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Fundamental Theorem of Calculus

  1. Mar 2, 2005 #1
    Could you check whether I am doing these questions right:

    1. [itex] \int_{0}^{4} (2+x) dx [/itex]. So I use the Fundamental Theorem of Calculus [itex] F(b)-F(a)[/itex] and receive: [itex] \frac{(x+2)^{2}}{2} = F(4) - F(0) = 16 [/itex]

    2. [itex] \int_{-1}^{1} (4t^{3} - 2t) dt = t^{4} - t^{2} = F(b)-F(a) = 0 [/itex]

    3. [itex] \int_{0}^{3} \frac{1}{\sqrt{1+x}} dx = \frac{-1}{2}(1+x)^\frac{-3}{2} = F(b) - F(a) = \frac{7}{16} [/itex]

    4. [itex] \int_{1}^{2}(\frac{1}{x^{2}} - \frac{1}{x^{3}}dx = \frac{x^{-1}}{-1} - \frac{x^{-2}}{-2} = F(b) - F(a) = \frac{1}{8}[/itex]
    5. How would you do this one: [itex] \frac{3+ \ln x}{x} dx [/itex]?
    6. [itex] \int^{1}_{-1} 3xe^{x^{2} -1} dx [/itex] Also how would I set this up? Would I let [itex] u = x^{2} - 1 [/itex]?

    If the marginal cost is [itex] \frac{dC}{dx} = 675 + 0.5x [/itex] how would C change when [itex] x [/itex] increases from 50 to 51? So [itex] C = 675x + \frac{1}{4} x^{2} [/itex]. So would I just compute [itex] F(51) - F(50) [/itex]?

    If you want to find the average value of [itex] f(x) = \frac{4}{\sqrt{x-1}}, [5,10] [/itex] would you use the formula [itex] \frac{1}{b-a}f'(x) [/itex]?




    Thanks :smile:
     
    Last edited: Mar 2, 2005
  2. jcsd
  3. Mar 2, 2005 #2
    1. correct
    2.correct
    3. Im not sure you integral is right. You might have mixed up derivarives with integration.

    You tell me the rest.

    Regards,

    Nenad
     
  4. Mar 2, 2005 #3
    evertyhing is good until you get to the ln x / x one
    i dont know of any method to do it, i simply stared at this one and came up with (ln x)^2 / 2

    and finally once again i stared keep the three out of it. Observe a chain rule that brought the x out in fdront since e always stays there anyway... Divide by that number in order to cancel its effect
     
  5. Mar 2, 2005 #4
    [tex]\int \frac{\ln{x}}{x}dx[/tex]

    can be evaluated with a simple u-substitution. Let u = ln(x), then du = dx/x, and the integral is [itex]\int u du[/itex].

    I'd be careful with your notation. The antiderivative does not equal F(b) - F(a), so statements like [itex]\int_{-1}^{1} (4t^{3} - 2t) dt = t^{4} - t^{2} = F(b)-F(a)[/itex] are simply not correct. [itex]F(t) = t^4 - t^2[/itex] and [itex]\int_{-1}^{1} (4t^{3} - 2t) dt = F(1) - F(-1)[/itex] are both true.

    --J
     
  6. Mar 2, 2005 #5
    how do you find the average value?
     
  7. Mar 2, 2005 #6
    The concept of "average" you're familiar with is most likely to add up the values of all your elements, and then divide by the number of elements. For continuous functions, "add up" is replaced by "integrate" and "number of elements" is replaced by "size of the interval", so your formula should look something like

    [tex]\frac{1}{b - a}\int_a^b f(x)dx[/tex]

    --J
     
  8. Mar 3, 2005 #7
    can you please show your steps for number 3. I'm getting a different answer when I evaluate it. I get an answer of 2


    Regards,

    Nenad
     
    Last edited: Mar 3, 2005
  9. Mar 3, 2005 #8

    dextercioby

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    The sollution provided by the OP for #3 is totally incorrect.
    [tex] \int_{0}^{3}\frac{dx}{\sqrt{1+x}}[/tex]
    U can integrate directly,or by a substitition:[itex] 1+x=u [/itex] and the final result is +2...

    Daniel.
     
  10. Mar 3, 2005 #9

    dextercioby

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    #6 has the answer "0",for the same reason as #2:symmetry (odd function integrated on a symmetric domain wrt origin).

    Daniel.
     
  11. Mar 3, 2005 #10

    Galileo

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    Here you mixed it up.

    Split the integral. Try [itex]u=\ln x[/itex].

    That would work.

    Yes, that is correct. Note though that you can only determine C up to a constant, since you only know the derivative of C. It doesn't matter since you need a difference in C(x).

    The average value over an interval [a,b] is:

    [tex]\frac{1}{b-a}\int_a^bf(x)dx[/tex].


    :smile:
     
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