Original subtitles

1

Today we're going to talk about how to evaluate the integral of an odd function to complete this problem

2

will confirm that our function is on and then simplify and evaluate the integral.

3

In this particular problem we've been asked to evaluate the integral of x squared sine X divided by

4

the quantity 1 plus X and the sixth on the range negative PI over 2 to PI over to.

5

So the first thing you want to notice about this integral is that you're being asked to evaluate the

6

definite integral on the range.

7

We'll call it negative a to positive A where A is a constant.

8

So whenever you have the range negative a positive way where these two are equal to one another but

9

the lower value here is the negative version of the upper value.

10

You should consider checking to see whether or not your function is even or odd.

11

The reason it's valuable to check to see whether the function is even or odd is because when you have

12

the range here negative a positive A If the function is odd you know immediately that the value of your

13

definite integral is zero.

14

If the function is even you'll still be able to simplify it before you actually evaluate the integral.

15

So because we have negative the positive A for our range here.

16

Let's go ahead and check to see whether or not we have a function that's even or odd.

17

So we'll call the function here f of x will say that f of x is equal to x squared sine X all divided

18

by the quantity 1 plus X to the sixth.

19

So given that that's our function we now want to evaluate to see whether or not this is an even or odd

20

function way that we do that is by plugging negative x in for x.

21

So plug negative x in for x everywhere we see X in our function.

22

Let's go ahead and say negative x squared times sign of negative x all divided by 1 plus negative x

23

to the 6.

24

Remember that when you're looking to see whether a function is even or odd.

25

You plug in negative x if the result you get back is f of x the same thing you started with then you

26

know that your function is even if the result you get back is negative f of x.

27

In other words the original function.

28

But just with one negative sign out in front then you know that your function is odd.

29

So we're going to be looking for either of those two things.

30

In this case negative x squared just gives us positive x squared that negative sign cancels sign of

31

negative x is actually the same thing as negative sign of X and we'll talk about why that's true in

32

one second and then in the denominator here we have 1 plus negative x rays to the sixth power.

33

Whenever we have negative x here raise to a to an even exponent.

34

We know that we're going to end up with those negative signs cancelling and we'll just have a positive

35

x to the sixth.

36

So now when we simplify this we can pull the negative sign out in front the negative sign associated

37

with the sign of X here we pull it out in front and we end up with negative x squared sine X all divided

38

by 1 plus X to the 6.

39

As you can see what we have here is f a vector.

40

It's exactly the same as our original function.

41

The only difference is that we have a negative sign here out in front which means we're looking at negative

42

f of x and as we know here that means that our function is on.

43

Now just a quick caveat here to return to the sign of negative x.

44

Part of our problem if you're ever unsure whether or not you can simplify a Trigon a metric identity

45

like this by pulling the negative sign out in front.

46

You can always graph the function to see whether or not it's appropriate to call the negative sign out

47

in front.

48

So for example we know that the graph of sign of X if we look at an x y coordinate plane we know that

49

the graph of sign of X if we have PI over to PI 3 PI over 2 etc..

50

The graph of sign of X looks roughly like this come down at PI.

51

Here we have three PI over two.

52

It comes up to me the x axis at 2 pi where this is 1.

53

This is negative 1.

54

So that's roughly the graph of sign of X to graph sign of negative x we can just take it one point at

55

a time.

56

So for example if we plug in zero for X to sign of negative x we get sign of negative zero.

57

That's the same thing a sign of zero sign of zero we know is just 0.

58

So we get that point.

59

If we plug in PI over to two sign of negative x we'll get sign of negative PI over two.

60

We know that sign of negative pyo or two is negative 1.

61

So we get a value here of negative 1.

62

If we plug in high to sign of negative x we'll get a sign of negative Pi which we know is 0 and we can

63

continue doing that.

64

And what we would see is that the graph of sign of negative x is this graph right here.

65

And what this tells us is that we have exactly the opposite here so whenever the graph of sign of x

66

is positive at this point let's say it's you know positive three fourths it's negative three fourths

67

here when we're at positive one where it negative 1.

68

So it's the same value except with sign of x.

69

It's the positive value with sign of negative x.

70

It's the negative value and that's how we know that it's OK to pull that negative sign out in front

71

of the sign of x so you can always test it that way.

72

But given now that our function is odd what we can deduce about the integral is that this whole integral

73

here is equal to zero.

74

Just by understanding that the function we had inside here was odd.

75

We know that the integral is equal to zero.

76

And the reason is because an odd function is symmetrical about the origin.

77

So if you imagine something like this then I don't know what the graph exactly of this function looks

78

like.

79

But let's take a graph that symmetrical about the origin and maybe it looks like this.

80

Right.

81

If it's symmetrical about the origin as long as you're evaluating on the range Let's call this negative

82

high over 2.

83

So say as long as you are evaluating on the range negative PI over to 2 positive high over 2 you know

84

that you're going to have the same negative area here to cancel out the positive area over here.

85

And so because these two are going to cancel one another you know immediately that the area from the

86

on this range is equal to zero because it's the same distance from this point to the y axis as it is

87

from the y axis to this point the area to the left of the y axis has to be the same.

88

But the negative version of this area on the right side of the y axis and they're going to cancel each

89

other.

90

Your integral will always be equal to zero.

91

So that's it because we're evaluating here on the range negative a positive a.

92

And because we've identified a function as odd the fact that we have all of those things means that

93

we know that the integral here is equal to zero.