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    In an integral, e.g

     \displaystyle\int 2y dy What does the  dy mean exactly?

    Wikipedia says:
    if x is a variable, then a change in the value of x is often denoted dx (or deltax when this change is considered to be small). The differential dx represents such a change, but is infinitely small. Although, as stated, it is not a precise mathematical concept, it is extremely useful intuitively, and there are a number of ways to make the notion mathematically precise.

    I kind of understand it when it is  dy/dx . But what does it mean in an integral? I got told it just meant "with respect to y" but in dy/dx, dy doesn't simply mean "with respect to y", so surely the meaning can't just change

    Thanks.
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    (Original post by H3rrW4rum)
    In an integral, e.g

     \displaystyle\int 2y dy What does the  dy mean exactly?

    Wikipedia says:
    if x is a variable, then a change in the value of x is often denoted dx (or deltax when this change is considered to be small). The differential dx represents such a change, but is infinitely small. Although, as stated, it is not a precise mathematical concept, it is extremely useful intuitively, and there are a number of ways to make the notion mathematically precise.

    I kind of understand it when it is  dy/dx . But what does it mean in an integral? I got told it just meant "with respect to y" but in dy/dx, dy doesn't simply mean "with respect to y", so surely the meaning can't just change

    Thanks.
    If you have \dfrac{dy}{dx} it can be read as "differentiate y, with respect to x," changing the variables how you will. With the integration stuff, you would read your example as "the integral of 2y, with respect to y." So in other words, treat it as you normally would. The only thing different is the variable.
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    It is basically means to integrate which is2(ysquared)/2 so just y squared.
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    (Original post by dknt)
    If you have \dfrac{dy}{dx} it can be read as "differentiate y, with respect to x," changing the variables how you will. With the integration stuff, you would read your example as "the integral of 2y, with respect to y." So in other words, treat it as you normally would. The only thing different is the variable.
    But hasn't dy changed from "differentiate y" to "with respect to y"?
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    It is just notation. Same as if you had say xsquared+2x dx, you would be integrating with respect to x.

    Simply put, at AS (which I presume you are doing) accept that dy/dx means differentiation so times the number in front of the x by the power then subtract 1 from the power

    And integration, as shown above, add 1 to the power then divide by the power.
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    (Original post by strawberryjellybaby)
    It is just notation. Same as if you had say xsquared+2x dx, you would be integrating with respect to x.

    Simply put, at AS (which I presume you are doing) accept that dy/dx means differentiation so times the number in front of the x by the power then subtract 1 from the power

    And integration, as shown above, add 1 to the power then divide by the power.
    Yeah I am doing C4 atm (differential equations) and basically wondered when at the start of the question you have the all-so-familiar dy/dx, which I know to be the derivative of y wrt x, then after a bit of work I am left with integrating wrt y. It's almost as if dy at the start is different to the dy at the end. I am sure it is obviously still "dy" but what confuses me is the notation of "derivative of y" changing to "wrt y".
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    Can you talk me through the notation here:

     \frac{dy}{dx} = 2xy

     \frac {1}{y} dy = 2x dx

     \displaystyle\int \frac {1}{y} dy =  \displaystyle\int 2x dx
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    In the first line, it just means "the gradient of y at a point (x,y) equals 2xy". As you probably know, the gradient of a line is the change in y divided by the change in x. The dy and dx represent infinitesimally small changes in y and x respectively; the ratio between these changes gives the gradient at a given point. This is expressed as a gradient function (in this case, 2xy), which is valid for any point on the curve. You aren't really dividing two values together on the left-hand side, that's just how it is conventionally represented.

    In the second line, it's not really referring to anything as such, but the \frac{dy}{dx} can be 'split' to prepare for integration. In the third line, each side is being integrated with respect to its own variable.
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    Suppose we have a curve y=f(x). Taking the integral corresponds to finding the area under that curve. We can approximate this by the trapezium rule:

    A \approx \displaystyle\sum_i^n \bar{y_i}\ \mathrm{d}x

    where dx is the strip width. If we let n \rightarrow \infty, then we have

    A = \displaystyle \lim_{n \rightarrow \infty} \displaystyle \left[\sum_i^n \bar{y_i}\ \mathrm{d}x\right]

    (\bar{y_i} tends to the single value y.) This is then given the alternative notation

    A = \displaystyle\int f(x)\ \mathrm{d}x

    So dx is the infinitesimal strip width.

    (Or something like that.)
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    (Original post by Cerdog)

    In the second line, it's not really referring to anything as such, but the \frac{dy}{dx} can be 'split' to prepare for integration. In the third line, each side is being integrated with respect to its own variable.
    Yeah it was kind of the second line that was confusing me because in FSMQ/C1/C2 I've seen dy in the following 2 forms:
     \frac {dy}{dx}
    and
      \displaystyle\int dy

    But I haven't seen it in the form shown in the second line before.
     \frac {1}{y} dy
    and I wondered what it meant in this part.

    Does it just mean "with respect to y" again?

    Sigh, sorry for all this. It doesn't help me whatsoever with the questions I just like to understand what I'm doing and why I'm doing it
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    (Original post by BJack)
    Suppose we have a curve y=f(x). Taking the integral corresponds to finding the area under that curve. We can approximate this by the trapezium rule:

    A \approx \displaystyle\sum_i^n \bar{y_i}\ \mathrm{d}x

    where dx is the strip width. If we let n \rightarrow \infty, then we have

    A = \displaystyle \lim_{n \rightarrow \infty} \displaystyle \left[\sum_i^n \bar{y_i}\ \mathrm{d}x\right]

    (\bar{y_i} tends to the single value y.) This is then given the alternative notation

    A = \displaystyle\int f(x)\ \mathrm{d}x

    So dx is the infinitesimal strip width.

    (Or something like that.)
    Pretty much this, although the "dx" you have should probably \Delta x to show it's a change in the width of the strips that we are summing. It is defined as \frac{b - a}{n} where b and a are the upper and lower limits respectively. As n goes to infinity this becomes "dx" to mean the infinitesimally small strips.
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    (Original post by H3rrW4rum)
    Yeah it was kind of the second line that was confusing me because in FSMQ/C1/C2 I've seen dy in the following 2 forms:
     \frac {dy}{dx}
    and
      \displaystyle\int dy

    But I haven't seen it in the form shown in the second line before.
     \frac {1}{y} dy
    and I wondered what it meant in this part.

    Does it just mean "with respect to y" again?
    I think in that line it just refers to a small change in y. On the other hand, it might not really mean anything; it could just be a convenient way to write things before integrating.
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    (Original post by Cerdog)
    I think in that line it just refers to a small change in y. On the other hand, it might not really mean anything; it could just be a convenient way to write things before integrating.
    No it definitely means something as it was present in the dy/dx in the line above.
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    (Original post by H3rrW4rum)
    No it definitely means something as it was present in the dy/dx in the line above.
    The term/symbol in general does mean something, but there might not be any sort of special meaning for this context. When you separate the variables, you aren't really splitting a fraction, but it can be treated as one for this action.
 
 
 
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