Study Guide Final - Summary Calculus II PDF

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Summarizes (almost all of) the entire course material in one PDF....

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Math 1320 Exam 1 Study Guide Note: This study guide is meant as a guide, NOT an exhaustive list of topics and problems that you may encounter on the exam. The topics are broken down individually, but you will often need to combine ideas from several sections to solve a problem. In addition to solving problems, you should also have a solid conceptual understanding of the topics. This study guide should not be your only material used when studying. Reviewing previous WebAssign assignments, quizzes, written homeworks, the textbook, practice exams, and your notes is all recommended, as is working new problems from the textbook to test your understanding of the material. Integration Methods Chapter 7 covers numerous methods for evaluating integrals. These methods are reviewed below, along with general strategies about when to use each method. You should know the following trigonometric identities and be able to use them to simplify or help evaluate integrals. The identities in the left column are called pythagorean identities, and the identities in the right column are double angle identities. sin2 x + cos2 x = 1 tan2 x + 1 = sec2 x 1 + cot2 x = csc2 x

1 cos2 x = (1 + cos(2x)) 2 1 2 sin x = (1 − cos(2x)) 2 sin(2x) = 2 sin x cos x

All the integrals in the table on page 495 of the textbook, with the exception of numbers 19 and 20, are assumed to be common knowledge. You may use the formulas without comment or justification while solving more complicated integrals. Z Z 1. Integration by Parts: The formula for integration by parts is udv = uv − vdu. This method is usually used to integrate a product, especially one of the form xn ex , xn ln xZ, xn sin x, Zor ex sin x. It can also be used to integrate a single function, as in the

cases

ln xdx,

arctan xdx.

Your solutions should include clearly labeled u, dv, du, andZ dv functions, ideally aru = ln x dv = x2 dx ranged in a 2 by 2 grid, such as for x2 ln xdx. du = x1 dx v = 13 x3 Sample problems:

(a)

Z

2

x sin x dx (b)

Z

1

1

e

ln x dx (c)

Z

ex cos x dx

Z

2. Trigonometric Integrals: The strategies for evaluating integrals of the form sinm x cosn xdx Z and tanm x secn xdx are given on pages 473 and 474, respectively, in the textbook. As a summary: Z To evaluate sinm x cosn xdx: if n or m is odd, use a pythagorean identity and the

u-substitution u = sin x or u = cos x to rewrite the integrand as a polynomial. If both n and m are even, use double angle identities to rewrite the integrand in terms of cos(2x). You may then need to apply one of these two strategies repeatedly to obtain the answer. Z To evaluate

tanm x secn xdx: if m is odd or n is even, use a pythagorean identity

and the u-substitution u = tan x or u = sec x. If m is even and n is odd, typically integration by parts and clever use of trig identities will be required. Z Z m n To evaluate cot x csc xdx, use the same methods as for tanm x secn xdx.

Sample problems: (a)

Z

sin3 x cos4 x dx (b)

Z

cos4 x dx (c)

Z

sec4 x tan2 x dx (d)

Z

sec3 x tan3 x dx

3. Trigonometric Substitution: When the integrand contains an expression of the form √ 2 2 ±a ± x , it is often advisable to use a trigonometric substitution to take advantage of the pythagorean identities. You should always look for a straightforward u-substitution before using a trig substitution. √ a2 − x2 x = a sin θ √ 2 2 √a + x x = a tan θ x2 − a2 x = a sec θ Note: Your final answer for indefinite integrals should be in terms of x, not θ. You may √ sometimes need topcomplete the square to put an integrand in the correct form: e.g. x2 + 6x + 13 = (x + 3)2 + 4. Trig substitution is also useful when powers of square roots are present, e.g. when the exponent of your quadratic polynomial is 3/2. Sample problems: (a)

Z

√ x2 4 − x2 dx (b)

Z

x2

√

1 dx (c) x2 − 9

Z

1 dx (x2 − 6x + 13)3/2

4. Partial Fractions: A rational function is a function of the form P (x)/Q(x) where Z P (x) P (x) and Q(x) are polynomials. To evaluate dx, first use polynomial long Q(x) division if deg P (x) > deg Q(x). After this, decompose the fraction using the method

2

of partial fractions. E.g. (note the difference between linear factors and irreducible quadratic factors): x3 (x2

A B C Dx + E F x + G Hx + I 1 . = + 2+ 3+ 2 + 2 + 2 2 2 (x + 4)2 + 1)(x + 4) x x x x +1 x +4

