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www.mathcentre.ac.uk/.../9.2%20Integration%20as%20the%20reverse%20of%20differentiation.mp4

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    By now you have had a lot
    of opportunity to practice
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    differentiating common
    functions.
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    And you have learned many
    techniques of Differentiation.
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    So for example, if I were to
    give you a function, let's call
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    this function capital F of X.
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    If it's a fairly simple
    function, you'll know
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    how to differentiate it.
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    And so by differentiating it,
    you'll be able to calculate its
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    derivative, which you'll
    remember we denote as DF DX. So
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    that's a process that you're
    very familiar with already.
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    Now in this unit we're going to
    refer to DF DX as little F of X.
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    So little F of X.
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    Is the derivative of big
    F of X. Let me write
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    that down little F of X.
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    Is the derivative.
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    As big banks.
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    And as I say, that's a process
    that you're very familiar with.
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    What we want to do now is try to
    work this process in reverse.
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    Work it backwards. In other
    words, we want to start with
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    little F. And try and
    come back this route.
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    And try and find the function or
    functions capital F which when
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    they are differentiated will
    give you little Earth. So we
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    want to carry out this process
    in reverse. We can think of this
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    as anti differentiation.
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    So we can think of this anti
    differentiation as
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    differentiation in reverse.
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    So big F of X we're going to
    call the anti derivative of
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    little F of X.
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    So by these two results in mind
    and try not to get confused with
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    a capital F in the lower case F
    little F is the derivative of
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    Big F. Because little F
    is DF DX.
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    And big F is the anti derivative
    of little laugh.
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    Now you may have already looked
    at a previous video called
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    integration as summation.
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    And in that video, the concept
    of an integral is defined in
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    terms of the sum of lots of
    rectangular areas under a curve.
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    To calculate an integral, you
    need to find the limit of a son.
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    And that's a very cumbersome
    and impractical process. What
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    we're going to learn about in
    this video is how to find
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    integrals, not by finding the
    limit of a sum, but instead
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    by using antiderivatives.
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    Let's start off
    with the example.
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    Suppose we start off with the
    function capital F of X.
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    Is equal to three
    X squared plus 7X.
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    Minus 2.
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    And what I'm going to do is
    something that you're already
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    very familiar with. We're
    going to differentiate it.
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    And if we differentiate it.
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    We get TF DX equals.
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    Turn by turn the derivative of
    three X squared is going to be.
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    2306 X.
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    And the derivative of Seven X is
    just going to be 7.
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    What about the minus two? Well,
    you remember that the derivative
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    of a constant is 0, so when we
    do this differentiation process,
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    the minus two disappears.
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    So our derivative DF DX is just
    six X +7.
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    So I'm going to write little F
    of X is 6 X +7.
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    And think about what will happen
    when we reverse the process when
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    we do anti differentiation.
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    Ask yourself what is an anti
    derivative of 6X plus 7?
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    Now we already have an answer to
    this question and anti
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    derivative of 6X plus 7.
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    Is 3 X squared plus 7X minus
    two, so the answer.
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    Is F of X is 3 X squared plus
    7X minus two, but I'm afraid
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    that's not the whole story.
    We've already seen that when we
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    differentiate it, three X
    squared plus 7X minus two, the
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    minus two disappeared.
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    In other words, whatever number
    had been in here, whether that
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    minus two had been minus 8 or
    plus 10 or 0, whatever it would
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    still have disappeared. We've
    lost some information during
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    this process of Differentiation,
    and when we want to reverse it
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    when we want to start with the
    six, XX have Seven, and working
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    backwards, we've really no idea
    what that minus two might have
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    been. It could have been another
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    number. So this leads us to the
    conclusion that once we found an
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    antiderivative like this one,
    three X squared plus, 7X minus
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    2. Than any constant added
    onto this will still be an
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    anti derivative.
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    If F of X.
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    Is an anti derivative.
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    A little F of X.
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    Then so too.
