Ch 11.2:  1(k,n,o), 2, 4, 5, 6, 8

1)  Evaluate the following using substitution:

    k)

[Graphics:Images/calculus_gr_1.gif]
[Graphics:Images/calculus_gr_2.gif]

    n)

[Graphics:Images/calculus_gr_3.gif]
[Graphics:Images/calculus_gr_4.gif]

    o)

[Graphics:Images/calculus_gr_5.gif]
[Graphics:Images/calculus_gr_6.gif]
[Graphics:Images/calculus_gr_7.gif]

3)  This question concerns the integral [Graphics:Images/calculus_gr_8.gif]:

    a)  Find [Graphics:Images/calculus_gr_9.gif] by using the substitution [Graphics:Images/calculus_gr_10.gif]:

[Graphics:Images/calculus_gr_11.gif]
[Graphics:Images/calculus_gr_12.gif]

    b)  Find [Graphics:Images/calculus_gr_13.gif] by using the substitution [Graphics:Images/calculus_gr_14.gif]:

[Graphics:Images/calculus_gr_15.gif]
[Graphics:Images/calculus_gr_16.gif]

    c)  Compare your answers to (a) and (b).  Are they the same?  If not, how do they differ?  Since both answers are antiderivatives of [Graphics:Images/calculus_gr_17.gif], they should differ only by a constant.  Is that true here?  If so, what is the constant?

    The answers are certainly not in the same form.  However, if we let the integration constants equal zero we can find out how much the answers differ from each other:

[Graphics:Images/calculus_gr_18.gif]

    So, with the use of the trig. identity [Graphics:Images/calculus_gr_19.gif] we find that the two answers differ by the constant [Graphics:Images/calculus_gr_20.gif].

    d)  Now calculate the value of the definite integral:

[Graphics:Images/calculus_gr_21.gif]

    using the two different u-values from (a) and (b):

[Graphics:Images/calculus_gr_22.gif]
[Graphics:Images/calculus_gr_23.gif]

    The answers match, as they must.

[Graphics:Images/calculus_gr_24.gif]

4)

    a)  Find all functions [Graphics:Images/calculus_gr_25.gif] that satisfy the differential equation:

[Graphics:Images/calculus_gr_26.gif]

    First we use separation of variables (put all terms involving x on one side of the equation and all term involving y on the other), then we integrate using substitution:

[Graphics:Images/calculus_gr_27.gif]
[Graphics:Images/calculus_gr_28.gif]
[Graphics:Images/calculus_gr_29.gif]

    b)  From among the functions [Graphics:Images/calculus_gr_30.gif] you found in part (a), select the one that satisfies [Graphics:Images/calculus_gr_31.gif].

    So using the initial condition [Graphics:Images/calculus_gr_32.gif] we can determine the integration constant [Graphics:Images/calculus_gr_33.gif]:

[Graphics:Images/calculus_gr_34.gif]

    So with the initial condition [Graphics:Images/calculus_gr_35.gif], the specific solution is:

[Graphics:Images/calculus_gr_36.gif]

    c)  From among the functions [Graphics:Images/calculus_gr_37.gif] you found in part (a), select the one that satisfies [Graphics:Images/calculus_gr_38.gif].

    So using the initial condition [Graphics:Images/calculus_gr_39.gif] we can determine the integration constant [Graphics:Images/calculus_gr_40.gif]:

[Graphics:Images/calculus_gr_41.gif]

    So with the initial condition [Graphics:Images/calculus_gr_42.gif], the specific solution is:

[Graphics:Images/calculus_gr_43.gif]
[Graphics:Images/calculus_gr_44.gif]

5)  Find a function [Graphics:Images/calculus_gr_45.gif] that solves the initial value problem:

[Graphics:Images/calculus_gr_46.gif]

    First we find the general solution using separation of variables (put all terms involving t on one side of the equation and all term involving y on the other), then we integrate using substitution:

[Graphics:Images/calculus_gr_47.gif]
[Graphics:Images/calculus_gr_48.gif]
[Graphics:Images/calculus_gr_49.gif]

    Using the initial condition [Graphics:Images/calculus_gr_50.gif] and the general solution we can determine the integration constant [Graphics:Images/calculus_gr_51.gif]:

[Graphics:Images/calculus_gr_52.gif]

    So with the initial condition [Graphics:Images/calculus_gr_53.gif], the specific solution is:

[Graphics:Images/calculus_gr_54.gif]
[Graphics:Images/calculus_gr_55.gif]

6)    (p351)

    a)  What is the average value of the function [Graphics:Images/calculus_gr_56.gif] on the interval [Graphics:Images/calculus_gr_57.gif]?

