# 𝘶-substitution: definite integrals (video)

U-substitution With Definite Integrals
U-substitution With Definite Integrals

## 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.

## Want to join the conversation?

• 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
• 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.
• 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)