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    Is F of X the anti derivative
    we've found plus any constant at
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    all we choose. See is what we
    call an arbitrary constant. Any
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    value of see any constant we can
    add on to the anti derivative
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    we've found and then we've got
    another antiderivative. So in
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    fact there are lots and lots of
    antiderivatives of 6X plus
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    Seven, just as another example,
    we could have had over here.
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    Three X squared plus 7X minus 8.
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    Or even just three X squared
    plus 7X where the constant term
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    was zero. So lots and lots of
    different antiderivatives for a
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    single term over here.
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    Let's do another example.
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    Suppose this time we look at.
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    A cubic function that suppose F
    of X is 4X cubed.
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    Minus Seven X squared.
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    Plus 12X
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    minus 4.
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    Again, you know how to
    differentiate this. You've
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    had a lot of practice
    differentiating functions
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    like this.
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    So the derivative DF
    DX is going to be.
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    Three 412 X squared.
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    The derivative of minus
    Seven X squared is going to
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    be minus 14X.
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    And the derivative
    of 12 X is just 12.
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    So think of our little F of X as
    being the function 12 X squared
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    minus 14X at 12.
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    Notice again the point that
    the minus four in the
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    differentiation process
    disappears.
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    So now ask the question, what's
    an anti derivative of this
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    function here little F.
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    Well, we've got one answer. It's
    on the page already. It's this
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    big F is an anti derivative of
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    Little F. But we've seen
    that any constant added on
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    here will still yield
    another antiderivative, so
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    we can write down lots of
    other ones, for example.
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    If we add on a constant.
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    And let's suppose we add on the
    constant 10 onto this one.
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    Then we'll get the function 4X
    cubed, minus Seven X squared.
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    Plus 12X plus six is also an
    anti derivative of Little F.
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    And the same goes by adding
    adding on any other constant at
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    all we choose. If we add on
    minus six to this.
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    Will get 4X cubed, minus Seven X
    squared plus 12X and adding on
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    minus six or just leave us this
    so that that is another
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    antiderivative. There's an
    infinite number of
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    antiderivatives of Little F.
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    Now in all the examples we've
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    looked at. I've in this in a
    sense, giving you the answer
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    right at the beginning because
    we started with big F, we
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    differentiated it to find little
    F, so we knew the answer all the
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    time. In practice, you won't
    know the answer all the time, so
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    one way that we can help
    ourselves by referring to a
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    table of antiderivatives. Now a
    table of antiderivatives will
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    look something like this one.
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    There's a copy of this in the
    notes accompanying the video and
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    a table of antiderivatives were
    list lots of functions.
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    F.
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    And then the anti derivative.
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    Big F in the next column and
    you'll see every anti
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    derivative will have a plus.
    See attached to it where C is
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    an arbitrary constant. Any
    constant you choose.
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    What I want to do now is I
    want to link the concept
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    of an antiderivative with
    integration as summation.
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    Let's think of a function.
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    Y equals F of X.
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    Let's suppose the graph of the
    function looks like this.
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    I'm going to consider
    functions which lie entirely
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    above the X axis, so I'm
    going to restrict our
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    attention to the region above
    the X axis were looking up
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    here.
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    And also I want to restrict
    attention to the right of the Y
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    axis, so I'm looking to the
    right of this line here.
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    Let's ask ourselves what is the
    area under the graph of Y equals
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    F of X?
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    Well, surely that depends on
    how far I want to move to the
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    right hand side. Let's
    suppose I want to move.
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    To the place where X
    has an X Coordinate X.
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    Then clearly, if X is a large
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    number. The area under this
    graph here will be large,
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    whereas if X is a small
    number, the area under the
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    graph will be quite small
    indeed. Effects is actually 0,
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    so this line is actually lying
    on the Y axis than the area
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    under the graph will be 0.
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    I'm going to do note the
    area under the graph by a,
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    so A is the area.
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    Under
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    why is F of X?
