• $\phantom{a}$ $\begin{array}[t]{ccccc} \dfrac{20x^4}{4x^3}&+&\dfrac{x^3}{4x^3}&+&\dfrac{2x^2}{4x^3} \\ \\ 5x&+&\dfrac{1}{4}&+&\dfrac{1}{2x} \end{array}$
• $\phantom{a}$ $\begin{array}[t]{ccccc} \dfrac{5x^4}{9x}&+&\dfrac{45x^3}{9x}&+&\dfrac{4x^2}{9x} \\ \\ \dfrac{5}{9}x^3&+&5x^2&+&\dfrac{4}{9}x \end{array}$
• $\phantom{a}$ $\begin{array}[t]{ccccc} \dfrac{20n^4}{10n}&+&\dfrac{n^3}{10n}&+&\dfrac{40n^2}{10n} \\ \\ 2n^3&+&\dfrac{n^2}{10}&+&4n \end{array}$
• $\phantom{a}$ $\begin{array}[t]{ccccc} \dfrac{3k^3}{8k}&+&\dfrac{4k^2}{8k}&+&\dfrac{2k}{8k} \\ \\ \dfrac{3}{8}k^2&+&\dfrac{k}{2}&+&\dfrac{1}{4} \end{array}$
• $\phantom{a}$ $\begin{array}[t]{ccccc} \dfrac{12x^4}{6x}&+&\dfrac{24x^3}{6x}&+&\dfrac{3x^2}{6x} \\ \\ 2x^3&+&4x^2&+&\dfrac{x}{2} \end{array}$
• $\phantom{a}$ $\begin{array}[t]{ccccc} \dfrac{5p^4}{4p}&+&\dfrac{16p^3}{4p}&+&\dfrac{16p^2}{4p} \\ \\ \dfrac{5}{4}p^3&+&4p^2&+&4p \end{array}$
• $\phantom{a}$ $\begin{array}[t]{ccccc} \dfrac{10n^4}{10n^2}&+&\dfrac{50n^3}{10n^2}&+&\dfrac{2n^2}{10n^2} \\ \\ n^2&+&5n&+&\dfrac{1}{5} \end{array}$
• $\phantom{a}$ $\begin{array}[t]{ccccc} \dfrac{3m^4}{9m^2}&+&\dfrac{18m^3}{9m^2}&+&\dfrac{27m^2}{9m^2} \\ \\ \dfrac{m^2}{3}&+&2m&+&3 \end{array}$

## Section 6.7. Integration by substitution

• Introduction
• 1.1 Approximating Areas
• 1.2 The Definite Integral
• 1.3 The Fundamental Theorem of Calculus
• 1.4 Integration Formulas and the Net Change Theorem
• 1.5 Substitution
• 1.6 Integrals Involving Exponential and Logarithmic Functions
• 1.7 Integrals Resulting in Inverse Trigonometric Functions
• Key Equations
• Key Concepts

## Review Exercises

• 2.1 Areas between Curves
• 2.2 Determining Volumes by Slicing
• 2.3 Volumes of Revolution: Cylindrical Shells
• 2.4 Arc Length of a Curve and Surface Area
• 2.5 Physical Applications
• 2.6 Moments and Centers of Mass
• 2.7 Integrals, Exponential Functions, and Logarithms
• 2.8 Exponential Growth and Decay
• 2.9 Calculus of the Hyperbolic Functions
• 3.1 Integration by Parts
• 3.2 Trigonometric Integrals
• 3.3 Trigonometric Substitution
• 3.4 Partial Fractions
• 3.5 Other Strategies for Integration
• 3.6 Numerical Integration
• 3.7 Improper Integrals
• 4.1 Basics of Differential Equations
• 4.2 Direction Fields and Numerical Methods
• 4.3 Separable Equations
• 4.4 The Logistic Equation
• 4.5 First-order Linear Equations
• 5.1 Sequences
• 5.2 Infinite Series
• 5.3 The Divergence and Integral Tests
• 5.4 Comparison Tests
• 5.5 Alternating Series
• 5.6 Ratio and Root Tests
• 6.1 Power Series and Functions
• 6.2 Properties of Power Series
• 6.3 Taylor and Maclaurin Series
• 6.4 Working with Taylor Series
• 7.1 Parametric Equations
• 7.2 Calculus of Parametric Curves
• 7.3 Polar Coordinates
• 7.4 Area and Arc Length in Polar Coordinates
• 7.5 Conic Sections
• A | Table of Integrals
• B | Table of Derivatives
• C | Review of Pre-Calculus

The interval of convergence is [ −1 , 1 ) . [ −1 , 1 ) . The radius of convergence is R = 1 . R = 1 .

∑ n = 0 ∞ x n + 3 2 n + 1 ∑ n = 0 ∞ x n + 3 2 n + 1 with interval of convergence ( −2 , 2 ) ( −2 , 2 )

Interval of convergence is ( −2 , 2 ) . ( −2 , 2 ) .

∑ n = 0 ∞ ( −1 + 1 2 n + 1 ) x n . ∑ n = 0 ∞ ( −1 + 1 2 n + 1 ) x n . The interval of convergence is ( −1 , 1 ) . ( −1 , 1 ) .

f ( x ) = 3 3 − x . f ( x ) = 3 3 − x . The interval of convergence is ( −3 , 3 ) . ( −3 , 3 ) .

1 + 2 x + 3 x 2 + 4 x 3 + ⋯ 1 + 2 x + 3 x 2 + 4 x 3 + ⋯

∑ n = 0 ∞ ( n + 2 ) ( n + 1 ) x n ∑ n = 0 ∞ ( n + 2 ) ( n + 1 ) x n

∑ n = 2 ∞ ( −1 ) n x n n ( n − 1 ) ∑ n = 2 ∞ ( −1 ) n x n n ( n − 1 )

p 0 ( x ) = 1 ; p 1 ( x ) = 1 − 2 ( x − 1 ) ; p 2 ( x ) = 1 − 2 ( x − 1 ) + 3 ( x − 1 ) 2 ; p 3 ( x ) = 1 − 2 ( x − 1 ) + 3 ( x − 1 ) 2 − 4 ( x − 1 ) 3 p 0 ( x ) = 1 ; p 1 ( x ) = 1 − 2 ( x − 1 ) ; p 2 ( x ) = 1 − 2 ( x − 1 ) + 3 ( x − 1 ) 2 ; p 3 ( x ) = 1 − 2 ( x − 1 ) + 3 ( x − 1 ) 2 − 4 ( x − 1 ) 3

p 0 ( x ) = 1 ; p 1 ( x ) = 1 − x ; p 2 ( x ) = 1 − x + x 2 ; p 3 ( x ) = 1 − x + x 2 − x 3 ; p n ( x ) = 1 − x + x 2 − x 3 + ⋯ + ( −1 ) n x n = ∑ k = 0 n ( −1 ) k x k p 0 ( x ) = 1 ; p 1 ( x ) = 1 − x ; p 2 ( x ) = 1 − x + x 2 ; p 3 ( x ) = 1 − x + x 2 − x 3 ; p n ( x ) = 1 − x + x 2 − x 3 + ⋯ + ( −1 ) n x n = ∑ k = 0 n ( −1 ) k x k

p 1 ( x ) = 2 + 1 4 ( x − 4 ) ; p 2 ( x ) = 2 + 1 4 ( x − 4 ) − 1 64 ( x − 4 ) 2 ; p 1 ( 6 ) = 2.5 ; p 2 ( 6 ) = 2.4375 ; p 1 ( x ) = 2 + 1 4 ( x − 4 ) ; p 2 ( x ) = 2 + 1 4 ( x − 4 ) − 1 64 ( x − 4 ) 2 ; p 1 ( 6 ) = 2.5 ; p 2 ( 6 ) = 2.4375 ;

| R 1 ( 6 ) | ≤ 0.0625 ; | R 2 ( 6 ) | ≤ 0.015625 | R 1 ( 6 ) | ≤ 0.0625 ; | R 2 ( 6 ) | ≤ 0.015625

∑ n = 0 ∞ ( 2 − x 2 n + 2 ) n . ∑ n = 0 ∞ ( 2 − x 2 n + 2 ) n . The interval of convergence is ( 0 , 4 ) . ( 0 , 4 ) .

∑ n = 0 ∞ ( −1 ) n x 2 n ( 2 n ) ! ∑ n = 0 ∞ ( −1 ) n x 2 n ( 2 n ) !

By the ratio test, the interval of convergence is ( − ∞ , ∞ ) . ( − ∞ , ∞ ) . Since | R n ( x ) | ≤ | x | n + 1 ( n + 1 ) ! , | R n ( x ) | ≤ | x | n + 1 ( n + 1 ) ! , the series converges to cos x cos x for all real x .

∑ n = 0 ∞ ( −1 ) n ( n + 1 ) x n ∑ n = 0 ∞ ( −1 ) n ( n + 1 ) x n

∑ n = 0 ∞ ( −1 ) n x 4 n + 2 ( 2 n + 1 ) ! ∑ n = 0 ∞ ( −1 ) n x 4 n + 2 ( 2 n + 1 ) !

