7.5, #46. The integrand is an improper rational function. Long division gives

x2
x2+6x+13
 = 1- 6x+13
x2+6x+13
 .
Completing the square gives x2+6x+13 = (x+3)2+22, so set w = x+3 and dw = dx. Then
6x+13
x2+6x+13
 = 6w-5
w2+22
 ,
so formula 25 applies with a = 2, b = 6, c = -5, giving
ó
õ 		x2
x2+6x+13
  dx = x-3ln ê
ê 		(x+3)2+22 ê
ê 		+ 5
2
 arctan x+3
2
 +C.


7.5, #50. The integrand is an improper rational function. Long division gives

t2+1
t2-1
 = 1+ 2
t2-1
 .
Since t2-1 = (t-1)(t+1), formula 26 with a = 1, b = -1 gives
ó
õ 		t2+1
t2-1
  dt = t+ln|t-1|-ln|t+1|+C.


7.5, #51. Formula 24 with a = 2 gives

ó
õ 		2

0 
 dx
4+x2
 = 1
2
 arctan x
2
 ê
ê 		2
0 
 = 1
2
 (arctan1-arctan0) = p
8
 = 0.3927.
The left and right Riemann sums with n = 100 are approximately 0.3939 and 0.3915. Since the integrand is decreasing on the interval [0,2], its value is between these two estimates.

7.5, #53. Since x2+2x+5 = (x+1)2+22, formula 24 with a = 2 gives

ó
õ 		1

0 
 dx
x2+2x+5
 = 1
2
 arctan x+1
2
 ê
ê
ê 		1

0 
 = 1
2
 (arctan1- arctan 1
2
 ) = 0.1609.
The left and right Riemann sums with n = 100 are approximately 0.1613 and 0.1605. Since the integrand is decreasing on the interval [0,1], its value is between these two estimates.


7.5, #64. (a) The average voltage over a second is

1
1-0
 ó
õ 		1

0 
 V0cos(120pt) dt = V0
120p
 sin(120pt) ê
ê
ê 		1

0 
 = 0.


(b) The average of V2 over a second is

1
1-0
 ó
õ 		1

0 
 V02cos2(120pt) dt
= V02
120p
 ó
õ 		120 p

0 
 cos2u du
= V02
240p
 (cosusinu+u) ê
ê
ê 		120p

0 
 = 1
2
 V20
using the substitution u = 120pt and du = 120p dt. Thus [`V] = V0/Ö2.


(c) If [`V] = 110, then V0 = 110Ö2 = 155.56 volts.



7.6, #1.

      n = 1    n = 2    n = 4
   LEFT    40.0000    40.7846    41.7116
   RIGHT    51.2250    46.3971    44.5179
   TRAP    45.6125    43.5909    43.1147
   MID    41.5692    42.6386    42.8795

7.6, #3.

   n     10     100     1000
   LEFT    5.4711    5.8116    5.8464
   RIGHT    6.2443    5.8890    5.8541
   TRAP    5.8577    5.8503    5.8502
   MID    5.8465    5.8502    5.8502

exÖx is increasing and concave-up on [1,2] since
(exÖx)¢ = ex æ
ç
è 		Öx+ 1
2Öx
 ö
÷
ø 		,    (exÖx)¢¢ = ex æ
ç
ç
ç
ç
è 		Öx+ 1
Öx
 - 1
4   __
Öx3
 
ö
÷
÷
÷
÷
ø
are positive on [1,2]. (Öx ³ 1 and 1/(4[Ö(x3)]) £ 1/4 for 1 £ x £ 2.) So LEFT and MID underestimate the integral, RIGHT and TRAP overestimate the integral.


7.6, #4.

   n    10    100    1000
   LEFT    3.0132    2.9948    2.9930
   RIGHT    2.9711    2.9906    2.9925
   TRAP    2.9922    2.9927    2.9927
   MID    2.9930    2.9927    2.9927

Ö{3+cosq} is decreasing and concave-down on [0,p/2] since
(   æ
Ö
3+cosq
 
 )¢
= -sinq
2   æ
Ö
3+cosq
 
,
(   æ
Ö
3+cosq
 
 )¢¢
= 2(3+cosq)1/2(-cosq)+sinq (3+cosq)-1/2(-sinq)
4(3+cosq)
are negative on [0,p/2]. (The second derivative is messy, but its sign is easy to determine. The numerator is always negative since square roots are non-negative; the denominator is positive for 0 £ q £ p/2.) So LEFT and MID overestimate the integral, RIGHT and TRAP underestimate the integral.

File translated from TEX by TTH, version 2.00.
On 15 Aug 1999, 15:32.