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𝘶-substitution: definite integral of exponential function

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
,
FUN‑6.D.2 (EK)
Finding the definite integral from 0 to 1 of x²⋅2^(x³). Created by Sal Khan.

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  • spunky sam blue style avatar for user Akshayan
    did Sal forget to change the boundaries in terms of u?
    (9 votes)
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    • blobby green style avatar for user Creeksider
      No, he worked the problem in a way that made it unnecessary to change the boundaries. One way to work these problems is to change the boundaries and then solve in terms of u. The other way, which Sal used here, is to treat it as an indefinite integral (no boundaries) when you do the u-substitution, but then after integrating, transform the result back from u to x. When you do that, you can evaluate the integral in terms of the original boundaries, because you've reversed the effect of the substitution. The reversal happens at in the video.
      (35 votes)
  • blobby green style avatar for user Luigi Diaz
    made u=2^x^3 and du=ln(2) (x^2) (2^x^3) dx, using the 12th basic differentiation rule in larson, ended up with 1/(3ln(2)) which i think is the same as 1/(ln(8))
    (17 votes)
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    • leaf blue style avatar for user Stefen
      Yes - the answer is equivalent.
      Good one!
      There are often lots of ways to solve a problem, and learning as many options as possible will help your understanding and creativity when faced with more challenging problems.
      (13 votes)
  • mr pink red style avatar for user spikehackerinc
    I must be honest, you lost me when you added the e term... Why couldn't you have just made u = x^3 which means du = 3x^2 so to get it in the form x^2 you divide both sides by 2 resulting in du/2 = x^2 . Then it is in a simpler form of the integral of 1/3 2^u du. Just saying I think this method would have been much easier to understand, as well as being easier when the exponential term is more complicated.
    (2 votes)
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    • blobby green style avatar for user Creeksider
      I'm not sure it makes much difference. You still have to convert 2^u to a power of e, so it's a question of whether you introduce e before doing the u-substitution or after. Anyway, the goal would be to gain enough experience dealing with exponents and logs so that all these steps come naturally without confusion.
      (31 votes)
  • leaf red style avatar for user alexis sorrell
    Just like others, I don't understand why the ln came in the the problem, just like the other problems I applied the same technique and i got that the antiderivative was (1/3)(2^x^3) +c, i dont understand why it was necessary to use the ln
    (4 votes)
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    • piceratops ultimate style avatar for user Just Keith
      You misused the power rule: the power rule is for x^n forms, NOT for n^x forms.
      Thus, if you have a variable in the exponent or a constant base, then the power rule does not apply.

      Thus,
      ∫ x^n dx = [x^(n+1)] ÷ (n+1) + C
      whereas
      ∫ n^x dx = ( n^x ) / ln (n) + C
      And, of course,
      ∫ x^x dx is an integral no mathematician has ever been able to solve apart from estimating it with a Taylor polynomial or some other approximation.
      (27 votes)
  • blobby green style avatar for user Sneha Srinivasan
    i did not understand how 2 = e ^ ln2.. can someone please expln... thanks
    (3 votes)
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  • leaf blue style avatar for user Kudrat
    If I do this problem in my calculator - I don' tget Ln8 but instead 2x^3/3ln2 , which is after making x^3=u and working from there. If I did it this way, I got the answer that my calculator gave me, but not what was during this whole video so I'm kind of really confused right now.
    (3 votes)
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    • piceratops ultimate style avatar for user Just Keith
      First, know that ln 8 = 3 ln 2, so that is just a basic log property, a different way of expressing the same thing.

      Second, you stopped at the indefinite integral. Note that to get the final answer, you need to apply the bounds of integration from x=0 to 1.
      (11 votes)
  • blobby green style avatar for user durkasmurka
    At , why isn't the chain rule applied when taking the derivative of x^3ln(2)?
    (3 votes)
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    • blobby green style avatar for user Creeksider
      The chain rule applies when one function applies to the output of another. Here we have only one function, x^3, multiplied by a constant, ln(2). I realize ln(2) looks like a function, but it's a constant like 7 or π. And if it were a function, we still wouldn't apply the chain rule, we'd apply the product rule, because then we'd have two functions multiplied together instead of one function applying to the output of another.
      (7 votes)
  • blobby green style avatar for user Giovana Werneck
    Uh? Wasn't this a definite integral in the beginning? I don't understand why it became indefinite afterwards.
    (2 votes)
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  • blobby green style avatar for user Jacob Kv
    HI :)
    Why is that normally when we have e.g. ln(8) inside the interval, we find the anti-derivative of it = 1/8 and take out of the interval. But Sal says that it is constant? why in this case is it so?
    how can we differentiate between one ln(8) which is not constant and thus applies to 1/x rule when taking out of the interval and one in which it is constant and thus we should keep it like we deal with constants e.g. 8,p, etc. ?
    Thanks for the answer in advance!
    (2 votes)
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    • piceratops ultimate style avatar for user Just Keith
      First, it is the derivative of ln x that equals 1/x, not the antiderivative. The antiderivative is:
      ∫ ln(x) dx = x ln(x) − x + C


      In a derivative or an integral, it is a constant if there is no variable, just a number. Thus,
      ∫ ln(8) dx = x ln(8) + C  and d/dx [ln(8)] = 0

