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Definite integral involving natural log

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.B (LO)
,
FUN‑6.C (LO)
Sal finds the definite integral of (6+x²)/x³ between 2 and 4. To do that, he has to use the integral of 1/x, which is ln(x).

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  • starky ultimate style avatar for user LBKA
    Would it be correct to consider (3/4)^2 a simplification of 9/16?
    (5 votes)
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  • leafers seedling style avatar for user Nathan Presta-Hislop
    integrals are the difference between high bound and low bound
    (4 votes)
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  • starky ultimate style avatar for user Aarav Shah
    Why is there no '+ C' in the final answer to the question? Did Sal forget or is there some specific reason to not include it?
    (3 votes)
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    • male robot hal style avatar for user Yana Veitsman
      Not like he didn't include something, but it just cancels out.
      If you remember the fundamental theorem, the integral from a to b of f(x)dx = F(b) - F(a). Thus, there will be +C wherever there is F(b) and because of that minus in front of F(a) - -C wherever there is F(a). In the end, they will be just C - C, which would equal to 0.
      (5 votes)
  • leaf orange style avatar for user Robert Tackett
    Could you explain or elaborate about the use the absolute value signs for f(x) = ln (|x|), based on the points I've made below?

    First:
    a) The function f(x) = ln (x) has it's domain defined on the the interval from (0, + infinite).
    b) The function f(x) = ln (-x) has it's domain defined on the interval from (- infinite, 0).
    The function f(0) = ln (0) is undefined.

    Likewise:
    c) The function f(x) = 1/x has it's domain defined from (- infinite, 0) and (0, + infinite), with the first interval returning negative values, and the second returning positive values.
    d) The function f(0) = 1/0 is undefined.

    So it seems using the integral of 1/x = the ln(|x|) [+ C ], could lead to misapplications of the integral, or misinterpretations of the answers:
    1a) For example, it seems it would be meaningless to take the definite integral of f(x) = 1/x dx between negative and positive bounds, say from - 1 to +1, because including 0 within these bounds would cross over x = 0 where both f(x) = 1/x and f(x) = ln (x) are both undefined.
    1b) But, it seems, integrating f(x) = 1/x by saying the integral is ln (|x|) [+ C] on an interval between an upper positive value "a" and a lower negative value "b" (which would thus include x = 0 as part of the bounded interval of the definite integral) could be "interpreted" as ln (|a|) - ln (|b|) and seem to return a "defined" answer, but the actual result would be undefined.

    2a) Second, it seems the area "under" the curve of f(x) = 1/x on the domain from
    (- infinite, 0) would be a negative value, as the area of the function would be "below" the x-axis, and the integral of f(x) = 1/x on this interval would be - ln (- x) [+ C], not simply ln (|-x|) [+ C].
    [Note: d/dx ( ln (-x)) would return (-1/(-x)) (or simply 1/x) by the chain rule, so couldn't the anti-derivative of 1/x on the domain interval (-infinite, 0) be rewritten as the anti-derivative of f(x) = 1/(-x) (-dx)?
    2b) Thus, it seems using ln (|x|) on this interval would not indicate that the area is below the x-axis.

    Yes, I know geometrically, area is always positive, but because of the above points ... well ... I'll just say I do have some confusion about the "validity" of using ln (|x|), at least using it "willy nilly". So, is it possible to "explain away" my confusion?

    Thanks
    (2 votes)
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    • leaf yellow style avatar for user Howard Bradley
      On your point 1:
      Yes, you're right it could be misapplied. One always has to be careful. In order for the Fundamental Theorem of Calculus to hold the integrand needs to be continuous over the interval. This is not the case for 1/x for any interval that goes from negative to positive. We say the integral does not converge.
      Try typing this into Wolfram Alpha: integral 1/x, x from -1 to 1

      On point 2.
      The area would be negative. Take the integral of 1/x for x = -e to -1.
      A = ln|-1| - ln|-e| = 0 - 1 = -1, which is negative.
      Or did I misunderstand the point you were making?
      (5 votes)
  • duskpin ultimate style avatar for user Lee
    One thing that has been confusing me is that the derivative of x^-2 is -1/x, which looks very close to 1/x (whose anti-derivative is ln|x|). But, if you saw an expression like the integral of -1/x, you could either treat the negative as a constant multiple -1 and the anti-derivative would be -ln|x|, OR you could treat it as part of the expression, and the anti-derivative would be x^-2.

