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## Applying properties of definite integrals

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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, 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.

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