulated units.
A calculator finds definite integrals numerically: write the integral as your setup, then the value to three decimal places.

each is Δx = (b − a)/n wide, with right ends x_i = a + i Δx it becomes ∑ for i = 1 to n of f(x_i) Δx the sums approach one number, the definite integral ∫ from a to b of f(x) dx the width gives b − a , the points give a , and the height gives f

definite integral: The limit of Riemann sums of a function $f$ on $[a, b]$ as the widths of the subintervals shrink to 0, written $\int_a^b f(x)\,dx$ . The function $f(x)$ is the integrand; $a$ and $b$ are the limits of integration. For a positive $f$ it is the area under the graph; for a rate, the accumulated change.
sigma notation: Shorthand for a sum: $\sum_{i=1}^{n} a_i$ means $a_1 + a_2 + \cdots + a_n$ . A Riemann sum with right ends is $\sum_{i=1}^{n} f(x_i)\,\Delta x$ , with $\Delta x = \frac{b - a}{n}$ and $x_i = a + i\,\Delta x$ .

6.4 The Fundamental Theorem and Accumulation Functions

An accumulation function g(x) = ∫ from a to x of f(t) dt starts at 0 and grows at the rate f(x) : for continuous f , g′(x) = f(x) , so g is an antiderivative of f . With an upper limit u(x) , the derivative is f(u(x)) u′(x) .

An accumulation function $g(x)$ $= \int_a^x f(t)\,dt$ is the area so far; $g(a) = 0$ .
Fundamental Theorem of Calculus: if $f$ is continuous, $g'(x) = f(x)$ . The constant lower limit drops out.
An antiderivative of $f$ is any $F$ with $F' = f$ . Every accumulation function of $f$ is one.
Upper limit $u(x)$ : $\frac{d}{dx}\int_a^{u(x)} f(t)\,dt$ $= f(u(x)) \cdot u'(x)$ . Forgetting $u'(x)$ is the classic slip.

g(x) is the area accumulated under f from a up to x , and g(a) = 0 the area grows by a thin strip about f(x) tall and h wide, so (g(x + h) − g(x))/h ≈ f(x) g′(x) = f(x) : the Fundamental Theorem of Calculus, so g is an antiderivative of f the chain rule adds a factor: d/dx ∫ from a to u(x) of f(t) dt = f(u(x)) u′(x)

antiderivative: A function $F$ whose derivative is $f$ : $F'(x) = f(x)$ . For example, $x^2 + 5$ is an antiderivative of $2x$ . Any two antiderivatives of the same function on an interval differ by a constant.
accumulation function: A function defined by a definite integral with a variable limit, such as $g(x)$ $= \int_a^x f(t)\,dt$ : the amount accumulated from $a$ up to $x$ . It is 0 at $x = a$ .
Fundamental Theorem of Calculus: The link between derivatives and integrals. Part 1: if $f$ is continuous, $\frac{d}{dx}\int_a^x f(t)\,dt$ $= f(x)$ , and with a function as the upper limit, $\frac{d}{dx}\int_a^{u(x)} f(t)\,dt$ $= f(u(x))\,u'(x)$ . Part 2 (topic 6.7): $\int_a^b f(x)\,dx$ $= F(b) - F(a)$ for any antiderivative $F$ of $f$ .

6.5 Behavior of Accumulation Functions

A definite integral is a signed area: regions below the axis subtract. For g(x) = ∫ from a to x of f(t) dt , read g 's values from signed areas, its direction from the sign of f , and its concavity from whether f rises or falls. Justify with g′ = f named, and use the candidates test for absolute extrema.

Signed area: $\int_a^b f(x)\,dx$ = (area above the axis) minus (area below).
$g' = f$ : $g$ increases where $f > 0$ , decreases where $f < 0$ , and has a relative extremum where $f$ changes sign.
$g'' = f'$ : $g$ is concave up where $f$ increases and concave down where $f$ decreases; inflection points sit where $f$ changes direction.
For an absolute extremum of g, compare g at the endpoints and at the sign changes of f (the candidates test).

g′(x) = f(x) and g″(x) = f′(x) g is increasing; where f changes sign, g has a relative extremum g is concave up; where f turns around, g has a point of inflection values of g are signed areas: add what is above, subtract what is below

signed area: The value of a definite integral read from a graph: areas of regions above the x-axis count as positive and areas below count as negative, so $\int_a^b f(x)\,dx$ = (area above) minus (area below).

