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

Let's take the
indefinite integral of the square root
of 7x plus 9 dx. So my first question
to you is, is this going to be a good case
for u-substitution? Well, when you look here,
maybe the natural thing to set to be equal
to u is 7x plus 9. But do I see its derivative
anywhere over here? Well, let's see. If we set u to be
equal to 7x plus 9, what is the derivative of u
with respect to x going to be? Derivative of u
with respect to x is just going to be equal to 7. Derivative of 7x is 7. Derivative of 9 is 0. So do we see a 7 lying
around anywhere over here? Well, we don't. But what could we do in order
to have a 7 lying around, but not change the
value of the integral? Well, the neat thing-- and we've
seen this multiple times-- is when you're
evaluating integrals, scalars can go in and outside
of the integral very easily. Just to remind ourselves, if I
have the integral of let's say some scalar a times
f of x dx, this is the same thing as a times
the integral of f of x dx. The integral of the
scalar times a function is equal to the scalar times
the integral of the functions. So let me put this
aside right over here. So with that in mind, can
we multiply and divide by something that will
have a 7 showing up? Well, we can multiply
and divide by 7. So imagine doing this. Let's rewrite our
original integral. So let me draw a
little arrow here just to go around that aside. We could rewrite our
original integral as being 9 to the integral
of times 1/7 times 7 times the square root of 7x plus 9 dx. And if we want to, we
could take the 1/7 outside of the integral. We don't have to,
but we can rewrite this as 1/7 times
the integral of 7, times the square
root of 7x plus 9 dx. So now if we set u
equal to 7x plus 9, do we have its
derivative laying around? Well, sure. The 7 is right over here. We know that du-- if we want to
write it in differential form-- du is equal to 7 times dx. So du is equal to 7 times dx. That part right over
there is equal to du. And if we want to care
about u, well, that's just going to be the 7x plus 9. That is are u. So let's rewrite this indefinite
integral in terms of u. It's going to be equal to
1/7 times the integral of-- and I'll just take the 7
and put it in the back. So we could just
write the square root of u du, 7 times dx is du. And we can rewrite this if we
want as u to the 1/2 power. It makes it a little bit
easier for us to kind of do the reverse power rule here. So we can rewrite this as equal
to 1/7 times the integral of u to the 1/2 power du. And let me just make it clear. This u I could have
written in white if I want it the same color. And this du is the same
du right over here. So what is the antiderivative
of u to the 1/2 power? Well, we increment
u's power by 1. So this is going to be
equal to-- let me not forget this 1/7 out front. So it's going to be 1/7 times--
if we increment the power here, it's going to be u to the 3/2,
1/2 plus 1 is 1 and 1/2 or 3/2. So it's going to
be u to the 3/2. And then we're going to
multiply this new thing times the reciprocal
of 3/2, which is 2/3. And I encourage you to verify
the derivative of 2/3 u to the 3/2 is
indeed u to the 1/2. And so we have that. And since we're
multiplying 1/7 times this entire indefinite
integral, we could also throw in a
plus c right over here. There might have
been a constant. And if we want, we can
distribute the 1/7. So it would get 1/7 times
2/3 is 2/21 u to the 3/2. And 1/7 times some
constant, well, that's just going to be some constant. And so I could write
a constant like that. I could call that c1 and
then I could call this c2, but it's really just
some arbitrary constant. And we're done. Oh, actually, no we aren't done. We still just have our
entire thing in terms of u. So now let's unsubstitute it. So this is going to be equal
to 2/21 times u to the 3/2. And we already know
what u is equal to. u is equal to 7x plus 9. Let me put a new color here
just to ease the monotony. So it's going to be
2/21 times 7x plus 9 to the 3/2 power plus c. And we are done. We were able to take a kind
of hairy looking integral and realize that even
though it wasn't completely obvious at first, that
u-substitution is applicable.

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