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Clippy's Calculus Thread!

  1. Apr 24, 2007 #1
    Instead of cluttering the forum with all my cries for help, I decided to condense them all into one!
    First, I know this is basic, but I just can't seem to figure it out:

    [tex]Int (x^2+1)/x^2 dx[/tex]

    [tex]x^3/x^3+x/(x^3/3)[/tex] ??

    Thanks in advance!
  2. jcsd
  3. Apr 24, 2007 #2


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    Ack, no... It's not correct. =.="

    Ok, what you should remember is:
    [tex]\int x ^ \alpha dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C , \ \ \ \alpha \in \mathbb{R} \ {-1}[/tex]
    [tex]\int dx = x + C[/tex]
    In case [tex]\alpha = -1[/tex], we have:
    [tex]\int x ^ {-1} dx = \int \frac{dx}{x} = \ln |x| + C[/tex]

    You should also note that:
    x-m = 1 / xm.
    So the first one goes like this:

    [tex]\int \frac{x ^ 2 + 1}{x ^ 2} dx = \int \frac{x ^ 2}{x ^ 2} dx + \int \frac{dx}{x ^ 2} = \int dx + \int x ^ {-2} dx = ...[/tex]
    Can you finish the problem now? :)
  4. Apr 24, 2007 #3

    Thanks for the wonderfully quick reply! The thing is, I have a terrible (TERRIBLE!) memory for anything regarding numbers, so I'll suddenly forget how to integrate x. I know, it's that bad. So...practice makes perfect!
  5. Apr 24, 2007 #4


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    Yup, it's correct. But don't forget the Constant of Integration (i.e, the '+ C' thingy). :)

    You can make it look a little bit nicer, by changing x-1 to 1 / x:
    [tex]\int \frac{x ^ 2 + 1}{x ^ 2} dx = x - \frac{1}{x} + C[/tex]
  6. Apr 24, 2007 #5
    You're a dote! Thank you!
    Hang around for some more fun integration.. bring a coffee, we can have a chat as well!
  7. Apr 28, 2007 #6
    This might be an odd question, but what exactly is LIAT? I keep seeing people writing it on the margin of their calculus- is it a method to remember what to check for first?

    My main problem right now is I can never remember which method to use- u-substitution, uv-intvdu, integrating by parts, integration by partial fractions.... any good tips as to how to know what to do first?

    I know how to do them, I just don't know when to do them! Do I simplify things first, because sometimes it appears you don't. Help!
  8. Apr 28, 2007 #7

    Gib Z

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    Well It actually should be LIATE. It stands for Logarithmic, Inverse Trig, Algebraic, Trig, Exponential. That is the order to go to pick your "u" when doing integration by parts. Which leads me to your next paragraph, [itex]uv-\int v du[/itex] IS integrating by parts.

    You use u substitution when you see a function and the functions derivative in the integrand. Eg [itex]\int 2x \sin (x^2) dx[/itex]. There is x^2 in the sin, and there is 2x as well. Then let u=x^2 and you can see the rest.

    Integrating by partial fractions only works when you have a rational function and the denominator must be factor-able. A rational function is the ratio of 2 polynomials eg [tex]\frac{x^2+2x+53}{(x^8+1)(x-1)}[/tex] is a rational function.

    Integrate by parts when you can't think of any other method that could help. EG [itex]\int e^x \sin x dx[/itex] No substitution will help, its not a rational function and any other methods don't work on it. So try integration by parts and this problem is such a common basic Integration by parts question that i think i even remember the answer..check me on this but its [tex]\frac{e^x(\cos x - \sin x)}{2}[/tex]
  9. May 6, 2007 #8
    These comments are so helpful!! It was confusing me before, but now the murky waters of calculus are starting to get a little clearer... but one more question, when do you integrate by parts? For example the example you listed, why couldn't we use [tex]u=e^x [/tex]and [tex] dv=sin x[/tex] to produce [tex]uv-intvdu[/tex]?
  10. May 6, 2007 #9


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    This is precisely what he said you must do.

