calculus

1.Evaluate the integral. (Use C for the constant of integration.)
integral ln(sqrtx)dx
2. Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the curves about the given axis.
y = 4ex, y = 4e−x, x = 1; about the y-axis
3. Calculate the average value of f(x) = 2x sec2(x) on the interval [0, π/4].
4. Evaluate the integral. (Use C for the constant of integration.)
integral 7 ln(cubertx)dx

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  1. integration by parts is just the chain rule in reverse:
    d(uv) = u dv + v du
    So, you have
    #1.
    ∫ ln √x dx
    So, let u = ln √x = 1/2 lnx
    dv = dx
    Then du = 1/(2x) dx and v = x
    ∫ ln √x dx = uv - ∫ v du
    = x ln √x - ∫ 1/2 dx
    = x ln √x - 1/2 x
    = 1/2 x (lnx - 1) + C

    #2. You have a curved triangular region with vertices at (0,4), (1,4e), (1,4/e)
    so, using shells of thickness dx,
    v = ∫[0,1] 2πrh dx
    where r = x and h = 4e^x - 4e^-x
    v = ∫[0,1] 2πx(4e^x - 4e^-x) dx

    you can check your answer using discs (washers) of thickness dy, but you have to split the region into two parts because the left boundary changes where the curves intersect at (0,4). You also have to express x as a function of y.
    v = ∫[4/e,4] π(R^2-r^2) dy + ∫[4,4e] π(R^2-r^2) dy
    where R=1 and r changes from -ln(y/4) to ln(y/4)
    v = ∫[4/e,4] π(1-(-ln(y/4))^2) dy + ∫[4,4e] π(1-(ln(y/4))^2) dy
    but that's a bit more work ...

    #3. the average value of f(x) on [a,b] is (∫[a,b] f(x) dx)/(b-a)That is, the area divided by the width gives the average height. So, in this case, the average is
    (∫2x sec^2(x) dx)/(π/4 - 0)
    Use integration by parts, with u = 2x and dv = sec^2(x) dx

    #4.
    ∫7 ln∛x dx = 7/3 ∫ lnx dx
    See #1.

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    oobleck
  2. It just occurred to me that I said chain rule, when it's really the product rule.
    Bu I'm sure you caught that.

    Looking ahead, differentiation under the integral sign is the chain rule in reverse.

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    oobleck

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