    Both seem valid, but how is it possible to have two answers? Is there any relationship between these two anti-derivatives?
    (1 vote)
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  • starky sapling style avatar for user kaferkin
    Is the "bar/line" notation for integrals at around to official? or is it something Sal came up with?
    (1 vote)
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  • stelly orange style avatar for user jungbeenjo
    What if the bounds become negative, then would the integral be undefined as x in ln(x) cannot be negative?
    (1 vote)
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  • blobby green style avatar for user mashrafe.03
    about the integration part involving the natural log-- if it was something like (1-x)^-1 instead of just x, then what would its anti-derivative be?
    (1 vote)
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  • aqualine sapling style avatar for user YODASMUSTACHE
    Why didn't the bounds change? For most definite integrals I've done, the bounds have changed. When do we determine whether or not to change bounds?
    (1 vote)
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  • duskpin ultimate style avatar for user Claire Miller
    I was wondering why the integral of 1/x is always ln(x). For any constant k, the derivate of ln(kx) equals 1/x, doesn't it? So it seems to me that the anti-derivative, or integral, of 1/x should include solutions such as ln(2x) and ln(-x/5) as well. Does anybody understand why this is not the case?
    (1 vote)
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Video transcript

- [Voiceover] Let's now take the definite integral from two to four of six plus x squared over x to the third power dx. At first this might seem pretty daunting. I have this rational expression. But if we just rewrite this, it might jump out at you how this could be a little bit simpler. So this is equal to the integral from two to four of six over x to the third power plus x squared over x to the third power dx. I just separated this numerator out. I just divided each of those terms by x to the third power. And this I could rewrite. This is equal to the integral from two to four of six x to the negative three power, that's that first term there. And x squared divided by x to the third, well that it going to be one over x. So plus one over x dx. Now this is going to be equal to, let's take the antiderivative of the different parts and we're going to evaluate that at four and we're going to evaluate that at two. And we're going to find the difference between the antiderivative valued at four and at two. Now what is the antiderivative of six x to the negative three? Well, here, once again we can just use, we could use the power rule for taking the antiderivative or it's the reverse of the derivative power rule. We know that if we're taking the integral of x to the n dx, the antiderivative of that is going to be x to the n plus one over n plus one. And if we were just taking an indefinite integral there would be sum pluc C. The reason why we don't put the plus Cs here is when you evaluate at both bounds of integration, the C would cancel out regardless of what it is. So we don't really think about the C much when we're taking definite integrals. But let's apply that to six x to the negative third power. So it's going to be, we're going to take x to the negative three plus one. So it's x to the negative two. And so we're gonna divide by negative two as well. And, of course, we had that six out front from the get go. So that's the antiderivative of six x to the negative three power. And what's the antiderivative of one over x? You might be tempted to use the same idea right over here. You might be tempted to say, all right, well the antiderivative of x to the negative one, which is the same thing as one over x would be equal to x to the negative one plus one over negative one plus one. But what is negative one plus one? It is zero. So this doesn't fit this property right over here. But lucky for us, there is another property. And we went the other way when we were first taking derivatives of natural log functions. The antiderivative of one over x or x to the negative one is equal to, sometimes you'll see it written as natural log of x plus C. And sometimes and I actually prefer this one, because you could actually evaluate it for negative values, is to say the absolute value, the natural log of the absolute value of x. And this is useful, because this is defined for negative values, not just positive values. The natural log of x is only defined for positive values of x, but when you take the absolute value, now it could be negative or positive values of x. And it works, the derivative of this is indeed one over x. Now it's not so relevant here, because our bounds of integration are both positive. But if both of our bounds of integration were negative, you could still do this by just reminding yourself that this is a natural log of absolute value of x. So this, we could say is plus the natural log of the absolute value of x. It's not a bad habit to do it, and if everthing's positive, well, the absolute value of x is equal to x. And so what is this going to be equal to? This is equal to, let's evaluate everything at four. And actually before I even evaluate it at four, what's six divided by negative two, that's negative three. So if we evaluate it at four, it's going to be negative three over four squared. Four to the negative two is one over four squared and then plus the natural log of the, we could say the absolute value of four, but the absolute value of four is just four. So the natural log of four. And from that we're going to subtract everything evaluated at two. So let's do that. So if we evaluate it at two, it's going to be negative three over two squared. So two to the negative two is one over two squared, over two squared, plus the natural log of, the absolute value of positive two is, once again, it's just two. And so what does this give us? So, let's try to simplify it a little bit. So this is negative 3/16. We'll do that same color. So this is going to be equal to negative, negative, sorry, not negative 3/16, it's gotta be very careful, oh, sorry, yes, it is negative 3/16. For some reason my brain started thinking four to the third power. Negative 3/16 plus natural log of four. And then this right over here is negative 3/4, negative 3/4, do that same color. This right over here is negative 3/4. We have this negative sign out front that we're going to have to distribute. So the negative of negative 3/4 is plus 3/4, plus 3/4, and then we're going to subtract, remember we're distributing this negative sign, the natural log, the natural log of two. And what does this equal to? All right, so this is going to be equal to, and I'm now going to switch to a neutral color. So let's add these two terms that don't involve a natural log. And let's see, if we have a common denominator three over four is the same thing, that is the same thing as we multiply the numerator and the denominator by four, that is 12 over 16. And so you have negative 3/16, negative 3/16 plus 12/16 will give you 9/16, 9/16. And then we're gonna have the ones that do involve the natural log. Natural log of four minus the natural log of two. So we could write this plus the natural log of four minus the natural log of two. And you might remember from your logarithm properties that this over here, this is the same thing as the natural log of four divided by two, this comes straight out of your logarithm properties. And so this is going to be the natural log of two, natural log of two. So, we deserve a little bit of a drum roll now. This is all going to be equal to, this is going to be equal to the natural log, nine over 16 plus the natural log of two, plus the natural log of two. And we are done.