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Final Exam Review Solutions - Calculus II | MATH 1132, Exams of Calculus

Material Type: Exam; Class: Calculus II; Subject: Mathematics; University: University of Connecticut; Term: Fall 2009;

Typology: Exams

2009/2010

Uploaded on 02/25/2010

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Math 1132Q: Fall 2009
Final Exam Review Solutions
1. Use the definition of the integral to evaluate Z2
0(3 x)dx
By the Theorem 4 of section 5.2 on page 368,
Z2
0(3 x)dx = lim
n→∞
n
X
i=1
(3 xi)∆x
where x=2
nand xi= 0 + ix=2i
n. Thus we have
Z2
0(3 x)dx = lim
n→∞
n
X
i=1
(3 2i
n)2
n= lim
n→∞
2
n
n
X
i=1
(3 2i
n) = lim
n→∞(6
n
n
X
i=1
14
n2
n
X
i=1
i)
= lim
n→∞(6
nn4
n2
n(n+ 1)
2) = lim
n→∞(6 2(n2+n)
n2) = lim
n→∞(6 22
n) = 4
We can check our result as follows
Z2
0(3 x)dx = 3xx2
2
2
0
= 6 2=4.
2. Find dg
dx , where g(x) = Z0
3x
u21
u2+ 1 du
Note that g(x) = Z3x
0
u21
u2+ 1 du, so
dg
dx =(3x)21
(3x)2+ 1 ·3 = 27x23
9x2+ 1
3. Zsin3(θ) cos5(θ)
Zsin3(θ) cos5(θ) =Zsin3(θ) cos4(θ)d(sin(θ)) = Zsin3(θ)(1 sin2(θ))2d(sin(θ))
=Zu3(1 u2)2du =Z(u32u5+u7)du =u4
4u6
3+u8
8+C
=sin4(θ)
4sin6(θ)
3+sin8(θ)
8+C
4. Zxsin2x dx
Zxsin2x dx =Zx·1cos 2x
2dx =1
2Z(xxcos(2x)) dx
=1
2(x2
2(x·sin(2x)
2Zsin(2x)
2dx)) (integration by parts)
=x2
4xsin(2x)
4cos(2x)
8+C
1
pf3
pf4
pf5

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Math 1132Q: Fall 2009 Final Exam Review Solutions

  1. Use the definition of the integral to evaluate

∫ (^2) 0

(3 − x) dx By the Theorem 4 of section 5.2 on page 368, ∫ (^2) 0

(3 − x) dx = limn→∞

∑^ n i=

(3 − xi)∆x

where ∆x =

n

and xi = 0 + i∆x = 2 i n

. Thus we have ∫ (^2) 0

(3 − x) dx = (^) nlim→∞

∑^ n i=

2 i n

n = limn→∞

n

∑^ n i=

2 i n ) = limn→∞(

n

∑^ n i=

n^2

∑^ n i=

i)

= (^) nlim→∞(

n n −

n^2

n(n + 1) 2 ) = lim n→∞(6 − 2(n^2 + n) n^2 ) = lim n→∞(6 − 2 −

n

We can check our result as follows ∫ (^2) 0 (3 − x) dx = 3x − x^2 2

∣∣ ∣∣ ∣

2

0

  1. Find dgdx , where g(x) =

∫ (^0) 3 x

u^2 − 1 u^2 + 1 du

Note that g(x) = −

∫ (^3) x 0

u^2 − 1 u^2 + 1 du, so

dg dx

(3x)^2 − 1 (3x)^2 + 1

27 x^2 − 3 9 x^2 + 1

∫ sin^3 (θ) cos^5 (θ) dθ ∫ sin^3 (θ) cos^5 (θ) dθ =

∫ sin^3 (θ) cos^4 (θ) d(sin(θ)) =

∫ sin^3 (θ)(1 − sin^2 (θ))^2 d(sin(θ))

=

∫ u^3 (1 − u^2 )^2 du =

∫ (u^3 − 2 u^5 + u^7 ) du = u^4 4

u^6 3

u^8 8

+ C

sin^4 (θ) 4

sin^6 (θ) 3

sin^8 (θ) 8

+ C

∫ x sin^2 x dx

∫ x sin^2 x dx =

∫ x · 1 − cos 2x 2 dx =

∫ (x − x cos(2x)) dx

=

x^2 2 − (x · sin(2x) 2

∫ (^) sin(2x) 2 dx)) (integration by parts)

= x^2 4

x sin(2x) 4

cos(2x) 8

+ C

∫ x^5 e−x 3 dx Let u = x^3 , so that du = 3x^2 dx, ∫ x^5 e−x 3 dx =

∫ ue−u^ du =

(−ue−u^ +

∫ e−u^ du) =

(−ue−u^ − e−u) + C

=

(−x^3 e−x 3 − e−x 3 ) + C

∫ (^3) x (^2) − 2 x^2 − 2 x − 8 dx

∫ (^3) x (^2) − 2 x^2 − 2 x − 8

dx =

∫ 3 + 6 x + 22 x^2 − 2 x − 8

dx = 3x +

∫ (^6) x + 22 (x − 4)(x + 2)

dx

= 3 x +

3(x − 4)

3(x + 2)

dx = 3x +

ln(x − 4) −

ln(x + 2) + C

  1. Find the volume of the solid obtained by rotating the region bounded by y = x and y =

x about the x-axis. The curve y = x and y =

x intersect at the points (0, 0) and (1, 1). The curve y =

x is on the top of the region and the curve y = x is on the bottom of the region.