      While,
      ∫ ln(8x) dx= x ln(8x) − x + C  and d/dx [ln(8x]) = 8/(8x) = 1/x 

      But note that
       d/dx [ ln(8x+3) ] = 8 / (8x+3)
      and ∫ ln(8x+3) dx = ⅛(8x+3) ln(8x+3) − ⅛(8x+3) + C
      (4 votes)
  • old spice man green style avatar for user Axel
    what i did was that i added 3 inside the integral and put 1/3 outside. And then i said that:
    u=x^3
    du=3x^2*dx

    then we have the integral: (2^u du) with the constant 1/3 in front of it. then after that we could just chage the basis to e^ln2*u. I dont see why this would be wrong?
    (2 votes)
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Video transcript

Sal: Let's see if we can calculate the definite integral from zero to one of x squared times two to the x to the third power d x. Like always I encourage you to pause this video and see if you can figure this out on your own. I'm assuming you've had a go at it. There's a couple of interesting things here. The first thing, at least that my brain does, it says, "I'm used to taking derivatives and anti-derivatives of e to the x, not some other base to the x." We know that the derivative with respect to x of e to the x is e to the x, or we could say that the anti-derivative of e to the x is equal to e to the x plus c. Since I'm dealing with something raised to, this particular situation, something raised to a function of x, it seems like I might want to put, I might want to change the base here, but how do I do that? The way I would do that is re-express two in terms of e. What would be two in terms of e? Two is equal to e, is equal to e raised to the power that you need to raise e to to get to two. What's the power that you have to raise two to to get to two? Well that's the natural log of two. Once again the natural log of two is the exponent that you have to raise e to to get to two. If you actually raise e to it you're going to get two. This is what two is. Now what is two to the x to the third? Well if we raise both sides of this to the x to the third power, we raise both sides to the x to the third power, two to the x to the third is equal to, if I raise something to an exponent and then raise that to an exponent, it's going to be equal to e to the x to the third, x to the third, times the natural log of two, times the natural log of two. That already seems pretty interesting. Let's rewrite this, and actually what I'm going to do, let's just focus on the indefinite integral first, see if we can figure that out. Then we can apply, then we can take, we can evaluate the definite ones. Let's just think about this, let's think about the indefinite integral of x squared times two to the x to the third power d x. I really want to find the anti-derivative of this. Well this is going to be the exact same thing as the integral of, I'll write my x squared still, but instead of two to the x to the third I'm going to write all of this business. Let me just copy and paste that. We already established that this is the same thing as two to the x to the third power. Copy and paste, just like that. Then let me close it with a d x. I was able to get it in terms of e as a base. That makes me a little bit more comfortable but it still seems pretty complicated. You might be saying, "Okay, look. "Maybe u substitution could be at play here." Because I have this crazy expression, x to the third times the natural log of two, but what's the derivative of that? Well that's going to be three x squared times the natural log of two, or three times the natural log of two times x squared. That's just a constant times x squared. We already have a x squared here so maybe we can engineer this a little bit to have the constant there as well. Let's think about that. If we made this, if we defined this as u, if we said u is equal to x to the third times the natural log of two, what is du going to be? du is going to be, it's going to be, well natural log of two is just a constant so it's going to be three x squared times the natural log of two. We could actually just change the order we're multiplying a little bit. We could say that this is the same thing as x squared times three natural log of two, which is the same thing just using logarithm properties, as x squared times the natural log of two to the third power. Three natural log of two is the same thing as the natural log of two to the third power. This is equal to x squared times the natural log of eight. Let's see, if this is u, where is du? Oh, and of course we can't forget the dx. This is a dx right over here, dx, dx, dx. Where is the du? Well we have a dx. Let me circle things. You have a dx here, you have a dx there. You have an x squared here, you have an x squared here. So really all we need is, all we need here is the natural log of eight. Ideally we would have the natural log of eight right over here, and we could put it there as long as we also, we could multiply by the natural log of eight as long as we also divide by a natural log of eight. We can do it like right over here, we could divide by natural log of eight. But we know that the anti-derivative of some constant times a function is the same thing as a constant times the anti-derivative of that function. We could just take that on the outside. It's one over the natural log of eight. Let's write this in terms of u and du. This simplifies to one over the natural log of eight times the anti-derivative of e to the u, e to the u, that's the u, du. This times this times that is du, du. And this is straightforward, we know what this is going to be. This is going to be equal to, let me just write the one over natural log of eight out here, one over natural log of eight times e to the u, and of course if we're thinking in terms of just anti-derivative there would be some constant out there. Then we would just reverse the substitution. We already know what u is. This is going to be equal to, the anti-derivative of this expression is one over the natural log of eight times e to the, instead of u, we know that u is x to the third times the natural log of two. And of course we could put a plus c there. Now, going back to the original problem. We just need to evaluate the anti-derivative of this at each of these points. Let's rewrite that. Given what we just figured out, let me copy and paste that. This is just going to be equal to, it's going to be equal to the anti-derivative evaluated at one minus the anti-derivative evaluated at zero. We don't have to worry about the constants because those will cancel out. So we are going to get, we are going to get one-- Let me evaluate it first at one. You're going to get one over the natural log of eight times e to the one to the third power, which is just one, times the natural log of two, natural log of two, that's evaluated at one. Then we're going to have minus it evaluated it at zero. It's going to be one over the natural log of eight times e to the, well when x is zero this whole thing is going to be zero. Well e to the zero is just one, and e to the natural log of two, well that's just going to be two, we already established that early on, this is just going to be equal to two. We are left with two over the natural log of eight minus one over the natural log of eight, which is just going to be equal to one over the natural log of eight. And we are, and we are done.