    This is not a particularly easy integration by part problem, but it's a classic and profs. love to put it in exams. At the "math-help center" at my university, when someone comes in for help about integrals, this problem is invariably in the list. :smile:
  11. May 6, 2007 #10
    Ahh...I get it now.Thanks! My mind is so slow today, it could mean I need more coffee, heh.

    I'm a pharmacy student and more into the physiology side of things than math... I like math, but I like it more when I know how to do it. ;)

    Question- what's it like to be good at math/physics?? I bet you guys get a lot of people coming up to you with random questions. I must memorise that integration by parts problem, who knows if it'll come up on the exam. I'll be back soon with (hopefully) the work!
    Another question here though-- I have a definite integrand with an inverse function and the worked problem shows that you inverse the boundaries- why/how do you do this?
  12. May 6, 2007 #11


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    It would be best to state the problem exactly as it is written in your book.
  13. May 9, 2007 #12
    I figured out what the problem was- the function was to the power of x, not 2. It was from a lecture example. No wonder I couldn't get the right answer. Talk about wasting time!!
    Another question for you to try and gently explain to my puny little brain: the whole P(t)=Poe^kt deal. See, the thing is, I 'understand' the equation but I'm having trouble figuring out exactly how everything relates to each other. Could you give me some pointers as to how to play around with it?
  14. May 9, 2007 #13


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    Not sure what you're asking but if you think of t as the time and P(t) as the number of individuals in a population of say, bacteria, then you notice that P(0)=Po (because e^0=1). So Po is the initial amount of people in the population. As to k, well, the bigger the k, the faster the population grows (because, for the sake of giving an exemple, if we look at a population growing as Poe^2t and one growing as Poe^3t, then at time t=1 say, the first population is at Poe^2 and the other is at Poe^3.... so the one with the greater k grows faster)
  15. May 14, 2007 #14
    Hey, thanks for the explanation...I understand it a bit better! I see from your title that you're all into physics, so can you tell me if I have the right end of the stick with this:


    Y=30 sin 40=19.28m
    X=30 cos 40=22.98

    Displacement in the X-direction:

    x=Vot+ 0.5 at^2, a=o, so cancel out (0.5 at)
    x= (22.98) (t)

    Need to find time when Y=-4, taking up as positive:

    -4=30(t)-0.5(8.314)(t^2)(multiply both sides by half)
    8+30t=8.314t^2 (t cancels out)
    38/8.314=t, t = 4.57

    FIND X WHEN t=-4:
    x=Vot= (22.98)(4.57)

    Therefore the launch site needs to be 105.02m away from the target in order to hit it.

    Does this seem correct to you?
    Last edited: May 14, 2007
  16. May 14, 2007 #15
    1) I hope that by Y and X you mean velocity along Y direction and velocity along X direction respectively.

    2) The value of g on the earth's surface is approximately taken as 9.8m/s^2. So unless you're shooting in some other celestial body.....

    Tut,tut... you need to use the vertical component of the velocity.

    No it doesn't. Look again.
    Once you find t after solving the quadratic equation, you may proceed as you have done.
    You can say that the target has to be so much distance away from the launch site, which would have been all that is required if it hadn't been for that pesky wall. So use the fact that the projectile reaches maximum height just at the top of the wall, and see if you can find how far the launch site should be from the wall.

  17. May 16, 2007 #16
    y=30sin 40= 19.28m

    (a) (b) (c)

    Use quadratic formula [- b +/- sqrt (b^2-4ac)] / 2a

    [-19.28 +/- sqrt (19.28^2-4 (-9.8)(8)]/2(-9.8)]

    and so on forth, until t is found to equal -0.3519 and 2.319. T must be positive!

    Therefore, we take 2.319 as t and input into dx=vot.



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