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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?
• This is a constant of integration and we see this when we integrate an INdefinite integral. We put the C there because we are given enough information to classify what the phase shift of the function will be because it is not DEFINED ( not a definite integral). For example we know that the derivative of x^2 + 2 = 2x, but what happens when we integrate 2x? we get x^2 but we dont know what the constant is unless we have more information. So, our final answer of the integral of 2x would equal x^2 + C.
• 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 ?
• 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.
• 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)
• How do we see when we need to apply swapping of bonds?.
• (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.
• 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.