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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 6

Lesson 11: Integrating using substitution- 𝘶-substitution intro
- 𝘶-substitution: multiplying by a constant
- 𝘶-substitution: defining 𝘶
- 𝘶-substitution: defining 𝘶 (more examples)
- 𝘶-substitution
- 𝘶-substitution: defining 𝘶
- 𝘶-substitution: rational function
- 𝘶-substitution: logarithmic function
- 𝘶-substitution warmup
- 𝘶-substitution: indefinite integrals
- 𝘶-substitution: definite integrals
- 𝘶-substitution with definite integrals
- 𝘶-substitution: definite integrals
- 𝘶-substitution: definite integral of exponential function

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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.

## Want to join the conversation?

- 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.(97 votes)
- 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.(32 votes)

- did Sal forget to change the boundaries in terms of u?(10 votes)
- 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 at6:00in the video.(37 votes)

- 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))(18 votes)
- 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)

- 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)
- 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.(28 votes)

- i did not understand how 2 = e ^ ln2.. can someone please expln... thanks(3 votes)
- ln 2 means “the natural logarithm of 2”. The natural logarithm is a logarithm with base e. So, another way to express ln 2 is log_e(2).

The logarithm tells us the power (exponent) that a number (base) needs to be raised to to equal a number (the argument). In the same way that log_10(1000) = 3 means that “the power that 10 is raised to to equal 1000 is 3”, ln 2 means “the power that e is raised to to equal 2”. So by raising e to the ln 2 power, you are raising e to the power that e is raised to to equal 2, thus e^(ln 2) = 2. If you’re struggling with the intuition behind logarithms, I recommend the tutorial at https://www.khanacademy.org/math/algebra2/exponential-and-logarithmic-functions/introduction-to-logarithms/v/logarithms to get a better intuition for what the logarithm is.(17 votes)

- 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)
- 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.(12 votes)

- At4:00, why isn't the chain rule applied when taking the derivative of x^3ln(2)?(3 votes)
- 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.(8 votes)

- Uh? Wasn't this a definite integral in the beginning? I don't understand why it became indefinite afterwards.(2 votes)
- When an integral is evaluated, first, the indefinite integral(antiderivative) must be found, then we apply the fundamental theorem of calculus to evaluate the definite integral(9 votes)

- 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?(4 votes)- That's legal.

Maybe what is confusing you is that you ended up with`3ln2`

in the denominator, instead of`ln8`

. You'd need to apply log properties:`3ln2 = ln2³ = ln8`

(3 votes)

- 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)- 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(5 votes)

## 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.