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    So I'm still getting a little confused when Integrating by parts. The problem I tend to have is when I have to denote what is  u and  \frac{dv}{dx}

    I can sometimes see what will work but I can never fully explain why. Sometimes I do it by trial and error.

    For example if I were to do...
     \int x e^x dx

    I would know to let  u = x and  \frac{dv}{dx} = e^x

    But like I said, there are times where I don't know. Is there anything I should be looking out for?
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    (Original post by Philip-flop)
    So I'm still getting a little confused when Integrating by parts. The problem I tend to have is when I have to denote what is  u and  \frac{dv}{dx}

    I can sometimes see what will work but I can never fully explain why. Sometimes I do it by trial and error.

    For example if I were to do...
     \int x e^x dx

    I would know to let  u = x and  \frac{dv}{dx} = e^x

    But like I said, there are times where I don't know. Is there anything I should be looking out for?
    Look out for functions you know how to integrate, and which ones you do not. In general, you want your u to be such that it differentiates down to a constant eventually, or quicker/more nicely than the other option.

    In other times, you may have \displaystyle \int \ln(x) .dx which may seem impossible at first, but this can be done by parts if you let u=\ln(x) and v'=1. Now remember that the second integral will be u'v and in this case it turns out to be a nice integral since u'v=1. So in essence, you are aiming to split your main integral into some expression plus some easier integral.


    With your example it is an obvious choice by the second sentence I wrote. If you were to let u=e^x and v'=x then obviously you will keep differentiating u which will not change whatsoever, and you will keep integrating v' which will only keep increasing the exponent of x, not making your integral any easier - and this can explicitly be seen from a single IBP iteration that \displaystyle \int x e^x .dx =\frac{1}{2}x^2e^x - \int \frac{1}{2}x^2e^x .dx is not anything easier than what you've started with in any way whatsoever.
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    (Original post by RDKGames)
    Look out for functions you know how to integrate, and which ones you do not. In general, you want your u to be such that it differentiates down to a constant eventually, or quicker/more nicely than the other option.

    In other times, you may have \displaystyle \int \ln(x) .dx which may seem impossible at first, but this can be done by parts if you let u=\ln(x) and v'=1. Now remember that the second integral will be u'v and in this case it turns out to be a nice integral since u'v=1. So in essence, you are aiming to split your main integral into some expression plus some easier integral.


    With your example it is an obvious choice by the second sentence I wrote. If you were to let u=e^x and v'=x then obviously you will keep differentiating u which will not change whatsoever, and you will keep integrating v' which will only keep increasing the exponent of x, not making your integral any easier - and this can explicitly be seen from a single IBP iteration that \displaystyle \int x e^x .dx =\frac{1}{2}x^2e^x - \int \frac{1}{2}x^2e^x .dx is not anything easier than what you've started with in any way whatsoever.
    Thanks a lot man!

    I'm still trying to get there with trying to predict the outcome of the two possible ways of IBP. But atm I'm still kind of doing things by trial and error when I come across more complicated equations. I think you saying...
    "you want your u to be such that it differentiates down to a constant eventually, or quicker/more nicely than the other option"
    ... is possibly the best tip I've been given when it comes to IBP.

    Yeah  \int ln(x) caught me out the first time I ever came across that!
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    (Original post by Philip-flop)
    "you want your u to be such that it differentiates down to a constant eventually, or quicker/more nicely than the other option"
    If you follow this advice and remember that questions with \ln are often exceptions then you should be fine in the C4 exam.
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    (Original post by notnek)
    If you follow this advice and remember that questions with \ln are often exceptions then you should be fine in the C4 exam.
    Yeah I've added that tip to my notes now
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    Ok so I seem to be having a mental block when it comes to integrating to find the area under a curve. Here is the question I'm stuck on (part b)...

    Name:  Edexcel C4 June 2010 Q7.png
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    Can someone explain the rules for integrating to find the area under a curve? I seem to be confused by this one. I thought that if I were to integrate this then I would end up with zero, but that is not the case!
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    (Original post by Philip-flop)
    Ok so I seem to be having a mental block when it comes to integrating to find the area under a curve. Here is the question I'm stuck on (part b)...

    Can someone explain the rules for integrating to find the area under a curve? I seem to be confused by this one. I thought that if I were to integrate this then I would end up with zero, but that is not the case!
    Yes that would be true if you were integrating the explicit equation on the xy-plane, however, the equation of this curve is NOT explicit. It is 125y^2=(x+4)(41-x)^2. Making it explicit is impossible without losing half of the graph as you need to square root both sides for y on its own. Either way, you CAN do this and end up with two separate equations which represent the two halves of the graph. Integrating one between the limits of x will give you the positive area, and integrating the other will give you the same area just negative (curve is symmetric on y=0 which is why the area would be the same). So yes, adding them would give you 0. But area can't be negative, so instead of adding on the negative area, add on the positive version of it.