    If we think of a rectangle with the same length and area as the function above over the interval, the corresponding height of the rectangle is the average value:

[Graphics:Images/calculus_gr_58.gif]

    The length of the interval is the length of the rectangle (2).  To get the area, we just compute the integral of the function over the x-interval:

[Graphics:Images/calculus_gr_59.gif]
[Graphics:Images/calculus_gr_60.gif]
[Graphics:Images/calculus_gr_61.gif]

    b)  Show that the average value of the function on the interval [Graphics:Images/calculus_gr_62.gif] is [Graphics:Images/calculus_gr_63.gif].  Sketch the graph of f(x) on this interval:

[Graphics:Images/calculus_gr_64.gif]

    As we can see from the graph the function is symmetrical and so has the average value of [Graphics:Images/calculus_gr_65.gif] over the interval [Graphics:Images/calculus_gr_66.gif].  But lets find the value analytically:

[Graphics:Images/calculus_gr_67.gif]
[Graphics:Images/calculus_gr_68.gif]
[Graphics:Images/calculus_gr_69.gif]

8)  

    a)  Use a computer graphing utility to confirm that:

[Graphics:Images/calculus_gr_70.gif]

[Graphics:Images/calculus_gr_71.gif]

[Graphics:Images/calculus_gr_72.gif]

[Graphics:Images/calculus_gr_73.gif]

[Graphics:Images/calculus_gr_74.gif]

[Graphics:Images/calculus_gr_75.gif]

    b)  Find a formula for:

[Graphics:Images/calculus_gr_76.gif]
[Graphics:Images/calculus_gr_77.gif]

    So our formula is:

[Graphics:Images/calculus_gr_78.gif]

    c)  What is the average value of [Graphics:Images/calculus_gr_79.gif] on the interval [Graphics:Images/calculus_gr_80.gif]?

[Graphics:Images/calculus_gr_81.gif]
[Graphics:Images/calculus_gr_82.gif]

    What is the average value of [Graphics:Images/calculus_gr_83.gif] on the interval [Graphics:Images/calculus_gr_84.gif] ([Graphics:Images/calculus_gr_85.gif])?

[Graphics:Images/calculus_gr_86.gif]
[Graphics:Images/calculus_gr_87.gif]

    d)  Explain your results in part (c) in terms of the graph of [Graphics:Images/calculus_gr_88.gif] you drew in part (a):

    Because the function is perfectly periodic with a period of [Graphics:Images/calculus_gr_89.gif], its average value will always be [Graphics:Images/calculus_gr_90.gif] over any interval that is a integer multiple of its period.

    e)  Here's a differential equation proof of the identity in part (a).  Let [Graphics:Images/calculus_gr_91.gif], and let [Graphics:Images/calculus_gr_92.gif].  Show that both of these functions satisfy the initial value problem::

[Graphics:Images/calculus_gr_93.gif]

    Hence conclude the two functions must be the same.

    So lets start by computing the second derivative of [Graphics:Images/calculus_gr_94.gif] and comparing it to the result of plugging [Graphics:Images/calculus_gr_95.gif] into the [Graphics:Images/calculus_gr_96.gif] equation:

[Graphics:Images/calculus_gr_97.gif]
[Graphics:Images/calculus_gr_98.gif]
[Graphics:Images/calculus_gr_99.gif]
[Graphics:Images/calculus_gr_100.gif]

    Now lets check that the initial conditions are satisfied:

[Graphics:Images/calculus_gr_101.gif]
[Graphics:Images/calculus_gr_102.gif]

    So [Graphics:Images/calculus_gr_103.gif] satisfies the given initial value problem.

[Graphics:Images/calculus_gr_104.gif]

    And for [Graphics:Images/calculus_gr_105.gif] lets start by computing the second derivative of [Graphics:Images/calculus_gr_106.gif] and comparing it to the result of plugging [Graphics:Images/calculus_gr_107.gif] into the [Graphics:Images/calculus_gr_108.gif] equation:

[Graphics:Images/calculus_gr_109.gif]
[Graphics:Images/calculus_gr_110.gif]
[Graphics:Images/calculus_gr_111.gif]
[Graphics:Images/calculus_gr_112.gif]

    Now lets check that the initial conditions are satisfied:

[Graphics:Images/calculus_gr_113.gif]
[Graphics:Images/calculus_gr_114.gif]

    So [Graphics:Images/calculus_gr_115.gif] satisfies the given initial value problem.

[Graphics:Images/calculus_gr_116.gif]

    But because an initial value problem only has one solution, both of these solutions must be equivalent.  So:

[Graphics:Images/calculus_gr_117.gif]


Converted by Mathematica      May 5, 2004