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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)

## Video transcript

let's see if we can calculate the definite integral from 0 to 1 of x squared times 2 times 2 to the X to the third power DX and like always I encourage you to pause this video and see if you can figure this out figure this out on your own so I'm assuming you've had a go at it and so there's a couple of interesting things here the first thing at least that my brain does it says well I'm used to taking derivatives and anti derivatives of e to the X not some other base to the X so 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 antiderivative of e to the X is equal to e to the X plus C so since I'm dealing with a something raised to a this particular situation something raised to a function of X it seems like I might want to put something I might want to change the base here but how do I do that well the way I would do that is reexpress to in terms of e so what would be to in terms of e well 2 is equal to e is equal to e raised to the power that you need to raise e to to get to 2 well what's the power that you have to raise e to to get to do well that's the natural log of 2 once again the natural log of 2 is the exponent that you have to raise e to to get to 2 so if you actually raise e to it you're going to get 2 so this is what 2 is now what is 2 to the X to the third well if we raise both sides of this to the X to the 3rd power if we raise both sides to the X to the 3rd power 2 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 3rd X to the 3rd times the natural log of 2 times the natural log of 2 so that already seems pretty interesting so let's let's rewrite this and actually what I'm going to do is let's just focus on the indefinite integral first see if we could figure that out then we can apply and then we could take we can evaluate the definite ones so let's just 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 DX so I really want to find the antiderivative of this well this is going to be the exact same thing as the integral of so I'll write my x squared still 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 this is the same thing as two to the X to the third power copy and paste and just like that and then let me just close it with a DX so 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 but you might be saying well okay look maybe u substitution could be at play here because I have this this kind of crazy expression X to the third time's the natural log of two but what's the derivative of that well that's going to be 3x squared times the natural log of two or three times the natural log of 2 times x squared well that's just a constant times x squared we already have an x squared here and so maybe we can engineer this a little bit to have the constant there as well so let's think about that so if we made this if we defined this as u so if we said u is equal to X to the third times the natural log of two what is d u going to be well D U is going to be it's going to be well natural log of 2 is just a constant so it's going to be 3x squared times the natural log of 2 and we can actually just change the order we're multiplying a little bit we could say that this is the same thing as x squared times 3 natural log of 2 which is the same thing just using logarithm properties as x squared times the natural log of 2 to the 3rd power 3 natural log of 2 is the same thing as the natural log of 2 to the 3rd power so this is equal to x squared times the natural log of 8 so let's see if this is U where is D u oh and of course we can't we can't forget the DX this is a DX right over here DX DX DX so where is the D you well we have a DX let me circle things so 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 so if we ideally we would have a natural log of eight right over here and we could put it there as long as we also we can multiply by a natural log of eight as long as we also divide by a natural log of eight and so we could we could do it like heat right over here we could write it divided by natural log of eight but we know that the antiderivative of some constant times a function is the same thing as a constant times the antiderivative of that function so we could just take that on the outside so it's one over the natural log of eight so let's write this in terms of U and D U this simplifies to 1 over the natural log of eight times the antiderivative of e e to the U e to the u that's the u D u this times this times that is d u D U and this is straightforward we know what this is going to be this is going to be equal to so let me just write the 1 over natural log of 8 out here 1 over natural log of 8 x times e to the U times e to the U e to the U and of course if we're thinking in terms of just antiderivative there would be some constant out there and then we would just reverse the substitution we already know what u is so this is going to be equal to the antiderivative of this expression is 1 over the natural log of 8 times e to the instead of U we know that U is X to the 3rd times the natural log of 2 and of course we could put a plus C there now going back to the original problem we just need to evaluate the antiderivative of this at each of these points so let's rewrite this so given what we just figured out so let me copy and paste that this is just going to be equal to it's going to be equal to the antiderivative evaluated at one minus the antiderivative evaluated zero we don't have to worry about the constants because those will cancel out and so we are going to get we are going to get one let me evaluate it first at one so you're going to get 1 over the natural log of 8 times e to the 1 to the third power which is just 1 times the natural log of 2 natural log of 2 that's it evaluated at 1 and then we're going to have minus it evaluated at 0 so it's going to be 1 over the natural log of 8 times e to the well when x is 0 this whole thing is going to be 0 well e to the 0 is just 1 and e to the natural log of 2 well that's just going to be 2 we already established that early on this is just going to be equal to 2 so we are left with 2 over the natural log of 8 minus 1 over the natural log of 8 which is just going to be equal to 1 over the natural log of 8 and we are and we are done
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