V =

∫ (^1) 0

A(x) dx =

∫ (^1) 0

π(x − x^2 ) dx = π ( x^2 2

x^3 3

∣∣ ∣∣ ∣

1

0

π 6

  1. A heavy rope, 50ft long, weighs 0.5lb/ft and hangs over the edge of a building 120ft high. How much work is done in pulling half the rope to the top of the building?

W = lim n→∞

∑^ n i=

  1. 5 x∗ i ∆x + 0. 5 ·

∫ (^25) 0

  1. 5 x dx +

x^2 4

∣∣ ∣∣ ∣

25

0

ft-lb

  1. What is the average value of h(x) = cos^4 (x) sin(x) on the interval [0, π]?

havg =

π

∫ (^) π 0

cos^4 (x) sin(x) dx = −

π

∫ (^) π 0

cos^4 (x) d(cos(x))

= −

π

cos^5 (x) 5

∣∣ ∣∣ ∣

π

0

5 π

∫ (^1) 0

x^2 + 1 dx Let x = tan θ, ∫ (^1) 0

x^2 + 1 dx =

∫ π 4 0

√ tan^2 (θ) + 1 sec^2 (θ) dθ =

∫ π 4 0 sec^3 (θ) dθ

sec(θ) tan(θ) + ln | sec(θ) + tan(θ)| 2

∣∣ ∣∣ ∣

π 4

0

2 + ln(

  1. Does the sequence converge? If so, find its limit. an = cos^2 n 2 n Note that 0 ≤ cos^2 n 2 n^

2 n and limn→∞

2 n^ = 0, by the Squeeze Theorem (of the section 11.1 on page 679), the sequence converges to 0.

  1. Do the following series converge? If you can find the sum, do so.

∑^ ∞ n=

1 + 2n 3 n =

∑^ ∞ n=

)n

∑^ ∞ n=

)n

∑^ ∞ n=

(−1)n^ n ln n

diverges by the Test for Divergence

  • ∑∞ n=1 ne−n^ converges by the Integral Test

∑^ ∞ n=

n^2 + 4 converges by the Comparison Test with the p-series (p = 2)

∑^ ∞ n=

n(n + 1) converges by the Comparison Test with the p-series (p = 2)

∑^ ∞ n=

n^2 + 1

diverges by the Comparison Test with the harmonic series (p = 1)

∑^ ∞ n=

(−1)n−^1 2 n n^4 diverges by the Test for Divergence

∑^ ∞ n=

sin

n

) diverges by the Limit Comparison Test with the harmonic series

∑^ ∞ n=

( n^2 + 1 2 n^2 + 1

)n converges by the Root Test

  1. Find the interval of convergence of the power series.

∑^ ∞ n=

(n!)(2x − 1)n

Let an = (n!)(2x − 1)n. Then ∣∣ ∣∣^ an+ an

∣∣ ∣∣ = (n + 1)| 2 x − 1 | → ∞, as n → ∞

By the Ratio Test, the series diverges when x 6 = 1/2. Thus the series converges only when x = 1/2.

∑^ ∞ n=

(x − 2)n nn Let an = (x − 2)n nn^

. Then √ n |an| = |x − 2 | n

→ 0 , as n → ∞ By the Root Test, the series converges for all values of x. So the interval of convergence is (−∞, ∞)

∑^ ∞ n=

(x − 2)n n^2 + 1 Let an = (x − 2)n n^2 + 1

. Then ∣∣ ∣∣^ an+ an

∣∣ ∣∣ = n

(n + 1)^2 + 1 |x − 2 | → |x − 2 |, as n → ∞

By the Ratio Test, the series converges when |x − 2 | < 1 ⇔ 1 < x < 3. When x = 1, the series is convergent by the Alternating Series Test, and when x = 3, the series is also convergent by the Comparison Test. So the interval of convergence is [1, 3].

  1. Find the Taylor series for f (x) = ex^ centered at a = 3.

∑^ ∞ n=

f (n)(a) n! (x − a)n^ =

∑^ ∞ n=

f (n)(3) n! (x − 3)n^ =

∑^ ∞ n=

e^3 n! (x − 3)n

  1. Find a power series representation for f (x) = ln(5 − x)

ln(5 − x) = −

5 − x dx = −

1 − x/ 5 dx = −

∫ (

∑^ ∞ n=

x 5 )n) dx

∑^ ∞ n=

∫ (^) xn 5 n^ dx = −

5 n+

∑^ ∞ n=

xn+ n + 1

  • C |x| < 5

5 n+

∑^ ∞ n=

xn n

  • C |x| < 5

Put x = 0 in this equation and obtain C = ln 5.Thus

f (x) = ln(5 − x) = −

5 n+

∑^ ∞ n=

xn n

  • ln 5 |x| < 5
  1. Find the Maclaurin series for f (x) = cos(3x) Note that f ′(x) = −3 sin(3x), f ′′(x) = − 32 cos(3x), f (3)(x) = 3^3 sin(3x) and f (4)(x) = 3^4 cos(3x), so f (0) = 1,f ′(0) = 0, f ′′(0) = − 32 , f (3)(0) = 0 and f (4)(0) = 3^4 , so the Maclaurin series is ∑^ ∞ n=

(−1)n^ 32 nx^2 n (2n)!