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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)

## 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 though 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 were if 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 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 essentially for this is corollary to the fundamental theorem where's the fundamental theorem part two or the second fundamental theorem of calculus is how we this is how we evaluate definite integrals now let's think about what the negative of this is so the negative of that from 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 from instead of A to B but from B to a B to a of f of T 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 there 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 instead of this it'll be the negative of the same definite integral with the boundary switch the negative of X with the boundaries of the upper boundaries X to the lower boundary is three of the square root of the absolute value of cosine I'm T DT which is equal to we can take the negative out front negative times 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 paste it so times the derivative with respect to X of all of 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 negative square root of the absolute value of cosine of not T anymore but X and we are done