To solve for the coefficients, multiply across by the denominator, collect like terms, and equate the coefficients of equal powers of x. One can then integrate each individual partial fraction to arrive at the final answer. Ax + B Ax B To integrate 2 , then use u-substitution on the first , write it as 2 + 2 2 2 x + a2 x +a x +a term and recognize the second term as the derivative of an arctan function. Note: If the irreducible quadratic in the denominator is not already of the form x2 +a2 , you will need to complete the square before integrating p (but after performing the partial fraction decomposition). Also, if there is a radical f (x) p involved in a rational function, it may be helpful to begin with a u-substitution of u = f (x). Sample problems:

(a)

Z

x2 − 2x − 1 dx (b) (x − 1)2 (x2 + 1)

Z √

x+9 dx x

5. Numerical Approximation of Integrals: You should understand the geometric interpretation of each of the integration rules: Left, Right, Midpoint, Trapezoid, Simpson. Facts to keep in mind: the Trapezoid rule is simply the average of the Left and Right approximations. Trapezoid and Midpoint tend to be better than Left and Right, with Midpoint usually more accurate than the Trapezoid, and Simpson’s Rule is better than all the rest. Note: Another geometric interpretation of the Midpoint rule is given at the bottom of page 509 in the book: it is the area of the trapezoid whose upper side is tangent to the graph at the midpoint of the relevant interval. 6. Improper Integrals: There are two types of improper integrals: integrating on an infinite interval (a, ∞), (−∞, a), or (−∞, ∞); and integrating across an infinite disZ 5 Z 4 1 dx √ continuity of the integrand, e.g. or dx. 4−x 2 x−4 0 These integrals are evaluated using LIMITS. E.g. Z t Z ∞ 1 1 dx dx = lim 2 t→∞ 3 x2 x 3 and

Z

0

4

dx √ = lim 4 − x t→4− 3

Z

t 0

√

dx . 4−x

The improper integral is convergent if the limit exists (that is, the limit is a finite number), and divergent if the limit does not exist (this includes the limit being infinite). The first step in any problem involving improper integrals is to rewrite the integral as a limit. By definition, Z

∞

f (x)dx =

−∞

Z

a

f (x)dx +

−∞

Z

∞

f (x)dx a

if the two integrals on the right side are convergent. If either one is divergent, the original improper integral is divergent. A similar statement holds for improper integrals over infinite discontinuities. Sample problems: (a)

Z

0

∞

x2 √ dx (b) 1 + x3

Z

9 0

√ 3

dx x−4

You can use the Comparison Theorem to determine if an improper integral is convergent or divergent. Comparison Theorem for improper integrals on infinite intervals: If f (x) > g (x) > 0 on [a, ∞), then Z ∞ Z ∞ g(x)dx is convergent. f (x)dx is convergent, then • If a a Z ∞ Z ∞ • If g(x)dx is divergent, then f (x)dx is divergent. a

a

To use this theorem, you need to produce a function to compare your integrand to such that you know the behavior of its improper integral. Common functions to use are x1p : Z ∞ 1 dx is convergent if p > 1 and divergent you may use without justification that xp 1 if p 6 1. When using the comparison theorem, you need to confirm both functions are > 0 and that the inequality holds on the range of integration. Not doing so will result in point deductions. Z ∞ x2 Sample problem: Use the comparison theorem to determine if dx is converx4 + 5 1 gent or divergent. Applications of Integration 1. Volumes by Shells: The volume of an infinitesimally-thin shell having thickness dr is given by 2πrh dr,

4

where r is the radius of the shell and h is its height. Let S be the solid generated when a region E in the xy-plane is rotated about a horizontal or vertical axis. Visualize E as composed of infinitesimally thin strips parallel to the axis of revolution; each strip generates a cylindrical shell the union of which constitutes S. Pick a strip, label its thickness dx or dy—whichever is appropriate. If dx is shell thickness, record a formula for shell volume as function of x; if dy is the thickness, record a formula for shell volume as a function of y . Add up shell volumes using a definite integral to obtain the volume of S . 2. Arc Length: The arc length differential is s s  2  2 dy dx dy ds = 1 + dx = 1 + dx dy .

q

 dy 2 You may view, e.g., 1 + dx dx, as the length of that infinitesimal portion of the curve y = f (x) lying between the vertical lines through x and x + dx. If f is a function with f ′ continuous on [a, b], then the length of the curve y = f (x) on [a, b] is L=

Z

b

Z

ds =

a

b a

p

1 + f ′ (x)2 dx.