∑ n = 1 ∞ ( −1 ) n n ! 1 · 3 · 5 ⋯ ( 2 n − 1 ) 2 n x n ∑ n = 1 ∞ ( −1 ) n n ! 1 · 3 · 5 ⋯ ( 2 n − 1 ) 2 n x n

y = 5 e 2 x y = 5 e 2 x

y = a ( 1 − x 4 3 · 4 + x 8 3 · 4 · 7 · 8 − ⋯ ) + b ( x − x 5 4 · 5 + x 9 4 · 5 · 8 · 9 − ⋯ ) y = a ( 1 − x 4 3 · 4 + x 8 3 · 4 · 7 · 8 − ⋯ ) + b ( x − x 5 4 · 5 + x 9 4 · 5 · 8 · 9 − ⋯ )

C + ∑ n = 1 ∞ ( −1 ) n + 1 x n n ( 2 n − 2 ) ! C + ∑ n = 1 ∞ ( −1 ) n + 1 x n n ( 2 n − 2 ) ! The definite integral is approximately 0.514 0.514 to within an error of 0.01 . 0.01 .

The estimate is approximately 0.3414 . 0.3414 . This estimate is accurate to within 0.0000094 . 0.0000094 .

## Section 6.1 Exercises

True. If a series converges then its terms tend to zero.

False. It would imply that a n x n → 0 a n x n → 0 for | x | < R . | x | < R . If a n = n n , a n = n n , then a n x n = ( n x ) n a n x n = ( n x ) n does not tend to zero for any x ≠ 0 . x ≠ 0 .

It must converge on ( 0 , 6 ] ( 0 , 6 ] and hence at: a. x = 1 ; x = 1 ; b. x = 2 ; x = 2 ; c. x = 3 ; x = 3 ; d. x = 0 ; x = 0 ; e. x = 5.99 ; x = 5.99 ; and f. x = 0.000001 . x = 0.000001 .

| a n + 1 2 n + 1 x n + 1 a n 2 n x n | = 2 | x | | a n + 1 a n | → 2 | x | | a n + 1 2 n + 1 x n + 1 a n 2 n x n | = 2 | x | | a n + 1 a n | → 2 | x | so R = 1 2 R = 1 2

| a n + 1 ( π e ) n + 1 x n + 1 a n ( π e ) n x n | = π | x | e | a n + 1 a n | → π | x | e | a n + 1 ( π e ) n + 1 x n + 1 a n ( π e ) n x n | = π | x | e | a n + 1 a n | → π | x | e so R = e π R = e π

| a n + 1 ( −1 ) n + 1 x 2 n + 2 a n ( −1 ) n x 2 n | = | x 2 | | a n + 1 a n | → | x 2 | | a n + 1 ( −1 ) n + 1 x 2 n + 2 a n ( −1 ) n x 2 n | = | x 2 | | a n + 1 a n | → | x 2 | so R = 1 R = 1

a n = 2 n n a n = 2 n n so a n + 1 x a n → 2 x . a n + 1 x a n → 2 x . so R = 1 2 . R = 1 2 . When x = 1 2 x = 1 2 the series is harmonic and diverges. When x = − 1 2 x = − 1 2 the series is alternating harmonic and converges. The interval of convergence is I = [ − 1 2 , 1 2 ) . I = [ − 1 2 , 1 2 ) .

a n = n 2 n a n = n 2 n so a n + 1 x a n → x 2 a n + 1 x a n → x 2 so R = 2 . R = 2 . When x = ± 2 x = ± 2 the series diverges by the divergence test. The interval of convergence is I = ( −2 , 2 ) . I = ( −2 , 2 ) .

a n = n 2 2 n a n = n 2 2 n so R = 2 . R = 2 . When x = ± 2 x = ± 2 the series diverges by the divergence test. The interval of convergence is I = ( −2 , 2 ) . I = ( −2 , 2 ) .

a k = π k k π a k = π k k π so R = 1 π . R = 1 π . When x = ± 1 π x = ± 1 π the series is an absolutely convergent p -series. The interval of convergence is I = [ − 1 π , 1 π ] . I = [ − 1 π , 1 π ] .

a n = 10 n n ! , a n + 1 x a n = 10 x n + 1 → 0 < 1 a n = 10 n n ! , a n + 1 x a n = 10 x n + 1 → 0 < 1 so the series converges for all x by the ratio test and I = ( − ∞ , ∞ ) . I = ( − ∞ , ∞ ) .

a k = ( k ! ) 2 ( 2 k ) ! a k = ( k ! ) 2 ( 2 k ) ! so a k + 1 a k = ( k + 1 ) 2 ( 2 k + 2 ) ( 2 k + 1 ) → 1 4 a k + 1 a k = ( k + 1 ) 2 ( 2 k + 2 ) ( 2 k + 1 ) → 1 4 so R = 4 R = 4

a k = k ! 1 · 3 · 5 ⋯ ( 2 k − 1 ) a k = k ! 1 · 3 · 5 ⋯ ( 2 k − 1 ) so a k + 1 a k = k + 1 2 k + 1 → 1 2 a k + 1 a k = k + 1 2 k + 1 → 1 2 so R = 2 R = 2

a n = 1 ( 2 n n ) a n = 1 ( 2 n n ) so a n + 1 a n = ( ( n + 1 ) ! ) 2 ( 2 n + 2 ) ! 2 n ! ( n ! ) 2 = ( n + 1 ) 2 ( 2 n + 2 ) ( 2 n + 1 ) → 1 4 a n + 1 a n = ( ( n + 1 ) ! ) 2 ( 2 n + 2 ) ! 2 n ! ( n ! ) 2 = ( n + 1 ) 2 ( 2 n + 2 ) ( 2 n + 1 ) → 1 4 so R = 4 R = 4

a n + 1 a n = ( n + 1 ) 3 ( 3 n + 3 ) ( 3 n + 2 ) ( 3 n + 1 ) → 1 27 a n + 1 a n = ( n + 1 ) 3 ( 3 n + 3 ) ( 3 n + 2 ) ( 3 n + 1 ) → 1 27 so R = 27 R = 27

a n = n ! n n a n = n ! n n so a n + 1 a n = ( n + 1 ) ! n ! n n ( n + 1 ) n + 1 = ( n n + 1 ) n → 1 e a n + 1 a n = ( n + 1 ) ! n ! n n ( n + 1 ) n + 1 = ( n n + 1 ) n → 1 e so R = e R = e

f ( x ) = ∑ n = 0 ∞ ( 1 − x ) n f ( x ) = ∑ n = 0 ∞ ( 1 − x ) n on I = ( 0 , 2 ) I = ( 0 , 2 )

∑ n = 0 ∞ x 2 n + 1 ∑ n = 0 ∞ x 2 n + 1 on I = ( −1 , 1 ) I = ( −1 , 1 )

∑ n = 0 ∞ ( −1 ) n x 2 n + 2 ∑ n = 0 ∞ ( −1 ) n x 2 n + 2 on I = ( −1 , 1 ) I = ( −1 , 1 )

∑ n = 0 ∞ 2 n x n ∑ n = 0 ∞ 2 n x n on ( − 1 2 , 1 2 ) ( − 1 2 , 1 2 )

∑ n = 0 ∞ 4 n x 2 n + 2 ∑ n = 0 ∞ 4 n x 2 n + 2 on ( − 1 2 , 1 2 ) ( − 1 2 , 1 2 )

| a n x n | 1 / n = | a n | 1 / n | x | → | x | r | a n x n | 1 / n = | a n | 1 / n | x | → | x | r as n → ∞ n → ∞ and | x | r < 1 | x | r < 1 when | x | < 1 r . | x | < 1 r . Therefore, ∑ n = 1 ∞ a n x n ∑ n = 1 ∞ a n x n converges when | x | < 1 r | x | < 1 r by the n th root test.

a k = ( k − 1 2 k + 3 ) k a k = ( k − 1 2 k + 3 ) k so ( a k ) 1 / k → 1 2 < 1 ( a k ) 1 / k → 1 2 < 1 so R = 2 R = 2

a n = ( n 1 / n − 1 ) n a n = ( n 1 / n − 1 ) n so ( a n ) 1 / n → 0 ( a n ) 1 / n → 0 so R = ∞ R = ∞

We can rewrite p ( x ) = ∑ n = 0 ∞ a 2 n + 1 x 2 n + 1 p ( x ) = ∑ n = 0 ∞ a 2 n + 1 x 2 n + 1 and p ( x ) = p ( − x ) p ( x ) = p ( − x ) since only even powers of x x remain, p ( x ) p ( x ) is an even function, for which, by definition p ( x ) = p ( –x ) p ( x ) = p ( –x ) .

If x ∈ [ 0 , 1 ] , x ∈ [ 0 , 1 ] , then y = 2 x − 1 ∈ [ −1 , 1 ] y = 2 x − 1 ∈ [ −1 , 1 ] so p ( 2 x − 1 ) = p ( y ) = ∑ n = 0 ∞ a n y n p ( 2 x − 1 ) = p ( y ) = ∑ n = 0 ∞ a n y n converges.

Converges on ( −1 , 1 ) ( −1 , 1 ) by the ratio test

Consider the series ∑ b k x k ∑ b k x k where b k = a k b k = a k if k = n 2 k = n 2 and b k = 0 b k = 0 otherwise. Then b k ≤ a k b k ≤ a k and so the series converges on ( −1 , 1 ) ( −1 , 1 ) by the comparison test.

