Math 31 – Calc II Exam 2 Flashcards

0.0(0)
Studied by 0 people
call kaiCall Kai
Locked
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/29

flashcard set

Earn XP

Description and Tags

Flashcards based on the Math 31 Calculus II Unit 2 lecture transcript covering integration techniques, convergence tests, and series definitions.

Last updated 11:42 PM on 7/20/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai
Chat

No analytics yet

Send a link to your students to track their progress

30 Terms

1
New cards

LIATE/LIPET

The priority order for picking uu in integration by parts: Logarithmic, Inverse trig, Algebraic/polynomial, Trigonometric, and Exponential.

2
New cards

Strategy for tanm(x)secn(x)dx\int \tan^m(x) \sec^n(x) dx (nn even)

Save one sec2(x)\sec^2(x) for dudu. Let u=tan(x)u = \tan(x) (du=sec2(x)dxdu = \sec^2(x) dx) and convert remaining sec2(x)\sec^2(x) factors using sec2(x)=1+tan2(x)\sec^2(x) = 1 + \tan^2(x).

3
New cards

Strategy for tanm(x)secn(x)dx\int \tan^m(x) \sec^n(x) dx (mm odd)

Save one sec(x)tan(x)\sec(x)\tan(x) for dudu. Let u=sec(x)u = \sec(x) (du=sec(x)tan(x)dxdu = \sec(x)\tan(x) dx) and convert remaining tan2(x)\tan^2(x) using tan2(x)=sec2(x)1\tan^2(x) = \sec^2(x) - 1.

4
New cards

Trig Substitution for a2x2\sqrt{a^2 - x^2}

Use x=asin(θ)x = a \sin(\theta), dx=acos(θ)dθdx = a \cos(\theta) d\theta, which simplifies to acos(θ)a \cos(\theta).

5
New cards

Trig Substitution for a2+x2\sqrt{a^2 + x^2}

Use x=atan(θ)x = a \tan(\theta), dx=asec2(θ)dθdx = a \sec^2(\theta) d\theta, which simplifies to asec(θ)a \sec(\theta).

6
New cards

Trig Substitution for x2a2\sqrt{x^2 - a^2}

Use x=asec(θ)x = a \sec(\theta), dx=asec(θ)tan(θ)dθdx = a \sec(\theta) \tan(\theta) d\theta, which simplifies to atan(θ)a \tan(\theta).

7
New cards

Completing the Square in Integration

A technique used when a quadratic under a radical has a linear term (e.g., 82xx28 - 2x - x^2) before applying a trig substitution.

8
New cards

Polynomial Long Division Requirement

Must be performed before partial fraction decomposition when the degree of the numerator is greater than or equal to the degree of the denominator (degree(numerator)degree(denominator)\text{degree(numerator)} \geq \text{degree(denominator)}).

9
New cards

Decomposition for Distinct Linear Factors

The form Ax+Bx3+Cx+3\frac{A}{x} + \frac{B}{x-3} + \frac{C}{x+3} for a denominator like x(x3)(x+3)x(x-3)(x+3).

10
New cards

Decomposition for Repeated Linear Factor (xa)2(x-a)^2

The form Axa+B(xa)2\frac{A}{x-a} + \frac{B}{(x-a)^2}.

11
New cards

Decomposition for Irreducible Quadratic Factor

For a denominator like (xa)(x2+b2)(x-a)(x^2+b^2), the form is Axa+Bx+Cx2+b2\frac{A}{x-a} + \frac{Bx+C}{x^2+b^2}.

12
New cards

Numerical Integration Rules

Trapezoidal Rule (TnT_n), Simpson’s Rule (SnS_n), and Midpoint Rule (MnM_n) used to approximate integrals over subintervals.

13
New cards

Simpson's Rule Special Requirement

The number of subintervals nn must be even.

14
New cards

Improper Integral Types

Type 1 integrals have infinite limits of integration; Type 2 integrals have a discontinuous integrand on the interval.

15
New cards

Improper Integral Convergence

Occurs when the relevant limit exists and equals a finite real number LL. If the limit does not exist or goes to ±\pm\infty, it diverges.

16
New cards

Sequence Convergence

A sequence {an}\left\{ a_n \right\} converges if limnan=L\lim_{n \to \infty} a_n = L for some finite real number LL.

17
New cards

Order of Growth Rates (Slowest to Fastest)

ln(n)<np(p>0)<n<nx(x>1)<xn<n!<nn\ln(n) < n^p \, (p>0) < n < n^x \, (x>1) < x^n < n! < n^n. This hierarchy is used for evaluating limits as nn \to \infty.

18
New cards

Infinite Series Sum Definition

an=limkSk\sum a_n = \lim_{k \to \infty} S_k, where Sk=n=1kanS_k = \sum_{n=1}^{k} a_n is the kk-th partial sum.

19
New cards

Absolute Convergence

A series an\sum a_n is absolutely convergent if the series of absolute values an\sum |a_n| converges.

20
New cards

Conditional Convergence

A series an\sum a_n converges, but an\sum |a_n| diverges.

21
New cards

Absolute Convergence Theorem

If a series is absolutely convergent, then it is convergent (Absolute convergenceconvergence\text{Absolute convergence} \Longrightarrow \text{convergence}).

22
New cards

n-th Term Test for Divergence

If limnan0\lim_{n \to \infty} a_n \neq 0, the series diverges. If the limit is 00, the test is inconclusive.

23
New cards

Geometric Series Test

A series arn\sum ar^n converges to S=a1rS = \frac{a}{1-r} if r<1|r| < 1 and diverges if r1|r| \geq 1.

24
New cards

p-Series Test

A series 1np\sum \frac{1}{n^p} converges if p>1p > 1 and diverges if p1p \leq 1 (e.g., the harmonic series 1n\sum \frac{1}{n} diverges).

25
New cards

Integral Test

If an=f(n)a_n = f(n) with ff being positive, continuous, and decreasing for x1x \geq 1, then an\sum a_n and 1f(x)dx\int_{1}^{\infty} f(x) dx converge or diverge together.

26
New cards

Direct Comparison Test

For 0anbn0 \leq a_n \leq b_n: if bn\sum b_n converges, an\sum a_n converges; if an\sum a_n diverges, bn\sum b_n diverges.

27
New cards

Limit Comparison Test

Given L=limnanbnL = \lim_{n \to \infty} \frac{a_n}{b_n} where an,bn>0a_n, b_n > 0, if 0<L<0 < L < \infty, both series behave the same.

28
New cards

Alternating Series Test Conditions

For (1)nbn\sum (-1)^n b_n with bn>0b_n > 0, the series converges if (1) bn+1bnb_{n+1} \leq b_n (decreasing) and (2) limnbn=0\lim_{n \to \infty} b_n = 0.

29
New cards

Ratio Test

A test where L=limnan+1anL = \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|. L<1L < 1 implies absolute convergence, L>1L > 1 divergence, and L=1L = 1 is inconclusive. Best for factorials and exponentials.

30
New cards

Root Test

A test where L=limnannL = \lim_{n \to \infty} \sqrt[n]{|a_n|}. Best used when the entire term is raised to the nn-th power; conclusions match the Ratio Test.