If the curve is x = g(y) on [c, d], g ′ continuous on [c, d], then L=

Z

d

ds =

Z dp

1 + g ′ (y )2 dy.

c

c

Sample problem: Set up the integral to find the length of the curve y = 1/x2 from x = 1 to x = 4. 3. Areas of Surfaces of Revolution: The surface area S of a solid of revolution formed by rotating a curve around an axis is given by the following formulas. For rotation about the x-axis: Z ∗ S= 2πyds; ∗

For rotation about the y-axis:

S=

Z

∗

2πxds, ∗

where ds is the arc length differential and the appropriate limits of integration are used. Combined with improper integration techniques, one can determine the surface area of infinite shapes, such as the horn of gabriel. Sample Problem: Find the surface area of the solid of revolution generated by rotating √ y = 3 x, 1 6 y 6 2, about the y-axis.

5

Sequences A sequence is an infinite ordered list of terms, usually written as (an ) or{an }, and assumed ∞ 1 is the list to start at n = 1 unless otherwise stated. For example, the sequence n n=1 1 1 1 1 1, , , , · · · , , · · · . n 2 3 4 You should be able to determine if a sequence is convergent or divergent, increasing or decreasing, bounded above or bounded below or both, and be able to show work justifying these statements. To show a sequence {an } is increasing (respectively decreasing), you might (a) show an < an+1 (resp. an > an+1 ) for all n, or (b) show f ′ (x) > 0 (resp. f ′ (x) < 0), where f (n) = an . Note: Because we generally only care about the “long term” behavior of sequences, to apply, say, the Monotone Sequence Theorem, it is enough to show that a sequence is ∞ eventually increasing or decreasing; that is, for some positive integer N , {an }n=N is increasing or decreasing. To confirm this it is enough to check one of the two statements above for all n > N or all x > N . Therem (Squeeze Theorem). If an 6 bn 6 cn for all n > N and lim an = lim cn = L, then n→∞ n→∞ lim bn = L. n→∞

Therem (Monotonic Sequence Theorem). Every bounded, monotonic sequence is convergent. Note that neither of these theorems can be used to say a sequence diverges. E.g. the sequence {(−1)n /n} is neither increasing or decreasing, but it converges to 0.

6

Series ∞ X P∞ The nth partial sum of the series an is sn = a1 + a2 + · · · an . The series n=1 an n=1

is the limit of the sequence of partial sums if that limit exists, in which case we say that the series P∞ converges. If the limit does not exist, we say that the series diverges. So the series n=1 an • is convergent if lim sn exists, i.e. if the sequence of partial sums is convergent. In n→∞ this case we say the series has sum s = lim sn . n→∞

• is divergent if lim sn does not exist (this includes the limit being ±∞). n→∞ P • is absolutely convergent, P or converges absolutely, if |an | converges. If this is the case, it is also true that an is convergent.

• is conditionally convergent, or converges conditionally, if the series is convergent but NOT absolutely convergent. Examples: P 1 • is convergent. 2n P n • is divergent. n+1

P (−1)n is absolutely convergent. n2 P (−1)n is conditionally convergent. • n There are two kinds of series which have easily computed sums. ∞ X ar n−1 = a + ar + ar2 + ar 3 + · · · , with first • A geometric series is of the form

•

n=1

term a 6= 0 and ratio of consecutive terms r. A geometric series diverges if |r| ≥ 1 and a if |r| < 1. converges with sum 1−r Remark: For r 6= 1, the geometric series test is a consequence of the following nice formula for partial sums of geometric series: n X 1 − rn (r 6= 1). ar i−1 = a(1 + r + r 2 + · · · + r n−1 ) = a 1−r i=1

• A telescoping series is one in which consecutive terms collapse when you write out the partial sums, leaving terms at the beginning and end.You may have  to use partial P 1 P 1 1 − fractions to see this cancellation. E.g. . = n n+1 n(n + 1) In other cases, we simply try to determine if the series is convergent or divergent and use partial sums or other methods to approximate the sum if the series is convergent. 7

Tests for Convergence and Divergence When showing a series is convergent or divergent using one of these tests, you should always clearly state what test you are using and make sure all the conditions for using the test are satisfied. More details are given under each test. 1. Geometric Series Test (GST): A geometric series converges with sum

∞ X n=1

a if |r| < 1. 1−r

2. Test for Divergence (TD): If lim an 6= 0, then n→∞

Remarks:

P

ar n−1 diverges if |r| ≥ 1 and

an diverges.

• The test for divergence can NEVER be used to say a series converges.

• If lim an = 0, we gain no information. For example, lim 1/n = 0, but n→∞

n→∞

diverges.

P

1/n

• You should always think about the test for divergence when looking at a series. You may save yourself quite a bit of work. P 1 converges for p > 1 and diverges for p 6 1. np P P 4. Comparison Test (CT): Suppose that an and bn are series with positive terms.

