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1) Find the area of the region bounded by the curves y=arcsin (x/4), y = 0, and x = 4 obtained by integrating with respect to y. Your work must include the definite integral and the antiderivative.

2)Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = -1.

3)The base of a solid in the xy-plane is the first-quadrant region bounded y = x and y = x^2. Cross sections of the solid perpendicular to the x-axis are squares. What is the volume, in cubic units, of the solid?

4)Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x3 - 9x on the interval [-1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.

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3 answers

  1. I've been trying to use the notes and past problems as a stepping stool to help me figure these out but I'm so lost. Especially number 4

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  2. Mean Value Theorem for Integrals. A variation of the mean value theorem which guarantees that a continuous function has at least one point where the function equals the average value of the function.

    Mean value= (f(1)+f(-1))/2
    = (-8+8)/2=0

    if f(x)=0=x^3-9x, the the one point is x=0

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  3. #1
    ∫[0,4] arcsin(x/4) dx = 2π-4

    To do this one, let
    x = 4sin(u)
    arcsin(x/4) = arcsin(sin(u)) = u
    dx = 4cos(u) du

    Now you have

    ∫[0,4] 4u cos(u) du
    That you can easily do using integration by parts. Note that in changing variables,

    ∫[0,4] f(x) dx = ∫[0,π/2] g(u) du

    #2
    Using discs of thickness dx,

    v = ∫[1,e] π(R^2-r^2) dx
    where R=2 and r=lnx+1
    v = ∫[1,e] π(4-(lnx + 1)^2) dx

    Using shells of thickness dy,

    v = ∫[0,1] 2πrh dy
    where r = y+1 and h=x-1=e^y-1
    v = ∫[0,1] 2π(y+1)(e^y-1) dy

    #3 Each square has base equal to the distance between the curves, so adding up all the areas,

    v = ∫[0,1] (x-x^2)^2 dx = 1/30

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