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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.
• 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.
• did Sal forget to change the boundaries in terms of u?
• 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.
• 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))
• 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.
• 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
• 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.
• i did not understand how 2 = e ^ ln2.. can someone please expln... thanks
• 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.
• 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.
• 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.
• At , why isn't the chain rule applied when taking the derivative of x^3ln(2)?
• 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.
• Uh? Wasn't this a definite integral in the beginning? I don't understand why it became indefinite afterwards.
• 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
• 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. ?
• 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
• 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?
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