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## Calculus 2

### Course: Calculus 2 > Unit 1

Lesson 8: Properties of definite integrals- Negative definite integrals
- Finding definite integrals using area formulas
- Finding definite integrals using area formulas
- Definite integral over a single point
- Integrating scaled version of function
- Switching bounds of definite integral
- Integrating sums of functions
- Worked examples: Finding definite integrals using algebraic properties
- Finding definite integrals using algebraic properties
- Definite integrals on adjacent intervals
- Worked example: Breaking up the integral's interval
- Worked example: Merging definite integrals over adjacent intervals
- Definite integrals over adjacent intervals
- Functions defined by integrals: switched interval
- Finding derivative with fundamental theorem of calculus: x is on lower bound
- Finding derivative with fundamental theorem of calculus: x is on both bounds
- Functions defined by integrals: challenge problem
- Definite integrals properties review

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# Finding derivative with fundamental theorem of calculus: x is on lower bound

AP.CALC:

FUN‑6 (EU)

, FUN‑6.A (LO)

, FUN‑6.A.1 (EK)

, FUN‑6.A.2 (EK)

Sometimes you need to swap the bounds of integration before applying the fundamental theorem of calculus. Created by Sal Khan.

## Want to join the conversation?

- Why is there no +C at the end?(40 votes)
- Only a indefinite integral has a the + C constant part while a definite integral does not have it as it cancels in the end on subtraction.(95 votes)

- I understand the mathematics behind this, but when I imagine the integral, it's the space under the graph. And no matter which direction I'm coming from (left or right on the x-axis), it should still sum up to the exact same amount, not to the negative. So my understanding of the integral is flawed somehow, I suppose?(12 votes)
- Define an integral to be "the area under the curve of a function between the curve and the x-axis, above the x-axis." Although this is not the most formal definition of an integral, it can be taken literally. When the curve of a function is above the x-axis, your area (integral) will be a positive value, as normal. But, when you have a portion of the curve that dips below the x-axis, the area literally "under" the curve extends indefinitely--it has no limit. Thus, when the curve goes below the x-axis, the "under" side of the curve is inverted (now facing upwards), and the "above" side of the x-axis is also inverted (now facing downwards). As a result, the area between a curve and the x-axis is multiplied by -1 when the curve is below the x-axis.

Another way to think of it is by thinking of the y-values as dimensions. The definition said "above" the x-axis, and when the curve is truly, literally above the x-axis, the y-values of the points on the curve are positive. Think of these as "positive dimensions." When the curve goes below the x-axis, the y-values of the points on the curve are negative, which can be thought of as "negative dimensions." Because the x-value dimensions are always positive values (they represent normal distances), the area of a portion of the curve under the x-axis will be the product of a negative y-dimension and a positive x-dimension, resulting in a "negative area."

Hope this helps!!(6 votes)

- What's the reason exactly why we flip the integral in the first place?(5 votes)
- Basically your bounds should move from some number to another that's bigger. This means you're moving from left to right on a graph, which is necessary especially in cases involving position, velocity, and acceleration functions because it represents the movement over time. If you didn't swap your bounds and add that extra negative outside the integral you'd yield a negative version of your answer; so in situations where you find the area under the curve of a velocity function, you can't have a negative area.(14 votes)

- Do we not use the chain rule at the end?(8 votes)
- the derivative of x is just one, so applying the chain rule will just be multiplication by 1.(7 votes)

- At 0.42, where does the formula he's using come from ? Is it explained in another video ?(8 votes)
- For anyone with the same question, it is explained in the next section/chapter: https://www.khanacademy.org/math/ap-calculus-bc/bc-integration-new/bc-6-7/v/connecting-the-first-and-second-fundamental-theorems-of-calculus(3 votes)

- Why do you not use the chain rule for sqrt( |cosx| ) ? Isn't that a function of a function?(5 votes)
- Yes,
`√(|cosx|)`

is a function of a function, but you are not differentiating that; you are differentiating the antiderivative of all that, by the time you get rid of the integral you have finished with the differentiation, so there is no need to try and use the chain rule.(4 votes)

- at1:00Sal drops a formula and calls it the "2nd fundamental Theorem of calculus", but I haven't seen it on the Integration playlists so far. I'm doing it in the exact order of the playlists, and my question is does he show how this formula and all works some time later?(3 votes)
- Yeah, that he does. That is covered in the next section, so his reference can be thought of as alluding to a consequence of the FToC. It also may be that some of the videos have been redone and small breaches of topic continuity have occured.