The approximation is more accurate near x = −1 . x = −1 . The partial sums follow 1 1 − x 1 1 − x more closely as N increases but are never accurate near x = 1 x = 1 since the series diverges there.

The approximation appears to stabilize quickly near both x = ± 1 . x = ± 1 .

The polynomial curves have roots close to those of sin x sin x up to their degree and then the polynomials diverge from sin x . sin x .

## Section 6.2 Exercises

1 2 ( f ( x ) + g ( x ) ) = ∑ n = 0 ∞ x 2 n ( 2 n ) ! 1 2 ( f ( x ) + g ( x ) ) = ∑ n = 0 ∞ x 2 n ( 2 n ) ! and 1 2 ( f ( x ) − g ( x ) ) = ∑ n = 0 ∞ x 2 n + 1 ( 2 n + 1 ) ! . 1 2 ( f ( x ) − g ( x ) ) = ∑ n = 0 ∞ x 2 n + 1 ( 2 n + 1 ) ! .

4 ( x − 3 ) ( x + 1 ) = 1 x − 3 − 1 x + 1 = − 1 3 ( 1 − x 3 ) − 1 1 − ( − x ) = − 1 3 ∑ n = 0 ∞ ( x 3 ) n − ∑ n = 0 ∞ ( −1 ) n x n = ∑ n = 0 ∞ ( ( −1 ) n + 1 − 1 3 n + 1 ) x n 4 ( x − 3 ) ( x + 1 ) = 1 x − 3 − 1 x + 1 = − 1 3 ( 1 − x 3 ) − 1 1 − ( − x ) = − 1 3 ∑ n = 0 ∞ ( x 3 ) n − ∑ n = 0 ∞ ( −1 ) n x n = ∑ n = 0 ∞ ( ( −1 ) n + 1 − 1 3 n + 1 ) x n

5 ( x 2 + 4 ) ( x 2 − 1 ) = 1 x 2 − 1 − 1 4 1 1 + ( x 2 ) 2 = − ∑ n = 0 ∞ x 2 n − 1 4 ∑ n = 0 ∞ ( −1 ) n ( x 2 ) 2 n = ∑ n = 0 ∞ ( - 1 + ( −1 ) · n + 1 1 2 n + 2 ) x 2 n 5 ( x 2 + 4 ) ( x 2 − 1 ) = 1 x 2 − 1 − 1 4 1 1 + ( x 2 ) 2 = − ∑ n = 0 ∞ x 2 n − 1 4 ∑ n = 0 ∞ ( −1 ) n ( x 2 ) 2 n = ∑ n = 0 ∞ ( - 1 + ( −1 ) · n + 1 1 2 n + 2 ) x 2 n

1 x ∑ n = 0 ∞ 1 x n = 1 x 1 1 − 1 x = 1 x − 1 1 x ∑ n = 0 ∞ 1 x n = 1 x 1 1 − 1 x = 1 x − 1

1 x − 3 1 1 − 1 ( x − 3 ) 2 = x − 3 ( x − 3 ) 2 − 1 1 x − 3 1 1 − 1 ( x − 3 ) 2 = x − 3 ( x − 3 ) 2 − 1

P = P 1 + ⋯ + P 20 P = P 1 + ⋯ + P 20 where P k = 10,000 1 ( 1 + r ) k . P k = 10,000 1 ( 1 + r ) k . Then P = 10,000 ∑ k = 1 20 1 ( 1 + r ) k = 10,000 1 − ( 1 + r ) −20 r . P = 10,000 ∑ k = 1 20 1 ( 1 + r ) k = 10,000 1 − ( 1 + r ) −20 r . When r = 0.03 , P ≈ 10,000 × 14.8775 = 148,775 . r = 0.03 , P ≈ 10,000 × 14.8775 = 148,775 . When r = 0.05 , P ≈ 10,000 × 12.4622 = 124,622 . r = 0.05 , P ≈ 10,000 × 12.4622 = 124,622 . When r = 0.07 , P ≈ 105,940 . r = 0.07 , P ≈ 105,940 .

In general, P = C ( 1 − ( 1 + r ) − N ) r P = C ( 1 − ( 1 + r ) − N ) r for N years of payouts, or C = P r 1 − ( 1 + r ) − N . C = P r 1 − ( 1 + r ) − N . For N = 20 N = 20 and P = 100,000 , P = 100,000 , one has C = 6721.57 C = 6721.57 when r = 0.03 ; C = 8024.26 r = 0.03 ; C = 8024.26 when r = 0.05 ; r = 0.05 ; and C ≈ 9439.29 C ≈ 9439.29 when r = 0.07 . r = 0.07 .

In general, P = C r . P = C r . Thus, r = C P = 5 × 10 4 10 6 = 0.05 . r = C P = 5 × 10 4 10 6 = 0.05 .

( x + x 2 − x 3 ) ( 1 + x 3 + x 6 + ⋯ ) = x + x 2 − x 3 1 − x 3 ( x + x 2 − x 3 ) ( 1 + x 3 + x 6 + ⋯ ) = x + x 2 − x 3 1 − x 3

( x − x 2 − x 3 ) ( 1 + x 3 + x 6 + ⋯ ) = x − x 2 − x 3 1 − x 3 ( x − x 2 − x 3 ) ( 1 + x 3 + x 6 + ⋯ ) = x − x 2 − x 3 1 − x 3

a n = 2 , b n = n a n = 2 , b n = n so c n = ∑ k = 0 n b k a n − k = 2 ∑ k = 0 n k = ( n ) ( n + 1 ) c n = ∑ k = 0 n b k a n − k = 2 ∑ k = 0 n k = ( n ) ( n + 1 ) and f ( x ) g ( x ) = ∑ n = 1 ∞ n ( n + 1 ) x n f ( x ) g ( x ) = ∑ n = 1 ∞ n ( n + 1 ) x n

a n = b n = 2 − n a n = b n = 2 − n so c n = ∑ k = 1 n b k a n − k = 2 − n ∑ k = 1 n 1 = n 2 n c n = ∑ k = 1 n b k a n − k = 2 − n ∑ k = 1 n 1 = n 2 n and f ( x ) g ( x ) = ∑ n = 1 ∞ n ( x 2 ) n f ( x ) g ( x ) = ∑ n = 1 ∞ n ( x 2 ) n

The derivative of f f is − 1 ( 1 + x ) 2 = − ∑ n = 0 ∞ ( −1 ) n ( n + 1 ) x n . − 1 ( 1 + x ) 2 = − ∑ n = 0 ∞ ( −1 ) n ( n + 1 ) x n .

The indefinite integral of f f is - 1 1 + x 2 = ∑ n = 0 ∞ ( −1 ) n x 2 n . - 1 1 + x 2 = ∑ n = 0 ∞ ( −1 ) n x 2 n .

f ( x ) = ∑ n = 0 ∞ x n = 1 1 − x ; f ′ ( 1 2 ) = ∑ n = 1 ∞ n 2 n − 1 = d d x ( 1 − x ) −1 | x = 1 / 2 = 1 ( 1 − x ) 2 | x = 1 / 2 = 4 f ( x ) = ∑ n = 0 ∞ x n = 1 1 − x ; f ′ ( 1 2 ) = ∑ n = 1 ∞ n 2 n − 1 = d d x ( 1 − x ) −1 | x = 1 / 2 = 1 ( 1 − x ) 2 | x = 1 / 2 = 4 so ∑ n = 1 ∞ n 2 n = 2 . ∑ n = 1 ∞ n 2 n = 2 .

f ( x ) = ∑ n = 0 ∞ x n = 1 1 − x ; f ″ ( 1 2 ) = ∑ n = 2 ∞ n ( n − 1 ) 2 n − 2 = d 2 d x 2 ( 1 − x ) −1 | x = 1 / 2 = 2 ( 1 − x ) 3 | x = 1 / 2 = 16 f ( x ) = ∑ n = 0 ∞ x n = 1 1 − x ; f ″ ( 1 2 ) = ∑ n = 2 ∞ n ( n − 1 ) 2 n − 2 = d 2 d x 2 ( 1 − x ) −1 | x = 1 / 2 = 2 ( 1 − x ) 3 | x = 1 / 2 = 16 so ∑ n = 2 ∞ n ( n − 1 ) 2 n = 4 . ∑ n = 2 ∞ n ( n − 1 ) 2 n = 4 .