6.6 Properties of Definite Integrals

Definite integrals add over adjacent intervals, change sign when the limits are reversed, are 0 over a single point, and split over sums and constant multiples (but not products). Use these with geometry, splitting at jumps, to find integrals from tables and graphs.

Additivity: $\int_a^b f(x)\,dx$ $+ \int_b^c f(x)\,dx$ $= \int_a^c f(x)\,dx$ , for any order of $a$ , $b$ , $c$ .
Reversed limits: $\int_b^a f(x)\,dx$ $= -\int_a^b f(x)\,dx$ , and $\int_a^a f(x)\,dx$ $= 0$ .
Linearity: $\int_a^b (kf(x) + g(x))\,dx$ $= k\int_a^b f(x)\,dx$ $+ \int_a^b g(x)\,dx$ , and $\int_a^b k\,dx$ $= k(b - a)$ .
A bounded function with a jump or a hole can still be integrated: split at the break and use geometry on each piece.

anything true of sums (splitting a sum, pulling out a constant) is true of integrals their integrals add: ∫ from a to b of f +∫ from b to c of f = ∫ from a to c of f every width is negative, so ∫ from b to a of f = −∫ from a to b of f , and ∫ from a to a of f = 0 the result is k times the width: ∫ from a to b of k dx = k(b − a)

properties of definite integrals: Rules for definite integrals: $\int_a^a f = 0$ ; $\int_b^a f$ $= -\int_a^b f$ ; additivity, $\int_a^b f$ $+ \int_b^c f$ $= \int_a^c f$ ; and linearity, $\int_a^b (kf + g)$ $= k\int_a^b f$ $+ \int_a^b g$ . Integrals split over sums and constant multiples, never over products.

6.7 The Fundamental Theorem and Definite Integrals

To evaluate ∫ from a to b of f(x) dx , find an antiderivative F and compute F(b) − F(a) . Read in context, the integral of a rate of change is the net change: f(b) = f(a) +∫ from a to b of f′(x) dx . Use signed areas from a graph of f′ , an antiderivative from a formula, or a calculator when no formula exists.

FTC, part 2: if $F' = f$ and $f$ is continuous on $[a, b]$ , then $\int_a^b f(x)\,dx$ $= F(b) - F(a)$ $= F(x)\Big|_a^b$ .
Net change: $\int_a^b f'(x)\,dx$ $= f(b) - f(a)$ , so $f(b)$ $= f(a)$ $+ \int_a^b f'(x)\,dx$ .
Going backward from a known value: $f(0)$ $= f(2) - \int_0^2 f'(x)\,dx$ .
With no elementary antiderivative, a calculator evaluates the integral; write the integral as the setup.

they differ by a constant: g(x) = F(x) − F(a) , since g(a) = 0 ∫ from a to b of f(x) dx = F(b) − F(a) , the Fundamental Theorem, part 2 ∫ from a to b of f′(x) dx = f(b) − f(a) : the integral of a rate is the net change f(b) = f(a) +∫ from a to b of f′(x) dx , with signed areas from the graph

net change theorem: The Fundamental Theorem read in context: the integral of a rate of change is the net change, $\int_a^b f'(x)\,dx$ $= f(b) - f(a)$ . So $f(b)$ $= f(a)$ $+ \int_a^b f'(x)\,dx$ : the value at the end is the starting value plus the accumulated change.

6.8 Antiderivatives and Indefinite Integrals

Every derivative rule read backward gives an antidifferentiation rule. The indefinite integral ∫ f(x) dx = F(x) + C is the whole family of antiderivatives, so it always carries +C . Rewrite the integrand into basic forms first, check by differentiating, and use a known point to find C .

Indefinite integral: $\int f(x)\,dx = F(x) + C$ , the family of all antiderivatives. No limits, a family, never a number.
Power rule backward: $\int x^n\,dx$ $= \frac{x^{n+1}}{n+1} + C$ for $n \ne -1$ ; and $\int \frac{1}{x}\,dx = \ln|x| + C$ .
Rewrite before integrating: roots as powers, products expanded, a fraction over a single term split up.
Check every answer by differentiating it. Find $C$ from a known point by substituting it.

if F′ = f , then (F + C)′ = f for every constant C differ by a constant, so the whole family is F(x) + C ∫ f(x) dx = F(x) + C : an indefinite integral always ends in +C substitute it to find C and pick out the one function

indefinite integral: The family of all antiderivatives of a function, written $\int f(x)\,dx = F(x) + C$ , where $F' = f$ and $C$ , the constant of integration, is any constant. It has no limits and is a family of functions, not a number.

