hi everyone today we're going to talk about how to evaluate the integral of an odd function
to complete this problem we'll confirm that our function is odd and then simplify and
evaluate the integral let's take a look in this particular problem we've been asked to
evaluate the integral of x squared sign x divided by the quantity one plus x to the
sixth on the range negative pi over two to pi over two so the first thing you want to
notice about this integral is that you're being asked to evaluate the definite integral
on the range we'll call it negative a to positive a where a is a constant so whenever you have
the range negative a to positive a where these two are equal to one another but the lower
value here is the negative version of the upper value you should consider checking to
see whether or not you're function is even or odd the reason it's valuable to check to
see whether the function is even or odd is because when you have the range here negative
a to positive a if the function is odd you know immediately that the value of your definite
integral is zero if the function is even you'll still be able to simplify it before you actually
evaluate the integral so because we have negative a to positive a for our range here let's go
ahead and check to see whether or not we have a function that's even or odd so we'll call
the function here f of x we'll say that f of x is equal to x squared sign x all divided
by the quantity one plus x to the sixth so given that that's our function we know want
to evaluate to see whether or not to see whether or not this is an even or odd function the
way that we do that is by plugging negative x in for x so we'll plug negative x in for
x everywhere where we see x in our function let's go ahead and say negative x squared
times sine of negative x all divided by one plus negative x to the sixth remember that
when you're looking to see whether a function is even or odd you plug in negative x if the
result you get back is f of x the same thing you started with then you know that your function
is even if the result you get back is negative f of x in other words the original function
but just with one negative sign out in front then you know that your function is odd so
we're going to be looking for either of those two things in this case negative x squared
just gives us positive x squared that negative sine cancels sine of negative x is actually
the same thing as negative sine of x and we'll talk about why that's true in one second and
then in the denominator here we have one plus negative x raised to the sixth power whenever
we have negative x here raised to a even exponent we know that we're going to end up with those
negative signs canceling and we'll just have a positive x to the sixth so now when we simplify
this we can pull the negative sign out in front the negative sign associated with the
sign of x here we pull it out in front and we end up with negative x squared sine of
x all divided by one plus x to the six and as you can see what we have here is f of x
it's exactly the same as our original function the only difference it that we have a negative
sign here out in front which means we're looking at negative f of x and as we know here that
means that our function odd now just a quick caveat here to return to the sine of negative
x part of our problem if you're ever unsure whether or not you can simplify a trigonometric
identity like this by pulling the negative sign out in front you can always graph the
function to see whether or not it's appropriate to pull the negative sign out in front so
for example we know that the graph of sine of x if we look at an x y coordinate plane
we know that the graph of sine of x if we have pi over two pi three pi over two etcetera
the graph of sine of x looks roughly like this we come down at pi here we have three
pi over two it comes up to meet the x axis at two pi where this is one and this is negative
one so that's roughly the graph of sine of x to graph sine of negative x we can just
take it one point at a time so for example if we plug in zero for x two sine of negative
x we get sine of negative zero that's the same thing as sine of zero sine of zero we
know is just zero so we get that point if we plug in pi over two to sine of negative
x we'll get sine of negative pi over two we know that sine of negative pi over two is
negative one so we get a value here of negative one if we plug in pi to sine of negative x
we'll get sine of negative pi which we know is zero and we can continue doing that and
what we would see is that the graph of sine of negative x is this graph right here and
what this tells us is that we have exactly the opposite here so whenever the graph of
sine of x is positive at this point let's say it's you know positive three fourths it's
negative three fourths here when we're at positive one we're at negative one so it's
the same value except with sine of x it's the positive value with sine of negative x
it's the negative value and that's how we know that it's ok to pull that negative sine
out in front of the sine of x so you can always test it that way but given now that our function
odd what we can deduce about the integral is that this whole integral here is equal
to zero just by understanding that the function that we had inside here was odd we know that
the integral was equal to zero and the reason is because an odd function is symmetrical
about the origin so if you imagine something like and I don't know what the graph exactly
of this function looks like but let's take a graph symmetrical about the origin and maybe
it looks like this right if it's symmetrical about the origin as long as your evaluating
along the range let's call this negative pi over two so we'll say as long as your evaluating
along the range negative pi over two to positive pi over two you know that you're going to
have the same negative area here to cancel out the positive area over here and so because
these two are going to cancel one another you know immediately that the area from the
on this range is equal to zero because it's the same distance from this point to the y
axis as it is from the y axis to this point the area to the left of the y axis has to
be the same but the negative version of this area on the right side of the y axis and they're
going to cancel each other your integral will always be equal to zero so that's it because
we're evaluating here on the range negative a to positive a and because we've identified
a function as odd the fact that we have both of those things means that we know that the
integral here is equal to zero so I hope you found that video helpful if you did like this
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