∫ ∑ ( 1 − x ) n d x = ∫ ∑ ( −1 ) n ( x − 1 ) n d x = ∑ ( −1 ) n ( x − 1 ) n + 1 n + 1 ∫ ∑ ( 1 − x ) n d x = ∫ ∑ ( −1 ) n ( x − 1 ) n d x = ∑ ( −1 ) n ( x − 1 ) n + 1 n + 1

− ∫ t = 0 x 2 1 1 − t d t = − ∑ n = 0 ∞ ∫ 0 x 2 t n d x − ∑ n = 0 ∞ x 2 ( n + 1 ) n + 1 = − ∑ n = 1 ∞ x 2 n n − ∫ t = 0 x 2 1 1 − t d t = − ∑ n = 0 ∞ ∫ 0 x 2 t n d x − ∑ n = 0 ∞ x 2 ( n + 1 ) n + 1 = − ∑ n = 1 ∞ x 2 n n

∫ 0 x 2 d t 1 + t 2 = ∑ n = 0 ∞ ( −1 ) n ∫ 0 x 2 t 2 n d t = ∑ n = 0 ∞ ( −1 ) n t 2 n + 1 2 n + 1 | t = 0 x 2 = ∑ n = 0 ∞ ( −1 ) n x 4 n + 2 2 n + 1 ∫ 0 x 2 d t 1 + t 2 = ∑ n = 0 ∞ ( −1 ) n ∫ 0 x 2 t 2 n d t = ∑ n = 0 ∞ ( −1 ) n t 2 n + 1 2 n + 1 | t = 0 x 2 = ∑ n = 0 ∞ ( −1 ) n x 4 n + 2 2 n + 1

Term-by-term integration gives ∫ 0 x ln t d t = ∑ n = 1 ∞ ( −1 ) n − 1 ( x − 1 ) n + 1 n ( n + 1 ) = ∑ n = 1 ∞ ( −1 ) n − 1 ( 1 n − 1 n + 1 ) ( x − 1 ) n + 1 = ( x − 1 ) ln x + ∑ n = 2 ∞ ( −1 ) n ( x − 1 ) n n = x ln x − x . ∫ 0 x ln t d t = ∑ n = 1 ∞ ( −1 ) n − 1 ( x − 1 ) n + 1 n ( n + 1 ) = ∑ n = 1 ∞ ( −1 ) n − 1 ( 1 n − 1 n + 1 ) ( x − 1 ) n + 1 = ( x − 1 ) ln x + ∑ n = 2 ∞ ( −1 ) n ( x − 1 ) n n = x ln x − x .

We have ln ( 1 − x ) = − ∑ n = 1 ∞ x n n ln ( 1 − x ) = − ∑ n = 1 ∞ x n n so ln ( 1 + x ) = ∑ n = 1 ∞ ( −1 ) n − 1 x n n . ln ( 1 + x ) = ∑ n = 1 ∞ ( −1 ) n − 1 x n n . Thus, ln ( 1 + x 1 − x ) = ∑ n = 1 ∞ ( 1 + ( −1 ) n − 1 ) x n n = 2 ∑ n = 1 ∞ x 2 n − 1 2 n − 1 . ln ( 1 + x 1 − x ) = ∑ n = 1 ∞ ( 1 + ( −1 ) n − 1 ) x n n = 2 ∑ n = 1 ∞ x 2 n − 1 2 n − 1 . When x = 1 3 x = 1 3 we obtain ln ( 2 ) = 2 ∑ n = 1 ∞ 1 3 2 n − 1 ( 2 n − 1 ) . ln ( 2 ) = 2 ∑ n = 1 ∞ 1 3 2 n − 1 ( 2 n − 1 ) . We have 2 ∑ n = 1 3 1 3 2 n − 1 ( 2 n − 1 ) = 0.69300 … , 2 ∑ n = 1 3 1 3 2 n − 1 ( 2 n − 1 ) = 0.69300 … , while 2 ∑ n = 1 4 1 3 2 n − 1 ( 2 n − 1 ) = 0.69313 … 2 ∑ n = 1 4 1 3 2 n − 1 ( 2 n − 1 ) = 0.69313 … and ln ( 2 ) = 0.69314 … ; ln ( 2 ) = 0.69314 … ; therefore, N = 4 . N = 4 .

∑ k = 1 ∞ x k k = − ln ( 1 − x ) ∑ k = 1 ∞ x k k = − ln ( 1 − x ) so ∑ k = 1 ∞ x 3 k 6 k = − 1 6 ln ( 1 − x 3 ) . ∑ k = 1 ∞ x 3 k 6 k = − 1 6 ln ( 1 − x 3 ) . The radius of convergence is equal to 1 by the ratio test.

If y = 2 − x , y = 2 − x , then ∑ k = 1 ∞ y k = y 1 − y = 2 − x 1 − 2 − x = 1 2 x − 1 . ∑ k = 1 ∞ y k = y 1 − y = 2 − x 1 − 2 − x = 1 2 x − 1 . If a k = 2 − k x , a k = 2 − k x , then a k + 1 a k = 2 − x < 1 a k + 1 a k = 2 − x < 1 when x > 0 . x > 0 . So the series converges for all x > 0 . x > 0 .

The solid curve is S 5 . The dashed curve is S 2 , dotted is S 3 , and dash-dotted is S 4

When x = − 1 2 , − ln ( 2 ) = ln ( 1 2 ) = − ∑ n = 1 ∞ 1 n 2 n . x = − 1 2 , − ln ( 2 ) = ln ( 1 2 ) = − ∑ n = 1 ∞ 1 n 2 n . Since ∑ n = 11 ∞ 1 n 2 n < ∑ n = 11 ∞ 1 2 n = 1 2 10 , ∑ n = 11 ∞ 1 n 2 n < ∑ n = 11 ∞ 1 2 n = 1 2 10 , one has ∑ n = 1 10 1 n 2 n = 0.69306 … ∑ n = 1 10 1 n 2 n = 0.69306 … whereas ln ( 2 ) = 0.69314 … ; ln ( 2 ) = 0.69314 … ; therefore, N = 10 . N = 10 .

6 S N ( 1 3 ) = 2 3 ∑ n = 0 N ( −1 ) n 1 3 n ( 2 n + 1 ) . 6 S N ( 1 3 ) = 2 3 ∑ n = 0 N ( −1 ) n 1 3 n ( 2 n + 1 ) . One has π − 6 S 4 ( 1 3 ) = 0.00101 … π − 6 S 4 ( 1 3 ) = 0.00101 … and π − 6 S 5 ( 1 3 ) = 0.00028 … π − 6 S 5 ( 1 3 ) = 0.00028 … so N = 5 N = 5 is the smallest partial sum with accuracy to within 0.001. Also, π − 6 S 7 ( 1 3 ) = 0.00002 … π − 6 S 7 ( 1 3 ) = 0.00002 … while π − 6 S 8 ( 1 3 ) = −0.000007 … π − 6 S 8 ( 1 3 ) = −0.000007 … so N = 8 N = 8 is the smallest N to give accuracy to within 0.00001.

## Section 6.3 Exercises

f ( −1 ) = 1 ; f ′ ( −1 ) = −1 ; f ″ ( −1 ) = 2 ; f ( x ) = 1 − ( x + 1 ) + ( x + 1 ) 2 f ( −1 ) = 1 ; f ′ ( −1 ) = −1 ; f ″ ( −1 ) = 2 ; f ( x ) = 1 − ( x + 1 ) + ( x + 1 ) 2

f ′ ( x ) = 2 cos ( 2 x ) ; f ″ ( x ) = −4 sin ( 2 x ) ; p 2 ( x ) = −2 ( x − π 2 ) f ′ ( x ) = 2 cos ( 2 x ) ; f ″ ( x ) = −4 sin ( 2 x ) ; p 2 ( x ) = −2 ( x − π 2 )

f ′ ( x ) = 1 x ; f ″ ( x ) = − 1 x 2 ; p 2 ( x ) = 0 + ( x − 1 ) − 1 2 ( x − 1 ) 2 f ′ ( x ) = 1 x ; f ″ ( x ) = − 1 x 2 ; p 2 ( x ) = 0 + ( x − 1 ) − 1 2 ( x − 1 ) 2

p 2 ( x ) = e + e ( x − 1 ) + e 2 ( x − 1 ) 2 p 2 ( x ) = e + e ( x − 1 ) + e 2 ( x − 1 ) 2

d 2 d x 2 x 1 / 3 = − 2 9 x 5 / 3 ≥ −0.00092 … d 2 d x 2 x 1 / 3 = − 2 9 x 5 / 3 ≥ −0.00092 … when x ≥ 28 x ≥ 28 so the remainder estimate applies to the linear approximation x 1 / 3 ≈ p 1 ( 27 ) = 3 + x − 27 27 , x 1 / 3 ≈ p 1 ( 27 ) = 3 + x − 27 27 , which gives ( 28 ) 1 / 3 ≈ 3 + 1 27 = 3. 037 ¯ , ( 28 ) 1 / 3 ≈ 3 + 1 27 = 3. 037 ¯ , while ( 28 ) 1 / 3 ≈ 3.03658 . ( 28 ) 1 / 3 ≈ 3.03658 .

Using the estimate 2 10 10 ! < 0.000283 2 10 10 ! < 0.000283 we can use the Taylor expansion of order 9 to estimate e x at x = 2 . x = 2 . as e 2 ≈ p 9 ( 2 ) = 1 + 2 + 2 2 2 + 2 3 6 + ⋯ + 2 9 9 ! = 7.3887 … e 2 ≈ p 9 ( 2 ) = 1 + 2 + 2 2 2 + 2 3 6 + ⋯ + 2 9 9 ! = 7.3887 … whereas e 2 ≈ 7.3891 . e 2 ≈ 7.3891 .

Since d n d x n ( ln x ) = ( −1 ) n − 1 ( n − 1 ) ! x n , R 1000 ≈ 1 1001 . d n d x n ( ln x ) = ( −1 ) n − 1 ( n − 1 ) ! x n , R 1000 ≈ 1 1001 . One has p 1000 ( 1 ) = ∑ n = 1 1000 ( −1 ) n − 1 n ≈ 0.6936 p 1000 ( 1 ) = ∑ n = 1 1000 ( −1 ) n − 1 n ≈ 0.6936 whereas ln ( 2 ) ≈ 0.6931 ⋯ . ln ( 2 ) ≈ 0.6931 ⋯ .