3. p-Series Test (PST): The series

P bn is convergent and an 6 bn for n > N , then an is also convergent. P P (ii) If bn is divergent and an > bn for n > N , then an is also divergent. (i) If

P

Remarks:

• This test is often used with series that are close to being p-series or geometric series. P • To receive full credit, identify the comparison series bn , verify one of the inequalP ities, andPsay why bn is convergent or divergent. For example, P your explanation can be “ bn is a p-series with p > 1, so it’s convergent,” or “ bn is a geometric series with |r| > 1, so it’s divergent.” P P • Make sure both the series your are testing an and the “comparison series” bn have positive terms (for n > N ). Sample problems: Determine if the series converges or diverges. (a)

∞ X n=1

∞ ∞ X X 1 n2 sin2 n (c) (b) n4 2 + 4n n3 − 8 n=4 n=2

8

P P 5. Limit Comparison Test (LCT): If an and bn are series with positive terms and an = c > 0, c finite, then either both series converge or both series diverge. lim n→∞ bn Remarks: • This test is very useful when the series is close to a p-series or geometric series but obtaining the correct inequality to use with the Comparison test is difficult. P • To receive full credit, you need to clearly identify the comparison series bn and say why it is convergent or divergent. Sample problems: Determine if the series converges or diverges. (a)

∞ X n=1

√

∞

∞

n=2

n=4

X n2 + 1 X n2 n+5 (b) (c) 2n2 + n + 1 n4 − 1 4n

6. Integral Test (IT): Suppose f is a continuous, positive, decreasing function on [N, ∞) ∞ X R∞ and an = f (n) for all integers n > N . Then an and N f (x)dx either both converge n=N

or both diverge. Remarks:

P • Given a series an , to receive full credit you need to define a function f (x) satisfying f (n) = an and then check that f is continuous, positive, and decreasing for n > N . The first two are often clear, but you need to confirm that they are true to use the test. To show f is decreasing, you can show f ′ (x) < 0. • This test is often useful when ln n is involved. You should usually try other tests before using the integral test. 7. Sample problems: Determine if the series converges or diverges. (a)

∞ X n=2

∞ ∞ X X 1 1 n (b) (c) 2 en n ln n n(ln n) n=2 n=1

Know the Remainder Estimate for the Integral Test: Suppose that fP (k) = ak , where f is a continuous, positive, decreasing function for x > n, and that ak is convergent. If Rn = s − sn , then Z Z ∞

∞

f (x) dx 6 Rn 6

f (x) dx.

n

n+1

8. Alternating Series Test (AST): An alternating series is a series of the form P or (−1)n bn . If an alternating series satisfies • bn+1 6 bn for n > N and

9

P (−1)n−1 bn

• lim bn = 0, n→∞

then the alternating series converges. Remarks: • To receive full credit, you need to confirm BOTH of the conditions.

• The alternating series test can NEVER be used to say a series diverges. If lim bn 6= n→∞ P P 0, then the alternating series (−1)n bn or (−1)n−1 bn diverges by the Test for Divergence.

• When you see an alternating series and are asked only about convergence or divergence, not absolute convergence, first try to use the alternating series test. Sample problems: Determine if the series converges or diverges. If it converges, determine if it is conditionally convergent or absolutely convergent. (a)

∞ X (−1)n n=2

n ln n

(b)

∞ X (−1)n−1 n2 n=2

4n2 − 3

You should know the : P P Alternating Series Error Estimate: If s = (−1)n−1 bn (or s = (−1)n bn ) is a sum of an alternating series that satisfies bn+1 6 bn for all n and limn→∞ bn = 0, then |Rn | = |s − sn | 6 bn+1 .

P i−1 1 For example, we know by the AST that the alternating harmonic series ∞ i=1 (−1) i converges. The Alternating Series Error Estimate above allows us to bound above Pn the error that results when we use the n-th partial sum sn = i=1 (−1)i−1 i1 of s = P ∞ 1 i−1 to approximate s: for any n > 1, we have i=1 (−1) i   n ∞ X  X 1 1 1   |s − sn | =  (−1)i−1 − (−1)i−1  6 .  i n+1 i i=1 i=1 9. Ratio Test: The   ratio test checks for absolute convergence. Given a series  an+1  . L = lim  n→∞ an  P (a) If L < 1, then the series an converges absolutely. P (b) If L > 1 or L = ∞, then the series an diverges. (c) If L = 1, the test is inconclusive.

Remarks: 10

P

an , let

• If you get L = 1, you need to use another test. You can check for absolute convergence by using the Comparison test, Limit comparison test, or Integral test P on ...