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AP Calc: FUN‑6 (EU), FUN‑6.C (LO), FUN‑6.C.1 (EK), FUN‑6.C.2 (EK)

Let's take the
derivative with respect to x of x to the n
plus 1-th power over n plus 1 plus some constant c. And we're going to
assume here, because we want this expression
to be defined, we're going to assume that
n does not equal negative 1. If it equaled negative
1, we'd be dividing by 0, and we haven't defined
what that means. So let's take the
derivative here. So this is going to be equal
to-- well, the derivative of x to the n plus 1
over n plus 1, we can just use the
power rule over here. So our exponent is n plus 1. We can bring it out front. So it's going to be n
plus 1 times x to the-- I want to use that same color. Colors are the hard part--
times x to the-- instead of n plus 1, we subtract
1 from the exponent. This is just the power rule. So n plus 1 minus
1 is going to be n. And then we can't forget that we
were dividing by this n plus 1. So we have divided by n plus 1. And then we have plus c. The derivative of a constant
with respect to x-- a constant does not change as x
changes, so it is just going to be 0, so plus 0. And since n is not
equal to negative 1, we know that this is
going to be defined. This is just going to
be something divided by itself, which is
just going to be 1. And this whole thing
simplifies to x to the n. So the derivative of
this thing-- and this is a very general terms--
is equal to x to the n. So given that, what is
the antiderivative-- let me switch colors here. What is the antiderivative
of x to the n? And remember, this
is just the kind of strange-looking
notation we use. It'll make more sense when we
start doing definite integrals. But what is the
antiderivative of x to the n? And we could say the
antiderivative with respect to x, if we want to. And another way of calling this
is the indefinite integral. Well, we know this
is saying x to the n is the derivative of what? Well, we just figured it out. It's the derivative
of this thing, and we've written it
in very general terms. We're actually encapsulating
multiple constants here. We could have x to
the n plus 1 over n plus 1 plus 0, plus 1, plus
2, plus pi, plus a billion. So this is going to be equal
to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this
as the reverse power rule. And it applies
for any n, as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing
would be undefined if n were equal to negative 1. So let's do a couple
of examples just to apply this-- you could
call it the reverse power rule if you want, or
the anti-power rule. So let's take the antiderivative
of x to the fifth power. What is the antiderivative
of x to the fifth? Well, all we have to
say is, well, look, the 5 is equal to the n. We just have to increment
the exponent by 1. So this is going to be equal
to x to the 5 plus 1 power. And then we divide
by that same value. Whatever the exponent was
when you increment it by 1, we divide by that same
value, divided by 5 plus 1. And of course, we
want to encapsulate all of the possible
antiderivatives, so you put the c
right over there. So this is going to be equal to
x to the sixth over 6 plus c. And you can verify. Take the derivative of
this using the power rule, you indeed get x to the fifth. Let's try another one. Let's try-- now
we'll do it in blue. Let's try the antiderivative
of-- let's make it interesting. Let's make it 5 times x to
the negative 2 power dx. So how would we evaluate this? Well, one simplification
you can do-- and I haven't rigorously
proven it to you just yet-- but we know that scalars
can go in and out of the derivative operator when
you're multiplying by a scalar. So this is, indeed, equal to
5 times the antiderivative of x to the negative
2 power, dx. And now we can just use,
I guess we could call it this anti-power rule,
so this is going to be equal to 5 times x to
the negative 2 power plus 1 over the negative 2 power plus
1 plus some constant right over here. And then we can rewrite this as
5 times negative 2 power plus 1 is x to the negative 1
over negative 2 plus 1 is negative 1,
plus some constant. And this is equal to 5 times
negative x to the negative 1 plus some constant. And then if we want, we
can distribute the 5. So this is equal to negative
5x to the negative 1. Now, we could write plus
5 times some constant, but this is just an
arbitrary constant. So this is still just
an arbitrary constant. So maybe we could
[INAUDIBLE] this. If you want it to show that
it's a different constant, you could say this
is c1, c1, c1. You multiply 5 times c1,
you get another constant. We could just call that c,
which is equal to 5 times c1. But there you have it. Negative 5x to the
negative 1 plus c. And once again, all of these,
try to evaluate the derivative, and you will see that you
get this business, right over there.