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Math Homework Solution
7.5, #46. The integrand is an improper rational function . Long division gives
Completing the square gives x^{2} + 6x + 13 = (x +
3)^{2} + 2^{2}, so set w = x + 3 and dw = dx.
Then
so formula 25 applies with a = 2, b = 6, c = 5, giving
7.5, #50. The integrand is an improper rational function . Long division gives
Since t^{2}  1 = (t  1)(t + 1), formula 26 with a = 1, b = 1 gives
7.5, #51. Formula 24 with a = 2 gives
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 x^{2} + 2x + 5 = (x + 1)^{2} + 2^{2}, formula 24 with a = 2 gives
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
(b) The average of V ^{2} over a second is
using the substitution u = 120πt and du = 120πdt. Thus
(c) If = 110, then V_{0} = = 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 
is increasing and concaveup on [1, 2] since
are positive on [1, 2]. (
and for )
So LEFT and MID
underestimate the integral, RIGHT and TRAP overestimate the
integral
7.6, #4.
n  20  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 
is decreasing and concavedown on [0, ] since
are negative on (The
second derivative is messy, but its sign is easy to determine.
The numerator is always negative since square roots are nonnegative; the
denominator
is positive for ) So LEFT and MID
overestimate the integral, RIGHT and
TRAP underestimate the integral.
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