    Anyway, this is all nice and all, but doing it parametrically avoids all this implicit/explicit stuff. The reason why you do NOT get a 0 doing it parametrically is because you are integrating \displaystyle \int_{t=0}^{3} y\frac{dx}{dt}.dt

    Now what is y\frac{dx}{dt}?? This is f(t)=t(9-t^2)(10t)

    Plotting this on a separate plane, gives the following graph below. And clearly, we would NOT get 0 as the integral as you would expect on the xy plane.

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    (Original post by RDKGames)
    Yes that would be true if you were integrating the explicit equation on the xy-plane, however, the equation of this curve is NOT explicit. It is 125y^2=(x+4)(41-x)^2. Making it explicit is impossible without losing half of the graph as you need to square root both sides for y on its own. Either way, you CAN do this and end up with two separate equations which represent the two halves of the graph. Integrating one between the limits of x will give you the positive area, and integrating the other will give you the same area just negative (curve is symmetric on y=0 which is why the area would be the same). So yes, adding them would give you 0. But area can't be negative, so instead of adding on the negative area, add on the positive version of it.


    Anyway, this is all nice and all, but doing it parametrically avoids all this implicit/explicit stuff. The reason why you do NOT get a 0 doing it parametrically is because you are integrating \displaystyle \int_{t=0}^{3} y\frac{dx}{dt}.dt

    Now what is y\frac{dx}{dt}?? This is f(t)=t(9-t^2)(10t)

    Plotting this on a separate plane, gives the following graph below. And clearly, we would NOT get 0 as the integral as you would expect on the xy plane.

    Ok the only bit I'm confused about is when you use the word explicit

    Besides that I think I am following what you mean Thank you so much!

    Also thanks for taking the time to include a graph, that part really helped with understanding!
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    (Original post by Philip-flop)
    Ok the only bit I'm confused about is when you use the word explicit

    Besides that I think I am following what you mean Thank you so much!
    Explicit is when y (or x) is on its own on one side and the other side is entirely in terms of some other variable(s).

    So a circle has equation y^2=1-x^2, agreed?

    This is an implicit equation and defines a full circle.

    Take square roots of both sides and you end up with a explicit equations y=\sqrt{1-x^2} OR y=-\sqrt{1-x^2}. Each of these do NOT define the full circle. SO a circle's equation cannot be expressed explicitly. The first eq. represents the top half of the circle, and the bottom represents the bottom half of a circle.
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    (Original post by RDKGames)
    Explicit is when y (or x) is on its own on one side and the other side is entirely in terms of some other variable(s).

    So a circle has equation y^2=1-x^2, agreed?

    This is an implicit equation and defines a full circle.

    Take square roots of both sides and you end up with a explicit equations y=\sqrt{1-x^2} OR y=-\sqrt{1-x^2}. Each of these do NOT define the full circle. SO a circle's equation cannot be expressed explicitly. The first eq. represents the top half of the circle, and the bottom represents the bottom half of a circle.
    Oh right! I get what you mean by that now.

    So in a way, explicit means that one side of the equation is defined singularly by a variable e.g.
     y = 3x^2 + 2x +1 or  x = 2y + 3 etc

    whereas inplicit isn't like that. Just like what you explained with an equation of a circle e.g.  y^2 = 1 - x^2 , but if we were to split this up to make an explicit equation then we would actually end up with two explicit equations, therefore only half of the circle.
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    (Original post by Philip-flop)
    Oh right! I get what you mean by that now.

    So in a way, explicit means that one side of the equation is defined singularly by a variable e.g.
     y = 3x^2 + 2x +1 or  x = 2y + 3 etc

    whereas inplicit isn't like that. Just like what you explained with an equation of a circle e.g.  y^2 = 1 - x^2 , but if we were to split this up to make an explicit equation then we would actually end up with two explicit equations, therefore only half of the circle.
    Yep. This is the problem with the question as you cannot get 0 even if you split it into two explicit equations and are careful. Though the explicit equations would be more challenging to integrate anyway as well.
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    RDKGames just out of curiosity, how did you figure out the cartesian equation to be 125y^2=(x+4)(41-x)^2 from the parametric equations in the question?
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    (Original post by Philip-flop)
    RDKGames just out of curiosity, how did you figure out the cartesian equation to be 125y^2=(x+4)(41-x)^2 from the parametric equations in the question?
    Sub t^2=\frac{x+4}{5} into y^2 and simplify.
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    (Original post by RDKGames)
    Sub t^2=\frac{x+4}{5} into y^2 and simplify.
    Oh yeah, haha. Thank you!!
 
 
 
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