this video is the third in a short series on anti-derivatives and indefinite integrals and
our goal this time is to learn how to calculate indefinite integrals of power functions
so there's the indefinite integral of the power function and we see that a power function means
x raised to some power and notice that whatever the answer is we have a plus c
attached to it because the indefinite integral is the most general antiderivative and it comes
with an arbitrary additive constant now we already know the rule for differentiating a power function
so when we take the derivative of x to the n we bring down the original exponent out in front
we subtract one from the original exponent and what we're trying to accomplish here is to do this
process backwards and we can figure this out by basically guessing and adjusting so we set up the
integral x to the n dx and the anti-derivative here means the thing you differentiate to get
what you see inside that integrand well to get an x to the n out of the power rule
i must be differentiating something times x to the n plus one that's the meat of the answer and
then we just go ahead and think about what would the derivative do to x to the n plus one it would
bring n plus one down in front and then subtract one from the exponent giving me an x to the n
in other words the derivative of this is n plus one x to the n so x to the n plus one
misses by a constant all we have to do is put that constant out in front in the denominator
as a one over n plus one now when we differentiate this thing the n plus ones are going to cancel and
i'll be left with just the x to the n and of course we don't want to forget our plus c so
there's our basic formula for using the power rule backwards in other words getting an indefinite
integral of a power function you add one to the exponent and then divide by the same resulting
number but we have to be careful here to note that there's one value of n that doesn't work
and that's negative one putting in n equals negative one into this denominator would cause
division by zero and the result would be undefined so we make a note on that this only works if n
is not equal to negative one and we'll get into the n equals negative one case on the next slide
first we're going to work a few examples of how to use the power rule in the first example we're
asked to find two functions capital f of x such that f prime of x is equal to x squared in other
words these capital f's are anti-derivatives of x squared the thing you differentiate to get x
squared when we apply the power rule to x squared we simply add 1 to the exponent and divide by
the resulting number so in my first answer for an anti-derivative i could just say it's one-third
x cubed we just added one to the exponent divided by the resulting number now what i've done here is
choose c equals zero for my first antiderivative the second one is found by just choosing some
other number for c and it can be whatever you want so i'm just going to make it a 5. so there's two
unique functions that differentiate to x squared in the second example we want the indefinite
integral of one over x to the fifth and the key to this is seeing it as a power function one over
x to the fifth is x to the negative five so we have the integral of x to the negative five dx
to get the antiderivative we just add one to the exponent that gives me negative four and we divide
by that number negative four so i have a negative one fourth out in front and we have our arbitrary
constant plus c now that's probably not the best way to write the answer it's a little nicer if we
put the x to the fourth back into the denominator and i get a negative one over four x to the fourth
plus c in the next example we want the indefinite integral of five dx and this is one that you could
probably just guess but we're asked to show it by using the power rule so the integral of
5 dx we have to see this as a power function but there's no x in there and the key is that 1 is x
to the 0 and 5 times 1 is the same thing as 5. so i have 5x to the 0 dx so the 5 is a spectator here
and then the anti-derivative of x to the 0 is found by adding 1 to the exponent and dividing by
the resulting number so x to the first over 1. of course the 1 doesn't do anything there and so we
end up with the most general antiderivative of 5x plus c finally in our last example we put together
several terms in a linear combination and we know the integral of a linear combination is just the
linear combination of the integrals in other words all these constant multiples are just spectators
so when we apply the power rule to the first term the 3 is a spectator we add 1 to the
exponent so i get x cubed and we divide by that resulting number 3 and the 3s are going to cancel
in the second term the negative 5 is a spectator and we view 1 over x cubed as x to the negative
3 that gives me an x to the negative 2 over negative 2 when i apply our power rule backwards
in the third term we just have the antiderivative of 2 which is 2x
then we just have to clean things up a little bit the first term simplifies to x cubed
in the second term we have a negative out in front but we have a negative factor in the
denominator so i get a positive result and that's a 5 over 2 out in front and then we
should move the x to the negative 2 back to the denominator so i end up with a 5 over 2 x squared
and my third term again was 2x and then my arbitrary constant plus c and we're done with
that one so what about that n equals negative 1 case so we're trying to imagine what do we
differentiate to get 1 over x and there's one sort of most obvious solution we remember that the
derivative of the natural log function gives us 1 over x but then we recall the natural log function
has a domain restriction where it's only defined for positive values of x so this is only true
if x is greater than zero what if x is less than zero in that case the natural log of x
is not defined but the natural log of negative x is defined and when we take the derivative
of that we have to use the chain rule so we get 1 over the argument of the natural log function
negative x multiplied by the derivative of the argument of the natural log function which is
negative 1. and the factors of negative 1 cancel out and we end up with exactly the same result 1
over x so this is the second candidate that works we differentiate it and we get 1 over x and this
only works if x is less than 0. now we're not worried about x actually being zero because that's
not even in the domain of the original function so we have a formula that works when x is bigger than
zero and a formula that works when x is less than zero and if we wanted to write these two functions
all at once we could just call it the natural log of the absolute value of x since the absolute
value function leaves things alone when they're positive but multiplies by negative 1 when they're
negative so our anti-derivative is essentially the natural log of the absolute value of x
but there's a subtle point lurking here about how to deal with the arbitrary constant part so i've
written the anti-derivative of one over x as the natural log absolute value of x plus c one that's
some arbitrary constant when x is bigger than zero and the natural log of absolute value of x plus c2
some other arbitrary constant when x is less than zero and this is the most general possible thing
that differentiates to one over x notice that you can pick different constants for the left and
right sides of the real number line when you're defining this anti-derivative now in practice
calculus books and calculus teachers are usually a little bit lazy about this and they say something
that's not 100 true they say the anti-derivative is the natural log of the absolute value of x
plus c where they only have one arbitrary constant and while that does differentiate to one over x
for the left side and the right side of the real number line it's not completely the most general
thing that does differentiate to one over x so this formula natural log absolute value of
x plus c that's kind of what i use by default unless we're doing some very unusual problem
where the distinction actually matters that you can pick different constants for the left and
right sides of the real number line so in the examples you'll see me use that lazy approach
our first example is to get the indefinite integral of 6 over x dx and of course the constant
is just a spectator here just to emphasize that i can factor it out of the integral and i get 6
integral of 1 over x dx and we write 6 natural log absolute value of x plus c where we know in the
back of our minds that we could be more general by writing the antiderivative as a piecewise defined
function with two different choices of arbitrary constants but people rarely ever do that the next
example is a cumulative sort of thing where we put together the power rule and this natural log
anti-derivative so i have the integral of x cubed minus seven over x plus x to the tenth minus three
and the integral of x cubed well we just add one to the exponent and divide by the result that
gives me 1 4 x to the fourth the negative 7 over x well the negative 7 is a spectator there so i have
negative 7 natural log absolute value of x the x to the 10th i add 1 to the exponent and divide by
the result so i get 1 over 11 x to the 11 and the negative 3 that integrates to negative 3x
and i put in my arbitrary constant plus c and we're done if you enjoyed this video
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