AP®︎ Calculus BC (2017 edition)
- 𝘶-substitution intro
- 𝘶-substitution: multiplying by a constant
- 𝘶-substitution: defining 𝘶
- 𝘶-substitution: defining 𝘶 (more examples)
- 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
𝘶-substitution: logarithmic function
Doing u-substitution with ln(x). Created by Sal Khan.
Want to join the conversation?
- How would you solve the integral of ln(x)(20 votes)
- whats the derivative of pi? or does it not have one?(0 votes)
- The derivative of any constant is zero. Therefore, the derivative of pi is 0.(61 votes)
- My calculator can't define ln|ln(x)|, is that normal?(4 votes)
- If it can't take it in one step, then simplify the process. First get ln(x), then get it's absolute value, and finally calculate ln of final value. You would need piece of paper to keep track of values.(8 votes)
- I don't understand why you can take out the pi outside of the integral, is it b/c it's a constant? What are other examples of things you can take out in fractions?(4 votes)
- Yes, it is because π is a constant.
Remember that an integral can be interpreted as a limit problem and the ability to factor out constants is a basic property of limits.
However, factoring out the constants is just a matter of convenience. It is not actually required so long as you know what you're doing.(7 votes)
- I don't get why we have to make it an absolute value towards the end around2:30?(3 votes)
- The antiderivative of 1/x is often given as ln(x) + C but it's more complete to give it as ln(|x|) + C. If you think of the graph of 1/x, you'll realize that it has two segments that are mirror images of each other, one in the first quadrant and the other in the third quadrant, and when we give the antiderivative without the absolute value we're really giving a partial solution that covers only the positive values of x, because ln(x) is undefined for negative x.
In this example, the original problem involves a logarithm, which can never apply to a negative number, but the integral is a log of a log, and the inner log could be negative, which means the more complete solution is one that includes the absolute value.(8 votes)
- Sal will you answer next questions, I hope.
What do dy and dx mean idividually?
What is definition of "dx"?
Is dy/dx a fraction or not fraction?(5 votes)
- nice question
dy/dx is not a fraction but in the field of calculus it is the rate of change of y with respect to x
dx is the slight change in the quantity of x
same is the case of dy(4 votes)
- The Integral of 2/(e^(2x)+4). I have used the substitution u=e^(2x)+4 and get as far as 1/u(u-4)du but I cannot get any further. The answer gives a log function. I hope this does not qualify as a homework question but I have tried everything can think of(3 votes)
- First, I applaude you on your substitution. That was very creative, and it took me a minute to catch up.
Now, the form into which you have gotten the integral involves an integration technique called integration by partial fraction decomposition. You can find partial fraction decomposition videos here: https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/partial-fraction-expansion/v/partial-fraction-expansion-1
Putting integrals in partial fraction form makes them much easier to integrate (using natural logarithms).(5 votes)
- Why does the 'du' go away after doing the integral?(2 votes)
- The common nomenclature of the integral includes both the integral sign at the left and the differential
duat the right. When you solve the integral both the integral sign at the left and the differential at the right go away. You must consider them to alway go together, so when one goes the other goes as well.(7 votes)
- AT2:15why isn't the derivative of 1/u not rewritten as u^-1 giving us -u^-2?(4 votes)
- At2:15Sal is taking the antiderivative not the derivative. So he gets ln | u |(2 votes)
- How would an integral with du *square root*u, be solved?(1 vote)
- To integrate this you would use the power rule. Add one to the power, the copy its reciprocal in front.
Integral sqrt(u) du
= Integral u^(1/2) du
= 2/3*u^(3/2) + C(4 votes)
We are faced with a fairly daunting-looking indefinite integral of pi over x natural log of x dx. Now, what can we do to address this? Is u-substitution a possibility here? Well for u-substitution, we want to look for an expression and its derivative. Well, what happens if we set u equal to the natural log of x? Now what would du be equal to in that scenario? du is going to be the derivative of the natural log of x with respect to x, which is just 1/x dx. This is an equivalent statement to saying that du dx is equal to 1/x. So do we see a 1/x dx anywhere in this original expression? Well, it's kind of hiding. It's not so obvious, but this x in the denominator is essentially a 1/x. And then that's being multiplied by a dx. Let me rewrite this original expression to make a little bit more sense. So the first thing I'm going to do is I'm going to take the pi. I should do that in a different color since I've already used-- let me take the pi and just stick it out front. So I'm going to stick the pi out in front of the integral. And so this becomes the integral of-- and let me write the 1 over natural log of x first. 1 over the natural log of x times 1/x dx. Now it becomes a little bit clearer. These are completely equivalent statements. But this makes it clear that, yes, u-substitution will work over here. If we set our u equal to natural log of x, then our du is 1/x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of 1/u. Natural log of x is u-- we set that equal to natural log of x-- times du. Now this becomes pretty straightforward. What is the antiderivative of all of this business? And we've done very similar things like this multiple times already. This is going to be equal to pi times the natural log of the absolute value of u so that we can handle even negative values of u. The natural log of the absolute value of u plus c, just in case we had a constant factor out here. And we're almost done. We just have to unsubstitute for the u. u is equal to natural log of x. So we end up with this kind of neat-looking expression. The anti of this entire indefinite integral we have simplified. We have evaluated it, and it is now equal to pi times the natural log of the absolute value of u. But u is just the natural log of x. And then we have this plus c right over here. And we could have assumed that, from the get go, this original expression was only defined for positive values of x because you had to take the natural log here, and it wasn't an absolute value. So we can leave this as just a natural log of x, but this also works for the situations now because we're doing the absolute value of that where the natural log of x might have been a negative number. For example, if it was a natural log of 0.5 or, who knows, whatever it might be. But then we are all done. We have simplified what seemed like a kind of daunting expression.