𝘶-substitution: definite integrals (video)

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AP®︎/College Calculus AB

• 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

𝘶-substitution: definite integrals

When using 𝘶-substitution in definite integrals, we must make sure we take care of the boundaries of integration.

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• There seem to be some questions that DON’T change the boundaries of integration. How do you tell when or when not to change the boundaries?(11 votes)
  • If you stick with the anti-derivative as a function of u, then you need to change the bounds of integration.
    If you “back substitute” for u, then that’s when you don’t change the bounds.
    The choice (back substitute or not) is up to you.
    The closing section of the video covers this:You can either keep it a definite integral [as a function of u] 5:32and then change your bounds of integration 5:34and express them in terms of u. 5:37That’s one way to do it. 5:39The other way is to try to evaluate 5:40the indefinite integral, 5:43use u-substitution as an intermediary step, 5:44then back-substitute back and then evaluate 5:47at your bounds. [values of x] 5:49(22 votes)
• I’m having difficulties conceptualizing the difference between definite and indefinite integrals, in particular the idea of the constant, +C, used in the indefinite integrals. At around, Sal integrates the u-substitution in the usual fashion and it makes sense that he uses the boundaries x = 2 to x = 1 because the problem is a definite integral. I guess my question is if you integrated the u-substitution as an indefinite integral you would get (u^4)/4 + C but the C goes away when you’ve constricted it to a set of boundaries. If there was a constant in your original problem, say a +4 for example, wouldn’t the constant have some sort of effect on your answer even after evaluating at the min and max boundaries? I’m not sure if I’m explaining the question effectively but I appreciate any clarification on the topic. 4:50(4 votes)
  • If there is a constant of integration then it gets added (with the upper bound) and then taken away again (with the lower bound)(2 votes)
• What happened to the 2x in the original integral?(3 votes)
  • 2𝑥 is the derivative of (𝑥² + 1)
    By letting 𝑢 = 𝑥² + 1 we get 𝑑𝑢∕𝑑𝑥 = 2𝑥 ⇒ 𝑑𝑢 = 2𝑥 ∙ 𝑑𝑥
    So, 2𝑥 ∙ 𝑑𝑥 in the original expression is substituted with 𝑑𝑢.(13 votes)
• What is the difference between definite and indefinite integrals?(3 votes)
  • A definite integral tells you the area under the curve between two points a & b, and indefinite integral gives you the general form of the anti-derivative of the function. Operationally the only difference is plugging in values once you’ve integrated. Sometimes it’s convenient to write down the indefinite form, and then plug in values a, b so that you can keep track if you need to change your bounds.(7 votes)
• I don’t understand why the integral of secant is tangent. I got a quiz asking this, which doesn’t explain about the trigonometric integral. May I get some explanation?(2 votes)
  • The integral of secant(x) is ln(secant(x) + tangent(x)), not tangent(x). I assume you are asking why the integral of secant(x)^2 is tangent(x). Seeing why the integral of secant(x)^2 equals tangent(x) is easier if you ask why is the derivative of tangent(x) equal to secant(x)^2.
    tan(x) = sin(x)/cos(x)
    1. (tan(x))’ = (sin(x)/cos(x))’ Use the derivative quotient rule
    2. (tan(x))’ = (cos(x)^2+sin(x)^2)/cos(x)^2 Split the fraction into two terms and simplify
    3. (tan(x))’ = 1+sin(x)^2/cos(x)^2 Substitute tan(x) for sin(x)/cos(x)
    4. (tan(x))’ = 1+tan(x)^2 Substitute the expression using the identity tan(x)^2+1=sec(x)^2
    5. (tan(x))’ = sec(x)^2(5 votes)
• if A definite integral tells us the area under the curve between two points a & b can we say that an indefinite integral tells us the area under the curve between +infinity and -infinity?(2 votes)
  • Yes, we can by defining the improper integral, then we can evaluate integrals from – infinity to +infinity. One way is to split it up to integral from 0 to +infinity + integral from -infinity to 0.
    video about improper integeral:https://www.khanacademy.org/math/ap-calculus-bc/bc-integration-new/bc-6-13/v/introduction-to-improper-integrals
    video about the idea you wrote:https://www.khanacademy.org/math/integral-calculus/ic-integration/ic-improper-integrals/v/improper-integral-with-two-infinite-bounds
    hope this helps(4 votes)
• What if substituting the bounds of integration makes the bottom one larger than the top?(3 votes)
  • What Tegh said, but also if the bounds do winde up like that keep in mind that it is the same thing as the negative of the bounds in the “right” order, though as long as you do things in the right order you will be doing that. It might just be hard to catch.(2 votes)
• Which method of u substitution is preferable in most circumstances?(3 votes)
  • It really depends on what the integrand. Some integrand will not require u-substitution.
    Sometimes you do not need to explicity substitute u but you can change variable you are integrating with respect to.
    integral(cos(x) sin(x)) dx = integral(sin(x)) dsin(x).
    Note integral(f(x)*g(x)) dx = integral(f(x)) dG(x)(2 votes)
• Why is the integral of 1/(1-u^2)^1/2= sin^-1(u^2)+C?(3 votes)
• In the second method of evaluating the integral (using indefinite integrals first and then “plugging in” the x values), why is there no +C at the end of the indefinite integral? Thanks!(1 vote)
  • We could have included the +C, but then it would have cancelled out after evaluating the function at the endpoints. We would have gotten
    (5^4)/4 +C -((2^4)/4 +C). Once you distribute the negative, you get a +C term and a -C term.
    So because the constant of integration always gets cancelled when evaluating definite integrals, we usually just ignore it entirely.(5 votes)