∫ 0 1 ( 1 − x 2 + x 4 2 − x 6 6 + x 8 24 − x 10 120 + x 12 720 ) d x ∫ 0 1 ( 1 − x 2 + x 4 2 − x 6 6 + x 8 24 − x 10 120 + x 12 720 ) d x

= 1 − 1 3 3 + 1 5 10 − 1 7 42 + 1 9 9 · 24 − 1 11 120 · 11 + 1 13 720 · 13 ≈ 0.74683 = 1 − 1 3 3 + 1 5 10 − 1 7 42 + 1 9 9 · 24 − 1 11 120 · 11 + 1 13 720 · 13 ≈ 0.74683 whereas ∫ 0 1 e − x 2 d x ≈ 0.74682 . ∫ 0 1 e − x 2 d x ≈ 0.74682 .

Since f ( n + 1 ) ( z ) f ( n + 1 ) ( z ) is sin z sin z or cos z , cos z , we have M = 1 . M = 1 . Since | x − 0 | ≤ π 2 , | x − 0 | ≤ π 2 , we seek the smallest n such that π n + 1 2 n + 1 ( n + 1 ) ! ≤ 0.001 . π n + 1 2 n + 1 ( n + 1 ) ! ≤ 0.001 . The smallest such value is n = 7 . n = 7 . The remainder estimate is R 7 ≤ 0.00092 . R 7 ≤ 0.00092 .

Since f ( n + 1 ) ( z ) = ± e − z f ( n + 1 ) ( z ) = ± e − z one has M = e 3 . M = e 3 . Since | x − 0 | ≤ 3 , | x − 0 | ≤ 3 , one seeks the smallest n such that 3 n + 1 e 3 ( n + 1 ) ! ≤ 0.001 . 3 n + 1 e 3 ( n + 1 ) ! ≤ 0.001 . The smallest such value is n = 14 . n = 14 . The remainder estimate is R 14 ≤ 0.000220 . R 14 ≤ 0.000220 .

Since sin x sin x is increasing for small x and since si n ″ x = − sin x , si n ″ x = − sin x , the estimate applies whenever R 2 sin ( R ) ≤ 0.2 , R 2 sin ( R ) ≤ 0.2 , which applies up to R = 0.596 . R = 0.596 .

Since the second derivative of cos x cos x is − cos x − cos x and since cos x cos x is decreasing away from x = 0 , x = 0 , the estimate applies when R 2 cos R ≤ 0.2 R 2 cos R ≤ 0.2 or R ≤ 0.447 . R ≤ 0.447 .

( x + 1 ) 3 − 2 ( x + 1 ) 2 + 2 ( x + 1 ) ( x + 1 ) 3 − 2 ( x + 1 ) 2 + 2 ( x + 1 )

Values of derivatives are the same as for x = 0 x = 0 so cos x = ∑ n = 0 ∞ ( −1 ) n ( x − 2 π ) 2 n ( 2 n ) ! cos x = ∑ n = 0 ∞ ( −1 ) n ( x − 2 π ) 2 n ( 2 n ) !

cos ( π 2 ) = 0 , − sin ( π 2 ) = −1 cos ( π 2 ) = 0 , − sin ( π 2 ) = −1 so cos x = ∑ n = 0 ∞ ( −1 ) n + 1 ( x − π 2 ) 2 n + 1 ( 2 n + 1 ) ! , cos x = ∑ n = 0 ∞ ( −1 ) n + 1 ( x − π 2 ) 2 n + 1 ( 2 n + 1 ) ! , which is also − cos ( x − π 2 ) . − cos ( x − π 2 ) .

The derivatives are f ( n ) ( 1 ) = e f ( n ) ( 1 ) = e so e x = e ∑ n = 0 ∞ ( x − 1 ) n n ! . e x = e ∑ n = 0 ∞ ( x − 1 ) n n ! .

1 ( x − 1 ) 3 = − ( 1 2 ) d 2 d x 2 1 1 − x = − ∑ n = 0 ∞ ( ( n + 2 ) ( n + 1 ) x n 2 ) 1 ( x − 1 ) 3 = − ( 1 2 ) d 2 d x 2 1 1 − x = − ∑ n = 0 ∞ ( ( n + 2 ) ( n + 1 ) x n 2 )

2 − x = 1 − ( x − 1 ) 2 − x = 1 − ( x − 1 )

( ( x − 1 ) − 1 ) 2 = ( x − 1 ) 2 − 2 ( x − 1 ) + 1 ( ( x − 1 ) − 1 ) 2 = ( x − 1 ) 2 − 2 ( x − 1 ) + 1

1 1 − ( 1 − x ) = ∑ n = 0 ∞ ( −1 ) n ( x − 1 ) n 1 1 − ( 1 − x ) = ∑ n = 0 ∞ ( −1 ) n ( x − 1 ) n

x ∑ n = 0 ∞ 2 n ( 1 − x ) 2 n = ∑ n = 0 ∞ 2 n ( x − 1 ) 2 n + 1 + ∑ n = 0 ∞ 2 n ( x − 1 ) 2 n x ∑ n = 0 ∞ 2 n ( 1 − x ) 2 n = ∑ n = 0 ∞ 2 n ( x − 1 ) 2 n + 1 + ∑ n = 0 ∞ 2 n ( x − 1 ) 2 n

e 2 x = e 2 ( x − 1 ) + 2 = e 2 ∑ n = 0 ∞ 2 n ( x − 1 ) n n ! e 2 x = e 2 ( x − 1 ) + 2 = e 2 ∑ n = 0 ∞ 2 n ( x − 1 ) n n !

x = e 2 ; S 10 = 34,913 4725 ≈ 7.3889947 x = e 2 ; S 10 = 34,913 4725 ≈ 7.3889947

sin ( 2 π ) = 0 ; S 10 = 8.27 × 10 −5 sin ( 2 π ) = 0 ; S 10 = 8.27 × 10 −5

The difference is small on the interior of the interval but approaches 1 1 near the endpoints. The remainder estimate is | R 4 | = π 5 120 ≈ 2.552 . | R 4 | = π 5 120 ≈ 2.552 .

The difference is on the order of 10 −4 10 −4 on [ −1 , 1 ] [ −1 , 1 ] while the Taylor approximation error is around 0.1 0.1 near ± 1 . ± 1 . The top curve is a plot of tan 2 x − ( S 5 ( x ) C 4 ( x ) ) 2 tan 2 x − ( S 5 ( x ) C 4 ( x ) ) 2 and the lower dashed plot shows t 2 − ( S 5 C 4 ) 2 . t 2 − ( S 5 C 4 ) 2 .

a. Answers will vary. b. The following are the x n x n values after 10 10 iterations of Newton’s method to approximation a root of p N ( x ) − 2 = 0 : p N ( x ) − 2 = 0 : for N = 4 , x = 0.6939... ; N = 4 , x = 0.6939... ; for N = 5 , x = 0.6932... ; N = 5 , x = 0.6932... ; for N = 6 , x = 0.69315... ; . N = 6 , x = 0.69315... ; . ( Note: ln ( 2 ) = 0.69314 ... ) ln ( 2 ) = 0.69314 ... ) c. Answers will vary.

ln ( 1 − x 2 ) x 2 → − 1 ln ( 1 − x 2 ) x 2 → − 1

cos ( x ) − 1 2 x ≈ ( 1 − x 2 + x 2 4 ! − ⋯ ) − 1 2 x → − 1 4 cos ( x ) − 1 2 x ≈ ( 1 − x 2 + x 2 4 ! − ⋯ ) − 1 2 x → − 1 4

## Section 6.4 Exercises

( 1 + x 2 ) −1 / 3 = ∑ n = 0 ∞ ( − 1 3 n ) x 2 n ( 1 + x 2 ) −1 / 3 = ∑ n = 0 ∞ ( − 1 3 n ) x 2 n

( 1 − 2 x ) 2 / 3 = ∑ n = 0 ∞ ( −1 ) n 2 n ( 2 3 n ) x n ( 1 − 2 x ) 2 / 3 = ∑ n = 0 ∞ ( −1 ) n 2 n ( 2 3 n ) x n

2 + x 2 = ∑ n = 0 ∞ 2 ( 1 / 2 ) − n ( 1 2 n ) x 2 n ; ( | x 2 | < 2 ) 2 + x 2 = ∑ n = 0 ∞ 2 ( 1 / 2 ) − n ( 1 2 n ) x 2 n ; ( | x 2 | < 2 )

2 x − x 2 = 1 − ( x − 1 ) 2 2 x − x 2 = 1 − ( x − 1 ) 2 so 2 x − x 2 = ∑ n = 0 ∞ ( −1 ) n ( 1 2 n ) ( x − 1 ) 2 n 2 x − x 2 = ∑ n = 0 ∞ ( −1 ) n ( 1 2 n ) ( x − 1 ) 2 n

x = 2 1 + x − 4 4 x = 2 1 + x − 4 4 so x = ∑ n = 0 ∞ 2 1 − 2 n ( 1 2 n ) ( x − 4 ) n x = ∑ n = 0 ∞ 2 1 − 2 n ( 1 2 n ) ( x − 4 ) n

x = ∑ n = 0 ∞ 3 1 − 3 n ( 1 2 n ) ( x − 9 ) n x = ∑ n = 0 ∞ 3 1 − 3 n ( 1 2 n ) ( x − 9 ) n

