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### Course: Class 12 math (India)>Unit 10

Lesson 3: Fundamental theorem of calculus: chain rule

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

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?
• 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.
• 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?
• 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!!
• What's the reason exactly why we flip the integral in the first place?
• 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.
• Do we not use the chain rule at the end?
• the derivative of x is just one, so applying the chain rule will just be multiplication by 1.
• At 0.42, where does the formula he's using come from ? Is it explained in another video ?
• 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?
• Why do you not use the chain rule for sqrt( |cosx| ) ? Isn't that a function of a function?
• 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.
• Does the chain rule not apply here?
• In the quesiton used for this video, the derivative of x is just one, so applying the chain rule will just be multiplication by 1.
• But what if the x bound, regardless of its position, has an exponent? Does the exponent affect the derivative of that integral.?
• 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)