6.9 Integrating by Substitution

Substitution undoes the chain rule. Choose u as an inside function whose derivative is a factor of the integrand, replace g′(x) dx by du , fix any constant factor, and integrate in u . In a definite integral, change the limits to values of u .

Look for an inside function $u = g(x)$ whose derivative $g'(x)$ is also a factor, up to a constant.
Write $du = g'(x)\,dx$ , rewrite the whole integral in $u$ (no $x$ left), integrate, then substitute back.
Definite integrals: new limits $u = g(a)$ and $u = g(b)$ ; keeping the $x$ -limits is the classic slip.
Linear inside: $\int f(kx)\,dx$ $= \frac{1}{k}F(kx) + C$ , such as $\int e^{3x}\,dx = \frac{1}{3}e^{3x} + C$ .

so ∫ f(g(x)) g′(x) dx = F(g(x)) + C when F′ = f the integral becomes ∫ f(u) du , a basic form adjust it: x dx = 1/2 du when du = 2x dx change the limits to u = g(a) and u = g(b) and never go back to x

u-substitution: Undoing the chain rule: when an integrand has the form $f(g(x))\,g'(x)$ , let $u = g(x)$ and $du = g'(x)\,dx$ , so $\int f(g(x))g'(x)\,dx$ $= \int f(u)\,du$ . In a definite integral the limits change to $u = g(a)$ and $u = g(b)$ .

6.10 Integrating by Long Division and Completing the Square

Rewrite before integrating. When the top's degree is at least the bottom's, divide to get a polynomial plus a remainder fraction. When the bottom is a quadratic with no real zeros and the top is constant, complete the square and use ∫ du/(u² + a²) = 1/a arctan u/a +C .

Long division first when the top's degree is at least the bottom's: $\frac{x^2 + 3x + 5}{x + 1}$ $= x$ $+ 2$ $+ \frac{3}{x + 1}$ .
Completing the square: $x^2$ $+ bx$ $+ c$ $= \left(x + \frac{b}{2}\right)^2$ $+ \left(c - \frac{b^2}{4}\right)$ .
$\int \frac{1}{u^2 + a^2}\,du$ $= \frac{1}{a}\arctan\frac{u}{a}$ $+ C$ : the factor $\frac{1}{a}$ is easy to drop.
If the top is the derivative of the bottom, skip both: substitute and get $\ln$ .

divide: a polynomial plus a remainder over the bottom it integrates to a logarithm, kln |x + 1| complete the square: (x − h)² + a² ∫ du/(u² + a²) = 1/a arctan u/a +C

long division: Rewriting an integrand so a basic rule applies. Polynomial long division turns a fraction whose top has degree at least the bottom's into a polynomial plus a remainder fraction. Completing the square turns a quadratic with no real zeros into $(x - h)^2 + a^2$ , which leads to an arctan form.

6.11 Integration by Parts

Integration by parts is the product rule run backward: ∫ u dv = uv − ∫ v du . Choose u to simplify when differentiated and dv to integrate easily, write out u , dv , du and v , and in a definite integral evaluate uv at both bounds. Repeated parts, or the tabular method, handles x^n times an exponential or a sine.

Formula: $\int u\,dv = uv - \int v\,du$ . Write $u$ , $dv$ , $du$ and $v$ before you start.
Put in $u$ what gets simpler when differentiated (a power of $x$ , $\ln x$ , an inverse trigonometric function); put in $dv$ what you can integrate ( $e^{kx}$ , $\sin kx$ , $\cos kx$ ).
A power $x^n$ times $e^{kx}$ , $\sin kx$ or $\cos kx$ takes $n$ rounds of parts; the tabular method does them all at once with alternating signs.
Definite integrals: $\int_a^b u\,dv$ $= uv\Big|_a^b - \int_a^b v\,du$ . The $uv$ term needs its bounds too.

integrating both sides gives uv = ∫ u dv + ∫ v du ∫ u dv = uv − ∫ v du : an integral traded for another the new integral ∫ v du is easier than the old one repeat parts, or lay the steps out in a table (the tabular method) evaluate uv at both bounds: ∫ from a to b of u dv = uv|_a^b − ∫ from a to b of v du

integration by parts: The product rule run backward: $\int u\,dv = uv - \int v\,du$ , where $u$ is a factor chosen to simplify when differentiated and $dv$ (with $dx$ ) a factor that can be integrated. For a definite integral, $uv$ is evaluated at both bounds. The tabular method lays out repeated rounds in two columns with alternating signs.