10 ( 1 + x 1000 ) 1 / 3 = ∑ n = 0 ∞ 10 1 − 3 n ( 1 3 n ) x n . 10 ( 1 + x 1000 ) 1 / 3 = ∑ n = 0 ∞ 10 1 − 3 n ( 1 3 n ) x n . Using, for example, a fourth-degree estimate at x = 1 x = 1 gives ( 1001 ) 1 / 3 ≈ 10 ( 1 + ( 1 3 1 ) 10 −3 + ( 1 3 2 ) 10 −6 + ( 1 3 3 ) 10 −9 + ( 1 3 4 ) 10 −12 ) = 10 ( 1 + 1 3.10 3 − 1 9.10 6 + 5 81.10 9 − 10 243.10 12 ) = 10.00333222... ( 1001 ) 1 / 3 ≈ 10 ( 1 + ( 1 3 1 ) 10 −3 + ( 1 3 2 ) 10 −6 + ( 1 3 3 ) 10 −9 + ( 1 3 4 ) 10 −12 ) = 10 ( 1 + 1 3.10 3 − 1 9.10 6 + 5 81.10 9 − 10 243.10 12 ) = 10.00333222... whereas ( 1001 ) 1 / 3 = 10.00332222839093 .... ( 1001 ) 1 / 3 = 10.00332222839093 .... Two terms would suffice for three-digit accuracy.

The approximation is 2.3152 ; 2.3152 ; the CAS value is 2.23 … . 2.23 … .

The approximation is 2.583 … ; 2.583 … ; the CAS value is 2.449 … . 2.449 … .

1 − x 2 = 1 − x 2 2 − x 4 8 − x 6 16 − 5 x 8 128 + ⋯ . 1 − x 2 = 1 − x 2 2 − x 4 8 − x 6 16 − 5 x 8 128 + ⋯ . Thus

∫ −1 1 1 − x 2 d x = x − x 3 6 − x 5 40 − x 7 7 · 16 − 5 x 9 9 · 128 + ⋯ | −1 1 ≈ 2 − 1 3 − 1 20 − 1 56 − 10 9 · 128 + error = 1.590 ... ∫ −1 1 1 − x 2 d x = x − x 3 6 − x 5 40 − x 7 7 · 16 − 5 x 9 9 · 128 + ⋯ | −1 1 ≈ 2 − 1 3 − 1 20 − 1 56 − 10 9 · 128 + error = 1.590 ... whereas π 2 = 1.570 ... π 2 = 1.570 ...

( 1 + 4 x ) 4 / 3 = ( 1 + 4 x ) ( 1 + 4 x ) 1 / 3 = ( 1 + 4 x ) ( 1 + 4 x 3 - 16 x 3 9 + 320 x 3 81 - 2560 x 4 243 ) = 1 + 16 3 x + 32 9 x 2 - 256 81 x 3 + 1280 243 x 4 - 10240 243 x 5 ( 1 + 4 x ) 4 / 3 = ( 1 + 4 x ) ( 1 + 4 x ) 1 / 3 = ( 1 + 4 x ) ( 1 + 4 x 3 - 16 x 3 9 + 320 x 3 81 - 2560 x 4 243 ) = 1 + 16 3 x + 32 9 x 2 - 256 81 x 3 + 1280 243 x 4 - 10240 243 x 5

( 1 + ( x + 3 ) 2 ) 1 / 3 = 1 + 1 3 ( x + 3 ) 2 − 1 9 ( x + 3 ) 4 + 5 81 ( x + 3 ) 6 − 10 243 ( x + 3 ) 8 + ⋯ ( 1 + ( x + 3 ) 2 ) 1 / 3 = 1 + 1 3 ( x + 3 ) 2 − 1 9 ( x + 3 ) 4 + 5 81 ( x + 3 ) 6 − 10 243 ( x + 3 ) 8 + ⋯

Twice the approximation is 1.260 … 1.260 … whereas 2 1 / 3 = 1.2599. ... 2 1 / 3 = 1.2599. ...

f ( 99 ) ( 0 ) = 0 f ( 99 ) ( 0 ) = 0

∑ n = 0 ∞ ( ln ( 2 ) x ) n n ! ∑ n = 0 ∞ ( ln ( 2 ) x ) n n !

For x > 0 , sin ( x ) = ∑ n = 0 ∞ ( −1 ) n x ( 2 n + 1 ) / 2 x ( 2 n + 1 ) ! = ∑ n = 0 ∞ ( −1 ) n x n ( 2 n + 1 ) ! . x > 0 , sin ( x ) = ∑ n = 0 ∞ ( −1 ) n x ( 2 n + 1 ) / 2 x ( 2 n + 1 ) ! = ∑ n = 0 ∞ ( −1 ) n x n ( 2 n + 1 ) ! .

e x 3 = ∑ n = 0 ∞ x 3 n n ! e x 3 = ∑ n = 0 ∞ x 3 n n !

sin 2 x = − ∑ k = 1 ∞ ( −1 ) k 2 2 k − 1 x 2 k ( 2 k ) ! sin 2 x = − ∑ k = 1 ∞ ( −1 ) k 2 2 k − 1 x 2 k ( 2 k ) !

tan −1 x = ∑ k = 0 ∞ ( −1 ) k x 2 k + 1 2 k + 1 tan −1 x = ∑ k = 0 ∞ ( −1 ) k x 2 k + 1 2 k + 1

sin −1 x = ∑ n = 0 ∞ ( 1 2 n ) x 2 n + 1 ( 2 n + 1 ) n ! sin −1 x = ∑ n = 0 ∞ ( 1 2 n ) x 2 n + 1 ( 2 n + 1 ) n !

F ( x ) = ∑ n = 0 ∞ ( −1 ) n x n + 1 ( n + 1 ) ( 2 n ) ! F ( x ) = ∑ n = 0 ∞ ( −1 ) n x n + 1 ( n + 1 ) ( 2 n ) !

F ( x ) = ∑ n = 1 ∞ ( −1 ) n + 1 x n n 2 F ( x ) = ∑ n = 1 ∞ ( −1 ) n + 1 x n n 2

x + x 3 3 + 2 x 5 15 + ⋯ x + x 3 3 + 2 x 5 15 + ⋯

1 + x − x 3 3 − x 4 6 + ⋯ 1 + x − x 3 3 − x 4 6 + ⋯

1 + x 2 + 2 x 4 3 + 17 x 6 45 + ⋯ 1 + x 2 + 2 x 4 3 + 17 x 6 45 + ⋯

Using the expansion for tan x tan x gives 1 + x 3 + 2 x 2 15 . 1 + x 3 + 2 x 2 15 .

1 1 + x 2 = ∑ n = 0 ∞ ( −1 ) n x 2 n 1 1 + x 2 = ∑ n = 0 ∞ ( −1 ) n x 2 n so R = 1 R = 1 by the ratio test.

ln ( 1 + x 2 ) = ∑ n = 1 ∞ ( −1 ) n − 1 n x 2 n ln ( 1 + x 2 ) = ∑ n = 1 ∞ ( −1 ) n − 1 n x 2 n so R = 1 R = 1 by the ratio test.

Add series of e x e x and e − x e − x term by term. Odd terms cancel and cosh x = ∑ n = 0 ∞ x 2 n ( 2 n ) ! . cosh x = ∑ n = 0 ∞ x 2 n ( 2 n ) ! .

The ratio S n ( x ) C n ( x ) S n ( x ) C n ( x ) approximates tan x tan x better than does p 7 ( x ) = x + x 3 3 + 2 x 5 15 + 17 x 7 315 p 7 ( x ) = x + x 3 3 + 2 x 5 15 + 17 x 7 315 for N ≥ 3 . N ≥ 3 . The dashed curves are S n C n − tan S n C n − tan for n = 1 , 2 . n = 1 , 2 . The dotted curve corresponds to n = 3 , n = 3 , and the dash-dotted curve corresponds to n = 4 . n = 4 . The solid curve is p 7 − tan x . p 7 − tan x .

By the term-by-term differentiation theorem, y ′ = ∑ n = 1 ∞ n a n x n − 1 y ′ = ∑ n = 1 ∞ n a n x n − 1 so y ′ = ∑ n = 1 ∞ n a n x n − 1 x y ′ = ∑ n = 1 ∞ n a n x n , y ′ = ∑ n = 1 ∞ n a n x n − 1 x y ′ = ∑ n = 1 ∞ n a n x n , whereas y ′ = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 y ′ = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 so x y ″ = ∑ n = 2 ∞ n ( n − 1 ) a n x n . x y ″ = ∑ n = 2 ∞ n ( n − 1 ) a n x n .

