AP®MathsCollege BoardCalculus BCRevision NotesUnit 6: Integration & Accumulation of ChangeUnit 6 OverviewUnit 6 Summary

Unit 6 Summary (College Board AP® Calculus BC): Revision Note

Written by: Roger B

Reviewed by: Jamie Wood

Updated on 14 May 2026

Integration & accumulation of change summary

Key definitions

$F (x)$ is an antiderivative of the function $f (x)$ if $F' (x) = f (x)$
An indefinite integral of a function $f$ is denoted $\int f (x) d x$
The accumulation of change of a function $f (x)$ starting at $x = a$ is $F (x) = \int_{a}^{x} f (t) d t$
The left Riemann sum to approximate $\int_{a}^{b} f (x) d x$ is $\sum_{i = 1}^{n} (x_{i} - x_{i - 1}) \cdot f (x_{i - 1})$
The right Riemann sum to approximate $\int_{a}^{b} f (x) d x$ is $\sum_{i = 1}^{n} (x_{i} - x_{i - 1}) \cdot f (x_{i})$
The midpoint Riemann sum to approximate $\int_{a}^{b} f (x) d x$ is $\sum_{i = 1}^{n} (x_{i} - x_{i - 1}) \cdot f (\frac{x_{i - 1} + x_{i}}{2})$
The trapezoidal sum to approximate $\int_{a}^{b} f (x) d x$ is $\sum_{i = 1}^{n} (x_{i} - x_{i - 1}) \cdot \frac{f (x_{i - 1}) + f (x_{i})}{2}$
The definite integral can be defined as a limit of Riemann sums
- $\int_{a}^{b} f (x) d x = \lim_{\max Δ x_{i} \to 0} \sum_{i = 1}^{n} f (x_{i}^{*}) Δ x_{i}$
The integral $\int_{a}^{b} f (x) d x$ is improper if one of the following is true:
- $f (x)$ is unbounded on $[a, b]$
- at least one of the limits of integration is infinite
The improper integrals are defined as follows:
- $\int_{a}^{\infty} f (x) d x = \lim_{k \to \infty} \int_{a}^{k} f (x) d x$
- $\int_{- \infty}^{b} f (x) d x = \lim_{h \to - \infty} \int_{h}^{b} f (x) d x$
- $\int_{a}^{c} f (x) d x = \lim_{k \to c^{-}} \int_{a}^{k} f (x) d x$
- $\int_{c}^{b} f (x) d x = \lim_{h \to c^{+}} \int_{h}^{b} f (x) d x$

Key theorems

The first fundamental theorem of calculus
- If $f$ is continuous on $[a, b]$ and $F$ is an antiderivative of $f$
- then $\int_{a}^{b} f (x) d x = F (b) - F (a)$
The second fundamental theorem of calculus
- If $f$ is continuous on an interval containing $a$ , then for values of $x$ in that interval
  - $F (x) = \int_{a}^{x} f (t) d t$ is an antiderivative of $f$
  - $\frac{d}{d x} (\int_{a}^{x} f (t) d t) = f (x)$

Key formulas

The integral of a function multiplied by a constant: If $k$ is a constant
- $\int k \cdot f (x) d x = k \cdot \int f (x) d x$
The integral of the sum or difference of two functions:
- $\int (f (x) \pm g (x)) d x = \int f (x) d x \pm \int g (x) d x$
The definite integral at a point:
- $\int_{a}^{a} f (x) d x = 0$
The change of limits of a definite integral:
- $\int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x$
Splitting a definite interval at a point: If $a < c < b$
- $\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$
Indefinite integrals of functions

$f (x)$	$\int f (x) d x$
$x^{n}$ where $n \neq - 1$	$\frac{1}{n + 1} x^{n + 1} + C$
$\frac{1}{x}$	$\ln \|x\| + C$
$e^{x}$	$e^{x} + C$
$a^{x}$ where $a > 0$ and $a \neq 1$	$\frac{1}{\ln a} a^{x} + C$
$\sin x$	$- \cos x + C$
$\cos x$	$\sin x + C$
$\sec^{2} x$	$\tan x + C$
$\frac{1}{\sqrt{1 - x^{2}}}$	$\arcsin x + C$
$\frac{1}{1 + x^{2}}$	$\arctan x + C$

If $f$ is continuous on an interval containing $a$ , then for values of $x$ in that interval
- $f (x) = f (a) + \int_{a}^{x} f (t) d t$
If $f$ is continuous on an interval containing $a$ and $g$ is differentiable, then for values of $x$ in that interval
- $\frac{d}{d x} (\int_{a}^{g (x)} f (t) d t) = f (g (x)) \cdot g' (x)$
The reverse chain rule is $\int f' (g (x)) \cdot g' (x) d x = f (g (x)) + C$
Integration by substitution can be used to write an integral $\int_{x_{1}}^{x_{2}} f (x) d x = \int_{u_{1}}^{u_{2}} g (u) d u$
Completing the square can be used to derive the following integrals
- $\int \frac{1}{\sqrt{q - {(x - p)}^{2}}} d x = \arcsin (\frac{x - p}{\sqrt{q}}) + C$
- $\int \frac{1}{q + {(x - p)}^{2}} d x = \frac{1}{\sqrt{q}} \arctan (\frac{x - p}{\sqrt{q}}) + C$
The integration by parts formula is $\int u \cdot \frac{d v}{d x} d x = u v - \int \frac{d u}{d x} \cdot v d x$
$\frac{a x + b}{(m x + n) (p x + q)}$ can be rewritten using partial fractions as $\frac{A}{m x + n} + \frac{B}{p q + q}$

Key facts

The criteria for under/overestimates can be summarized in the table below

Method	Condition in $f$	Result
Left Riemann	increasing	underestimate
Left Riemann	decreasing	overestimate
Right Riemann	increasing	overestimate
Right Riemann	decreasing	underestimate
Trapezoidal	concave up	overestimate
Trapezoidal	concave down	underestimate

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