6.12 Linear Partial Fractions

When a denominator factors into distinct linear factors, split the fraction into A/(x − a) + B/(x − b) . Clear the denominator, substitute the zeros to find A and B , and integrate each piece to a natural log. Divide first if the top's degree is not below the bottom's.

Factor the denominator into distinct linear factors; write one fraction $\frac{A}{x - a}$ per factor.
Clear the denominator and substitute each zero to find the constants. Check by adding the pieces back.
$\int \frac{A}{x - a}\,dx$ $= A\ln|x - a| + C$ ; combine logs only when it helps.
Top degree at least the bottom's: divide first. Top equal to the derivative of the bottom: substitute instead.

the fraction can be written as A/(x − a) + B/(x − b) a polynomial identity that holds for every x one term vanishes each time, giving A and then B it integrates to Aln |x − a| do long division first, then split the remainder

partial fractions: Splitting a rational function whose denominator factors into distinct linear factors into simple fractions: $\frac{p(x)}{(x - a)(x - b)}$ $= \frac{A}{x - a}$ $+ \frac{B}{x - b}$ when the top's degree is below the bottom's. Each piece integrates to a natural log, $\int \frac{A}{x - a}\,dx$ $= A\ln|x - a| + C$ .

6.13 Improper Integrals

An improper integral has an infinite limit of integration or an integrand that is unbounded on the interval. Replace the bad end by a variable, integrate, and take the limit: a finite limit means the integral converges to that value; otherwise it diverges. Write the limit in every line, and break the interval at any interior point where the integrand blows up.

$\int_a^\infty f(x)\,dx$ $= \lim_{b \to \infty} \int_a^b f(x)\,dx$ . It converges if this limit is a finite number and diverges otherwise.
Unbounded integrand at an end: $\int_0^1 f(x)\,dx$ $= \lim_{a \to 0^+} \int_a^1 f(x)\,dx$ . At an interior point, write it as two integrals that meet there.
$\int_1^\infty \frac{1}{x^p}\,dx$ converges (to $\frac{1}{p - 1}$ ) when $p > 1$ and diverges when $p \le 1$ .
Keep $\lim$ in every line until you take it; never substitute $\infty$ as a number.

the integral is improper: it is not a definite integral in the old sense an ordinary definite integral, ∫ from 1 to b of f(x) dx if the limit is a finite number, the integral converges to it; otherwise it diverges write it as two integrals that meet there, each one improper

improper integral: An integral with an infinite limit of integration, such as $\int_1^\infty \frac{1}{x^2}\,dx$ , or with an integrand unbounded on the interval, such as $\int_0^1 \frac{1}{\sqrt{x}}\,dx$ . It is defined as a limit of definite integrals: $\int_a^\infty f(x)\,dx$ $= \lim_{b \to \infty} \int_a^b f(x)\,dx$ .
convergence: An improper integral converges when the limit of the definite integrals that defines it is a finite number; that number is its value. It diverges when the limit is infinite or does not exist. Example: $\int_1^\infty \frac{1}{x^2}\,dx$ converges to 1; $\int_1^\infty \frac{1}{x}\,dx$ diverges.

6.14 Choosing an Antidifferentiation Technique

To choose a technique, ask in order: is it a basic form after rewriting? Is there an inside function with its derivative as a factor? Is it a fraction to divide? Is it a constant over an irreducible quadratic? Then integrate, check by differentiating, and change the limits whenever you substitute in a definite integral. On the BC exam, two more questions follow: is it a product with a factor that simplifies when differentiated (integration by parts)? Does the bottom factor into distinct linear factors (linear partial fractions)?

Rewrite first: roots as powers, products expanded, fractions over one term split.
Substitution needs the inside's derivative as a factor (up to a constant). $\int x e^{x^2}\,dx$ yes; $\int e^{x^2}\,dx$ no.
For fractions, compare: top is a multiple of the bottom's derivative ( $\ln$ ); top degree too high (divide); constant over an irreducible quadratic (arctan).
Always check by differentiating. If no method works, the definite integral may be a calculator question.

use the basic rules, after rewriting roots as powers and splitting fractions over one term substitute u for the inside and change any limits divide first, then integrate the pieces complete the square and use the arctan form

antidifferentiation technique: Choosing how to find an antiderivative: try a basic rule (after rewriting powers or splitting over a single term), then substitution (an inside function whose derivative is a factor), then long division (top degree at least bottom degree), then completing the square (a quadratic with no real zeros on the bottom). Check by differentiating.