The probability is p = 1 2 π ∫ ( a − μ ) / σ ( b − μ ) / σ e − x 2 / 2 d x p = 1 2 π ∫ ( a − μ ) / σ ( b − μ ) / σ e − x 2 / 2 d x where a = 90 a = 90 and b = 100 , b = 100 , that is, p = 1 2 π ∫ −1 1 e − x 2 / 2 d x = 1 2 π ∫ −1 1 ∑ n = 0 5 ( −1 ) n x 2 n 2 n n ! d x = 2 2 π ∑ n = 0 5 ( −1 ) n 1 ( 2 n + 1 ) 2 n n ! ≈ 0.6827 . p = 1 2 π ∫ −1 1 e − x 2 / 2 d x = 1 2 π ∫ −1 1 ∑ n = 0 5 ( −1 ) n x 2 n 2 n n ! d x = 2 2 π ∑ n = 0 5 ( −1 ) n 1 ( 2 n + 1 ) 2 n n ! ≈ 0.6827 .

As in the previous problem one obtains a n = 0 a n = 0 if n n is odd and a n = − ( n + 2 ) ( n + 1 ) a n + 2 a n = − ( n + 2 ) ( n + 1 ) a n + 2 if n n is even, so a 0 = 1 a 0 = 1 leads to a 2 n = ( −1 ) n ( 2 n ) ! . a 2 n = ( −1 ) n ( 2 n ) ! .

y ″ = ∑ n = 0 ∞ ( n + 2 ) ( n + 1 ) a n + 2 x n y ″ = ∑ n = 0 ∞ ( n + 2 ) ( n + 1 ) a n + 2 x n and y ′ = ∑ n = 0 ∞ ( n + 1 ) a n + 1 x n y ′ = ∑ n = 0 ∞ ( n + 1 ) a n + 1 x n so y ″ − y ′ + y = 0 y ″ − y ′ + y = 0 implies that ( n + 2 ) ( n + 1 ) a n + 2 − ( n + 1 ) a n + 1 + a n = 0 ( n + 2 ) ( n + 1 ) a n + 2 − ( n + 1 ) a n + 1 + a n = 0 or a n = a n − 1 n − a n − 2 n ( n − 1 ) a n = a n − 1 n − a n − 2 n ( n − 1 ) for all n · y ( 0 ) = a 0 = 1 n · y ( 0 ) = a 0 = 1 and y ′ ( 0 ) = a 1 = 0 , y ′ ( 0 ) = a 1 = 0 , so a 2 = 1 2 , a 3 = 1 6 , a 4 = 0 , a 2 = 1 2 , a 3 = 1 6 , a 4 = 0 , and a 5 = − 1 120 . a 5 = − 1 120 .

a. (Proof) b. We have R s ≤ 0.1 ( 9 ) ! π 9 ≈ 0.0082 < 0.01 . R s ≤ 0.1 ( 9 ) ! π 9 ≈ 0.0082 < 0.01 . We have ∫ 0 π ( 1 − x 2 3 ! + x 4 5 ! − x 6 7 ! + x 8 9 ! ) d x = π − π 3 3 · 3 ! + π 5 5 · 5 ! − π 7 7 · 7 ! + π 9 9 · 9 ! = 1.852... , ∫ 0 π ( 1 − x 2 3 ! + x 4 5 ! − x 6 7 ! + x 8 9 ! ) d x = π − π 3 3 · 3 ! + π 5 5 · 5 ! − π 7 7 · 7 ! + π 9 9 · 9 ! = 1.852... , whereas ∫ 0 π sin t t d t = 1.85194... , ∫ 0 π sin t t d t = 1.85194... , so the actual error is approximately 0.00006 . 0.00006 .

Since cos ( t 2 ) = ∑ n = 0 ∞ ( −1 ) n t 4 n ( 2 n ) ! cos ( t 2 ) = ∑ n = 0 ∞ ( −1 ) n t 4 n ( 2 n ) ! and sin ( t 2 ) = ∑ n = 0 ∞ ( −1 ) n t 4 n + 2 ( 2 n + 1 ) ! , sin ( t 2 ) = ∑ n = 0 ∞ ( −1 ) n t 4 n + 2 ( 2 n + 1 ) ! , one has S ( x ) = ∑ n = 0 ∞ ( −1 ) n x 4 n + 3 ( 4 n + 3 ) ( 2 n + 1 ) ! S ( x ) = ∑ n = 0 ∞ ( −1 ) n x 4 n + 3 ( 4 n + 3 ) ( 2 n + 1 ) ! and C ( x ) = ∑ n = 0 ∞ ( −1 ) n x 4 n + 1 ( 4 n + 1 ) ( 2 n ) ! . C ( x ) = ∑ n = 0 ∞ ( −1 ) n x 4 n + 1 ( 4 n + 1 ) ( 2 n ) ! . The sums of the first 50 50 nonzero terms are plotted below with C 50 ( x ) C 50 ( x ) the solid curve and S 50 ( x ) S 50 ( x ) the dashed curve.

∫ 0 1 / 4 x ( 1 − x 2 − x 2 8 − x 3 16 − 5 x 4 128 − 7 x 5 256 ) d x ∫ 0 1 / 4 x ( 1 − x 2 − x 2 8 − x 3 16 − 5 x 4 128 − 7 x 5 256 ) d x

= 2 3 2 −3 − 1 2 2 5 2 −5 − 1 8 2 7 2 −7 − 1 16 2 9 2 −9 − 5 128 2 11 2 −11 − 7 256 2 13 2 −13 = 0.0767732 ... = 2 3 2 −3 − 1 2 2 5 2 −5 − 1 8 2 7 2 −7 − 1 16 2 9 2 −9 − 5 128 2 11 2 −11 − 7 256 2 13 2 −13 = 0.0767732 ...

whereas ∫ 0 1 / 4 x − x 2 d x = 0.076773 . ∫ 0 1 / 4 x − x 2 d x = 0.076773 .

T ≈ 2 π 10 9.8 ( 1 + sin 2 ( θ / 12 ) 4 ) ≈ 6.453 T ≈ 2 π 10 9.8 ( 1 + sin 2 ( θ / 12 ) 4 ) ≈ 6.453 seconds. The small angle estimate is T ≈ 2 π 10 9.8 ≈ 6.347 . T ≈ 2 π 10 9.8 ≈ 6.347 . The relative error is around 2 2 percent.

∫ 0 π / 2 sin 4 θ d θ = 3 π 16 . ∫ 0 π / 2 sin 4 θ d θ = 3 π 16 . Hence T ≈ 2 π L g ( 1 + k 2 4 + 9 256 k 4 ) . T ≈ 2 π L g ( 1 + k 2 4 + 9 256 k 4 ) .

ROC: 1 ; 1 ; IOC: ( 0 , 2 ) ( 0 , 2 )

ROC: 12 ; 12 ; IOC: ( −16 , 8 ) ( −16 , 8 )

∑ n = 0 ∞ ( −1 ) n 3 n + 1 x n + 2 ; ∑ n = 0 ∞ ( −1 ) n 3 n + 1 x n + 2 ; ROC: 3 ; 3 ; IOC: ( −3 , 3 ) ( −3 , 3 )

integration: ∑ n = 0 ∞ ( −1 ) n 2 n + 1 ( 2 x ) 2 n + 1 ∑ n = 0 ∞ ( −1 ) n 2 n + 1 ( 2 x ) 2 n + 1

p 4 ( x ) = ( x + 3 ) 3 − 11 ( x + 3 ) 2 + 39 ( x + 3 ) − 41 ; p 4 ( x ) = ( x + 3 ) 3 − 11 ( x + 3 ) 2 + 39 ( x + 3 ) − 41 ; exact

∑ n = 0 ∞ ( −1 ) n ( 3 x ) 2 n 2 n ! ∑ n = 0 ∞ ( −1 ) n ( 3 x ) 2 n 2 n !

∑ n = 0 ∞ ( −1 ) n ( 2 n ) ! ( x − π 2 ) 2 n ∑ n = 0 ∞ ( −1 ) n ( 2 n ) ! ( x − π 2 ) 2 n

∑ n = 1 ∞ ( −1 ) n n ! x 2 n ∑ n = 1 ∞ ( −1 ) n n ! x 2 n

F ( x ) = ∑ n = 0 ∞ ( −1 ) n ( 2 n + 1 ) ( 2 n + 1 ) ! x 2 n + 1 F ( x ) = ∑ n = 0 ∞ ( −1 ) n ( 2 n + 1 ) ( 2 n + 1 ) ! x 2 n + 1

2.5 % 2.5 %

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• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 2
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-2/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-2/pages/chapter-6

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## 6.7E: Exercises for Section 6.7

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In exercises 1 - 3, find the derivative $$\dfrac{dy}{dx}$$.

1) $$y=\ln(2x)$$

2) $$y=\ln(2x+1)$$

3) $$y=\dfrac{1}{\ln x}$$

In exercises 4 - 5, find the indefinite integral.

4) $$\displaystyle ∫\frac{dt}{3t}$$

5) $$\displaystyle ∫\frac{dx}{1+x}$$

In exercises 6 - 15, find the derivative $$\dfrac{dy}{dx}.$$ (You can use a calculator to plot the function and the derivative to confirm that it is correct.)

6) [T] $$y=\dfrac{\ln x}{x}$$

7) [T] $$y=x\ln x$$

8) [T] $$y=\log_{10}x$$

9) [T] $$y=\ln(\sin x)$$

10) [T] $$y=\ln(\ln x)$$

11) [T] $$y=7\ln(4x)$$

12) [T] $$y=\ln\big((4x)^7\big)$$

13) [T] $$y=\ln(\tan x)$$

14) [T] $$y=\ln(\tan 3x)$$

15) [T] $$y=\ln(\cos^2x)$$

In exercises 16 - 25, find the definite or indefinite integral.