The connection between the 1st and 2nd FToC is covered here:

https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/definite_integrals/v/connecting-the-first-and-second-fundamental-theorems-of-calculus(5 votes)

- But what if the x bound, regardless of its position, has an exponent? Does the exponent affect the derivative of that integral.?(3 votes)
- Yes, it affects the derivative in that we have to use the chain rule.

𝑑∕𝑑𝑥[ ∫[𝑥^𝛼, 𝑏] 𝑓(𝑡)𝑑𝑡 ] = 𝑑∕𝑑𝑥[ 𝐹(𝑏) − 𝐹(𝑥^𝛼) ]

= 𝑑∕𝑑𝑥[ −𝐹(𝑥^𝛼) ] = −𝑓(𝑥^𝛼) ∙ 𝛼𝑥^(𝛼 − 1)(4 votes)

- How do we see when we need to apply swapping of bonds?.(4 votes)
- (1) As the video illustrates at the beginning, this is sometimes a necessary manipulation in applying the Fundamental Theorem of Calculus (derivative of the integral with a variable bound). The natural direction has the constant as the lower bound, the variable (or variable quantity) as the upper bound. If the variable quantity is the lower bound, then it is most easily interpreted by this principle.

(2) In one class of problems you are given the value of certain integrals (or can figure them out using geometric formulas from the graph). If the integral you are evaluating goes from right to left, then you need to understand to reverse the areas you get when going left to right.(2 votes)

- I am super confused! I don't understand why swapping the bounds results in switching the signs from positive to negative...I know area can be "negative" when the curve is under x-axis, but I still don't see why the total area will be different from approaching from a to b and approaching from b to a.(4 votes)
- This takes some getting used to. Yet it's a direct consequence of the fundamental theorem of calculus. We find the definite integral by evaluating the antiderivative at the lower bound and at the upper bound, and subtracting the first from the second. If you reverse the bounds, you're subtracting what was previously the second from what was previously the first, so the result has to be the negative of the first result. In simplest terms, we've gone from a - b to b - a.(1 vote)

## Video transcript

We want to find the
derivative with respect to x of all of this
business right over here. And you might guess-- and this
is definitely a function of x. x is one of the
boundaries of integration for this definite integral. And you might say,
well, it looks like the fundamental theorem
of calculus might apply, but I'm used to seeing
the x, or the function x, as the upper bound,
not as the lower bound. How do I deal with this? And the key realization
is to realize what happens when you switch
bounds for a definite integral. And I'll do a little bit
of an aside to review that. So if I'm taking the definite
integral from a to b of f of t, dt, we know that this is capital
F, the antiderivative of f, evaluated at b minus the
antiderivative of F evaluated at a. This is corollary to
the fundamental theorem, or it's the fundamental
theorem part two, or the second fundamental
theorem of calculus. This is how we evaluate
definite integrals. Now, let's think about what
the negative of this is. So the negative of that--
of a to b of f of t, dt, is just going to be equal
to the negative of this, which is equal to-- so it's the
negative of f of b minus f of a, which is equal to capital
F of a minus capital F of b. All I did is distribute
the negative sign and then switch the two terms. But this right over here is
equal to the definite integral from, instead of a to b, but
from b to a of f of t, dt. So notice, when
you put a negative, that's just like
switching the signs or switching the boundaries. Or if you switch
the boundaries, they are the negatives of each other. So we can go back to
our original problem. We can rewrite this as being
equal to the derivative with respect to x of--
instead of this, it'll be the negative of the
same definite integral but with the boundaries switched-- the
negative of x with the upper boundary is x, the lower
bound is 3 of the square root of the absolute value
of cosine t, dt, which is equal to--we can
take the negative out front-- negative times the derivative
with respect to x of all of this business. I should just copy
and paste that, so I'll just copy and paste. Let me-- and paste it. So times the derivative with
respect to x of all that, and now the fundamental theorem
of calculus directly applies. This is going to be equal to--
we deserve a drum roll now. This is going to be
equal to the negative-- can't forget the negative. And the fundamental
theorem of calculus tells us that that's
just going to be this function as
a function of x. So it's going to be
negative square root of the absolute value of
cosine of not t anymore, but x. And we are done.