Callahan
    Drill Sheet 5 (Odd numbers only if that is enough, but even numbers too if you need more practice.)
    6.3: 17, 19
    6.4: 11, 12
    6.4: 13, 14  (p351 Average of a Functions)

6.3: 17)

    a)  Sketch the graphs of and over the interval .

    b)  Find by visualizing the integral as a signed area.

    As you can see on the lower graph the negative and positive areas between the curve and the x-axis are exactly equal.  So, the integral must equal zero by symmetry.

    c)  Find .  Why does have the same value?

    Like the values of the lower curve, the values of the upper curve average to the value around which the cosine oscillates.  This integral would have a different value if the limits of integration were not a multiple of the period (2π).  Also by the rule for addition of integrals:

19)

    a)  On what interval does the graph of the function lie above the x-axis?

    b)  Sketch the graph of on the interval a you determined in part (a).

    c)  What is the area of the region that lies above the x-axis and below the graph of ?

6.4: 11)  Rules for finding antiderivatives:  Ch 11 (p609)

    a)

    b)

    To find our antiderivative we just think of the integrand as a derivative of a function :

    Guess cosine:

    Now simply set our derivative we found equal to and solve for :

    And substituting we find our antiderivative :

    And so our integral is:

    c)

    To find our antiderivative we just think of the integrand as a derivative of a function :

    Make our guess:

    Now we simply set the derivative we found equal to and solve for :

    And substituting we find our antiderivative :

    And so our integral is:

    d)

    To find our antiderivative we just think of the integrand as a derivative of a function :

    Guess :

    Now we just set the derivative we found equal to and solve for :

    And substituting we find our antiderivative :

    And so our integral is:

    e)

    f)

    g)

    To find our antiderivative we just think of the integrand as a derivative of a function :

    Guess :

    Now we just set the derivative we found equal to and solve for :

    And substituting we find our antiderivative :

    And so our integral is:

    h)

12)  Find a formula for the solution of each of the following initial value problems:

    a)  

    To find a formula for the solution, we integrate both sides of the differential equation:

    To find our antiderivative we just think of the integrand as a derivative of a function :

    Make our guess:

    Now we simply set the derivative we found equal to and solve for :

    And substituting we find our antiderivative :

    And so our integral is (because this is an indefinite integral we add a constant to be determined by the initial condition):

    Now using the initial condition we can determine the value of :

    So our specific solution for this initial condition is:

    b)  

    c)  

    d)  

    To find a formula for the solution, we integrate both sides of the differential equation:

    To find our antiderivative we just think of the integrand as a derivative of a function :

    Make our guess:

    Now we simply set the derivative we found equal to and solve for :

    And substituting we find our antiderivative :

    And so our integral is (because this is an indefinite integral we add a constant to be determined by the initial condition):

    Now using the initial condition we can determine the value of :

    So our specific solution for this initial condition is:

13)  Find the average value of each of the following functions over the indicated interval:    (p351)

    a)   over

    b)   over

    To find our antiderivative we just think of the integrand as a derivative of a function :

    Make our guess at that function:    (This one would take some trial and error!)    (p364)

    Now we simply set the derivative we found equal to and solve for :

    And substituting we find our antiderivative :

    And so our average value integral is:

    c)   over

14)

    a)  What is the average value of the functions over the interval ?

    To find our antiderivative we just think of the integrand as a derivative of a function :

    Make our guess:

    Now we simply set the derivative we found equal to and solve for :

    And substituting we find our antiderivative :

    And so our average value integral is:

    b)  For which value of will that average be zero?