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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.

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  • blobby green style avatar for user Jason
    Why is there no +C at the end?
    (39 votes)
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  • blobby green style avatar for user Randi Edwards
    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)
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    • leaf red style avatar for user Kyle Gatesman
      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)
  • blobby green style avatar for user Renata Hryndzio
    What's the reason exactly why we flip the integral in the first place?
    (5 votes)
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    • purple pi purple style avatar for user Ricardo Mendez
      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)
  • piceratops ultimate style avatar for user Alex Drozd
    Do we not use the chain rule at the end?
    (8 votes)
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  • piceratops ultimate style avatar for user Raphaël Dunant
    At 0.42, where does the formula he's using come from ? Is it explained in another video ?
    (7 votes)
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  • blobby green style avatar for user sdags asdga
    Why do you not use the chain rule for sqrt( |cosx| ) ? Isn't that a function of a function?
    (5 votes)
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  • leaf red style avatar for user John Takatz
    But what if the x bound, regardless of its position, has an exponent? Does the exponent affect the derivative of that integral.?
    (3 votes)
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  • starky tree style avatar for user mfundohmbambo
    How do we see when we need to apply swapping of bonds?.
    (4 votes)
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    • aqualine tree style avatar for user Ted Fischer
      (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)
  • leaf green style avatar for user Shengyu Huang
    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)
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    • blobby green style avatar for user Creeksider
      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)
  • piceratops ultimate style avatar for user Kevin George Joe
    at Sal 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?
    (2 votes)
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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.