16) $$\displaystyle ∫^1_0\frac{dx}{3+x}$$

17) $$\displaystyle ∫^1_0\frac{dt}{3+2t}$$

18) $$\displaystyle ∫^2_0\frac{x}{x^2+1}\, dx$$

19) $$\displaystyle ∫^2_0\frac{x^3}{x^2+1}\,dx$$

20) $$\displaystyle ∫^e_2\frac{dx}{x\ln x}$$

21) $$\displaystyle ∫^e_2\frac{dx}{(x\ln x)^2}$$

22) $$\displaystyle ∫\frac{\cos x}{\sin x}\, dx$$

23) $$\displaystyle ∫^{π/4}_0\tan x\,dx$$

24) $$\displaystyle ∫\cot(3x)\,dx$$

25) $$\displaystyle ∫\frac{(\ln x)^2}{x}\, dx$$

In exercises 26 - 35, compute $$\dfrac{dy}{dx}$$ by differentiating $$\ln y$$.

26) $$y=\sqrt{x^2+1}$$

27) $$y=\sqrt{x^2+1}\sqrt{x^2−1}$$

28) $$y=e^{\sin x}$$

29) $$y=x^{−1/x}$$

30) $$y=e^{ex}$$

31) $$y=x^e$$

32) $$y=x^{(ex)}$$

33) $$y=\sqrt{x}\sqrt[3]{x}\sqrt[6]{x}$$

34) $$y=x^{−1/\ln x}$$

35) $$y=e^{−\ln x}$$

In exercises 36 - 40, evaluate by any method.

36) $$\displaystyle ∫^{10}_5\dfrac{dt}{t}−∫^{10x}_{5x}\dfrac{dt}{t}$$

37) $$\displaystyle ∫^{e^π}_1\dfrac{dx}{x}+∫^{−1}_{−2}\dfrac{dx}{x}$$

38) $$\dfrac{d}{dx}\left[\displaystyle ∫^1_x\dfrac{dt}{t}\right]$$

39) $$\dfrac{d}{dx}\left[\displaystyle ∫^{x^2}_x\dfrac{dt}{t}\right]$$

40) $$\dfrac{d}{dx}\Big[\ln(\sec x+\tan x)\Big]$$

In exercises 41 - 44, use the function $$\ln x$$. If you are unable to find intersection points analytically, use a calculator.

41) Find the area of the region enclosed by $$x=1$$ and $$y=5$$ above $$y=\ln x$$.

42) [T] Find the arc length of $$\ln x$$ from $$x=1$$ to $$x=2$$.

43) Find the area between $$\ln x$$ and the $$x$$-axis from $$x=1$$ to $$x=2$$.

44) Find the volume of the shape created when rotating this curve from $$x=1$$ to $$x=2$$ around the $$x$$-axis, as pictured here.

45) [T] Find the surface area of the shape created when rotating the curve in the previous exercise from $$x=1$$ to $$x=2$$ around the $$x$$-axis.

If you are unable to find intersection points analytically in the following exercises, use a calculator.

46) Find the area of the hyperbolic quarter-circle enclosed by $$x=2$$ and $$y=2$$ above $$y=1/x.$$

47) [T] Find the arc length of $$y=1/x$$ from $$x=1$$ to $$x=4$$.

48) Find the area under $$y=1/x$$ and above the $$x$$-axis from $$x=1$$ to $$x=4$$.

In exercises 49 - 53, verify the derivatives and antiderivatives.

49) $$\dfrac{d}{dx}\Big[\ln(x+\sqrt{x^2+1})\Big]=\dfrac{1}{\sqrt{1+x^2}}$$

50) $$\dfrac{d}{dx}\Big[\ln\left(\frac{x−a}{x+a}\right)\Big]=\dfrac{2a}{(x^2−a^2)}$$

51) $$\dfrac{d}{dx}\Big[\ln\left(\frac{1+\sqrt{1−x^2}}{x}\right)\Big]=−\dfrac{1}{x\sqrt{1−x^2}}$$

52) $$\dfrac{d}{dx}\Big[\ln(x+\sqrt{x^2−a^2})\Big]=\dfrac{1}{\sqrt{x^2−a^2}}$$

53) $$\displaystyle ∫\frac{dx}{x\ln(x)\ln(\ln x)}=\ln|\ln(\ln x)|+C$$

## Description

Integration - The Accumulation of Change Homework for AP Calculus BC Bundle: This is an eleven-lesson unit on Integration - The Accumulation of Change for students enrolled in AP Calculus BC, Calculus 2 . Your AP Calculus BC students will have a set of homework assignments , corresponding daily content quizzes and a complete solution set for lessons covering the topics and concepts for Integration and Accumulation of Change. Teachers also have the benefit of the current Topics, Learning Objectives, and Essential Knowledge for the Fall 2019 AP Calculus ® CED Binder updates.

The entire bundle has 42 pages of homework practice and 19 pages of daily quizzes. There are two quizzes on a sheet of paper to minimize paper usage. Each quiz should be able to be completed in under 10 minutes of class time.

**(2/18/2022) Adjusted the key for 6.4 #19 and 6.7 #21

College Board® Recommended Sequence:

✎ Approximating Area With Riemann Sums (3 pages)

✎ Sigma Notation and Riemann Sums (2 pages)

✎ Accumulation Functions (4 pages)

✎ Applying Properties of Definite Integrals (4 pages)

✎ Antiderivatives and Indefinite Integrals Basic Rules (3 pages)

✎ Fundamental Theorem of Calculus (4 pages)

✎ Integration by Substitution (5 pages)

✎ Integration of Transcendental Functions – BC Version (4 pages)

✎ Integration by Parts BC Topic (4 pages)

✎ Partial Fractions and Long Division BC Topic (4 pages)

✎ Improper Integrals BC Topic (3 pages)

✎ Advanced Integration Mixed Practice (4 pages)

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You may also be interested in:

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Calculus Integration BC Guided Notes

Calculus Integration BC SMART Board Lessons

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This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer. Question: 6.7 Integration by Substitution Homework Problems 1-10, Find the indefinite integral. Sx (x² - 1)² ax 5. t 2 (t-4t+3) ๐ - 4143)3 dt. I newd help with these 2 problems.

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4. Section 6.7. Integration by substitution

Then work the Interactive Examples to practice the concepts and techniques before you start the Exercises. TOPICS IN THIS SECTION. 1 Overview. 2 Theorem 1: Integrals of 1/x. 3 An example. 4 A rate of change problem. 5 Theorem 2: Integrals of exponential functions. 6 Finding an area. 7 Theorem 3: Integrals of the hyperbolic cosine and sine.

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6. Calculus AB Homework 6.7 U-Substitution

Download Packet: https://goo.gl/xeg85Y=====AP Calculus AB / IB Math SLUnit 6: IntegrationLesson 7: U-Substitution=====...

7. 1.6: Substitution

In this section we examine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative. At first, the approach to the substitution procedure may not appear very obvious.

8. PDF Section 6.8 Integration by substitution

Then we substitute the formula u = x2 + 1 for the final answer: (x2 + 1)5(2x) dx = Z u5 du =. 6u6 1 + C = 1 6(x2. . 1)6+ C. As with any indefinite integral, we can check Example 1 by differentiating the result. This requires the Chain Rule because the technique of substitution is derived from the Chain Rule.

9. PDF Homework 01: Integration by Substitution

Homework 01: Integration by Substitution Instructor: Joseph Wells Arizona State University Due: (Wed) January 22, 2014/ (Fri) January 24, 2014 Instructions: Complete ALL the problems on this worksheet (and staple on any additional pages used). Show ALL your work in the spaces provided. If you do not show your work, you will not

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In exercises 4 - 5, find the indefinite integral. 4) ∫ dt 3t ∫ d t 3 t. 5) ∫ dx 1 + x ∫ d x 1 + x. Answer. In exercises 6 - 15, find the derivative dy dx. d y d x. (You can use a calculator to plot the function and the derivative to confirm that it is correct.) 6) [T] y = ln x x y = ln x x. 7) [T] y = x ln x y = x ln x.

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Find step-by-step solutions and answers to Applied Calculus - 9781305085312, as well as thousands of textbooks so you can move forward with confidence. ... Integration by Substitution. Page 378: Review Exercises and Chapter Test. Exercise 1. Exercise 2. Exercise 3. Exercise 4. ... Integration Using Tables. Section 6.3: Improper Integrals ...

12. Unit 6

Unit 6 - Integration and Accumulation of Change 6.1 Exploring Accumulation of Change 6.2 Approximating Areas with Riemann Sums 6.3 Riemann Sums, Summation Notation, and Definite ... 6.9 Integrating Using Substitution 6.10 Integrating Functions Using Long Division and Completing the Square 6.11 Integrating Using Integration by Parts (BC topic) 6 ...

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Section 5.3 : Substitution Rule for Indefinite Integrals. For problems 1 - 16 evaluate the given integral. Evaluate each of the following integrals. Here is a set of practice problems to accompany the Substitution Rule for Indefinite Integrals section of the Integrals chapter of the notes for Paul Dawkins Calculus I course at Lamar University.

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Study the section. Then work the Interactive Examples to practice the concepts and techniques before you start the Exercises.

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