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Homework Help: Problems with def. integrals/fundamental theorem

  1. Aug 4, 2008 #1
    1. Find [tex]\frac{dy}{dx}[/tex] and [tex]\frac{d^{2}y}{dx^{2}}[/tex] if [tex]\int^{3x}_{1} \frac{1}{t^{2}+t+1}\,dt[/tex]

    I expect that I'd make [tex]u=3x[/tex], then [tex]du=3dx[/tex]. I think when I differentiate, I'd end up with [tex]\frac{dy}{dx}=\frac{1}{3t^{2}+3t+3}[/tex]. I think that [tex]\frac{d^{2}y}{dx^{2}}[/tex] would just be the derivative of [tex]\frac{dy}{dx}[/tex]

    2. Find and classify the relative maxima and minima of [tex]f(x)[/tex], if
    f(x) = [tex]\int^x_0 \frac{t^{2}-{4}}{{1}+{cos}^{2}{t}}\,dt[/tex]

    I think to find max and min, I just need to find the second derivative and solve for zero right? Is the first derivative [tex]\frac{x^{2}-{4}}{{1}+{cos}^{2}{x}}[/tex]?
    Last edited: Aug 4, 2008
  2. jcsd
  3. Aug 4, 2008 #2
    Assuming that y is the integral expression in #1, then I guess you could use substitution, but it'd be easier just to apply the chain rule. But either way, you need to substitute u = 3x in place of t so dy/dx should be [1/(9x^2 + 3x +1)] * 3. I think it's a lot less confusing if you applied the FTC (remember the parameter t should disappear) by substituting 3x in place of t and then applied the chain rule to the upper limit of integration.

    The second derivative is then just the derivative of 3/(9x^2 + 3x +1), which is easy.

    For #2, you applied the FTC correctly but your subsequent reasoning is a bit flawed. To find the relative min and max, you need to find the zeros of f'(x), which you determined correctly. You could use the second derivative test to verify whether each critical point (zeros of f'(x)) is a relative max or min (or neither). Suppose x = a is a zero of f'(x). If f''(a) < 0, then f(x) has a relative max at x = a and if f''(a) > 0, then f(x) has a relative min at x = a. Of course just looking at sign changes prevents you from having to take the second derivative of f(x).
    Last edited: Aug 4, 2008
  4. Aug 4, 2008 #3
    So I've got the first one now, it's quite easy due to the help i've recieved.

    However, on the second one. I have my x values for the extrema, but I can't figure out the y values because I'm not sure how to take the integral of [tex]\frac{t^2-4}{1 + cos^{2}t}[/tex]. I just need to take the def. integral from (0,2) and from (0,-2). Also, I need that integral because I can't figure out how I'd plug in values around 2 and -2 to figure whether they were max/min
  5. Aug 4, 2008 #4


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    You don't need to plug values in if you can figure out whether the second derivative is positive or negative at those x values.
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