Video transcript

– [Instructor] What we’re going to do in this video is get some practice applying u-substitution to definite integrals. So let’s say we have the integral, so we’re gonna go from x equals one to x equals two, and the integral is two x times x squared plus one to the third power dx. So I already told you that we’re gonna apply u-substitution, but it’s interesting to be able to recognize when to use it. And the key giveaway here is, well I have this x squared plus one business to the third power, but then I also have the derivative of x squared plus one which is two x right over here. So I can do the substitution. I could say, u is equal to x squared plus one, in which case the derivative of u with respect to x is just two x plus zero or just two x. I could write that in differential form. A mathematical hand waving way of thinking about it is multiplying both sides by dx. And so you get du is equal to two x dx. And so at least this part of the integral I can rewrite. So let me at least write, so this is going to be, I’ll write the integral. We’re going think about the bounds in a second. So we have u to the third power, u, the same orange color, u to the third power. That’s this stuff right over here. And then two x times dx. Remember you could just view this as two x times x squared plus one to the third power times dx. So two x times dx, well two x times dx, that is du. So that and that together, that is du. Now an interesting question, because this isn’t an indefinite integral, we’re not just trying to find the antiderivative. This is a definite integral. So what happens to our bounds of integration? Well there’s two ways that you can approach this. You can change your bounds of integration. Because this one is x equals one to x equals two. But now we’re integrating with respect to u. So one way if you want to keep your bounds of integration, or if you wanna keep this a definite integral I guess you could say, you would change your bounds from u is equal to something to u is equal to something else. And so your bounds of integration, let’s see when x is equal to one, what is u? Well when x is equal to one, you have one squared plus one, so you have two, u is equal to two in that situation. When x is equal to two, what is u? Well you have two squared which is four plus one, which is five, so u is equal to five. And you won’t typically see someone writing u equal two or u equals five. It’s often just from two to five because we’re integrating with respect to u, you assume it’s u equals two to u equals five. And so we could just rewrite this as being equal to the integral from two to five of u to the third du. But it’s really important to realize why we changed our bounds. We are now integrating with respect to u, and the way we did it is we used our substitution right over here. When x equals one, u is equal to two. When x is equal to two, two squared plus one, u is equal to five. And then we can just evaluate this right over here. Let’s see this is going to be equal to, the antiderivative of u to the third power is u to the fourth over four. We’re gonna evaluate that at five and two. And so this is going to be five to the fourth over four minus two to the fourth over four. And then we can simplify this if we like, but we’ve just evaluated this definite integral. Now another way to do it is to think about the, is to try to solve the indefinite integral in terms of x and use u-substitution as an intermediate. So one way to think about this is to say, well let’s just try to evaluate what the indefinite integral of two x times x squared plus one to the third power dx is. And then whatever this expression ends up being algebraically, I’m going to evaluate it at x equals two and at x equals one. And so then you use the u-substitution right over here, and you would get. This would simplify using the same substitution as the integral of u, u to the third power, the du, u to the third power du, and once again you’re going to evaluate this whole thing from x equals two and then subtract from that it evaluated at x equals one. And so this is going to be equal to, well let’s see this is u to the fourth power over four. And once again evaluating it at x equals two and then subtracting from that at x equals one. And then now we can just back-substitute. We can say hey look, u is equal to x squared plus one, so this is the same thing as x squared plus one to the fourth power over four, and we’re now going to evaluate that at x equals two and at x equals one. And you will notice that you will get the exact same thing. When you put x equals two here, you get two squared plus one which is five to the fourth power over four, that right over there. And then minus one squared plus one is two to the fourth power over four, that right over there. So either way you’ll get the same result. You can either keep it a definite integral and then change your bounds of integration and express them in terms of u. That’s one way to do it. The other way is to try to evaluate the indefinite integral, use u-substitution as an intermediary step, then back-substitute back and then evaluate at your bounds.