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    And as I've just said, the area
    will depend upon the value that
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    we choose for X. In other
    words, A is a function of XA
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    depends upon X, so we write
    that like this, a is a of X and
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    that shows the dependence of
    the area on the value we choose
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    for X. As I said, LG X larger
    area small acts smaller area.
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    Ask yourself. What is the height
    of this line here?
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    Well, this point here.
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    Lies. On the curve Y equals F of
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    X. So the Y value at that point
    is simply the function evaluated
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    at this X value.
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    So the Y value here is just F of
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    X. In other words, the
    height of this line or the
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    length of this line is just
    F of X. The function
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    evaluated at this point.
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    Now I'd like you to think what
    happens if we just increase X by
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    a very small amount.
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    So I want to just move this
    point a little bit further
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    to the right.
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    And see what happens.
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    I'm going to increase X by a
    little bit and that little bit
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    that distance in here. I'm going
    to call Delta X. Delta X stands
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    for a small change in X or we
    call it a small increment in X.
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    By increasing axle little bit.
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    What we're doing is we're adding
    a little bit more to this area.
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    This is the additional
    contribution to the area. This
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    shaded region in here.
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    And that's a little bit extra
    area, so I'm going to call that.
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    Delta A, That's an incrementing
    area changing area.
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    I'm going to write down
    an expression for Delta
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    a try and work it out.
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    Let's try and see what it is,
    but I can't get it exactly. But
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    if I note that the height of
    this line is F of X.
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    And the width of this column in
    here is Delta X. Then I can get
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    an approximate value for this
    area by assuming that it's a
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    rectangular section. In other
    words, I'm going to ignore this
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    little bit at the top in there.
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    If I assume it's a rectangular
    section, then the area here this
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    additional area Delta A.
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    Is F of X multiplied by Delta X.
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    Now, that's only approximately
    true, so this is really an
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    approximately equal to symbol.
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    This additional area is
    approximately equal to
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    F of X Times Delta X.
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    Let me now divide both sides by
    Delta X. That's going to give me
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    Delta a over Delta. X is
    approximately equal to F of X.
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    How can we make it more
    accurate? Well, one way we can
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    make this more accurate is by
    choosing this column to be even
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    thinner by letting Delta X be
    smaller, because then this
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    additional contribution that
    I've got in here that I didn't
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    count is reducing its reducing
    in size. So what I want to do is
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    I want to let Delta X get even
    smaller and smaller and in the
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    end I want to take the limit.
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    As Delta X tends to zero of
    Delta over Delta X.
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    And when I do that, this
    approximation in here will
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    become exact and that will give
    us F of X.
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    Now, if you've studied the unit
    on differentiation from first
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    principles, you realize that
    this in here is the definition
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    of the derivative of A with
    respect to X, which we write as
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    DADX. So we have the result that
    DADX is a little F of X.
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    Let's explore this a
    little bit further.
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    We have the ADX is a little
    F of X.
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    What does this mean? Well,
    it means that little laugh
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    is the derivative of A.
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    But it also means if we think
    back to the discussion that we
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    had at the beginning of this
    video, that a must be the anti
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    derivative of F. A is
    an anti derivative.
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    F. In other words, the area is
    F of X. The anti derivative of
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    little F was any constant.
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    That's going to be an important
    result. What it's saying is that
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    if we have a function, why is F
    of X and we want to find the
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    area under the graph? What we do
    is we calculate an anti
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    derivative of Little F which is
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    big F. And use this expression.
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    To find the area.
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    There's one thing we don't know
    in this expression at the
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    moment, and it's this CC.
    Remember, is an arbitrary
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    constant, but we can get a value
    for. See if we just look back
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    again. So the graph I drew.
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    And ask yourself what will be
    the area under this curve when X
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    is chosen to be zero? Well, if
    you remember, we said that if
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    this vertical line here had been
    on the Y axis.
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    Then the area under the
    curve would have been 0.
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    So this gives us a
    condition. It tells us that
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    when X is 0, the area is 0.
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    When X is 0.
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    Area is 0. What does this mean?
    While the area being 0?
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    X being 0.
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    Plus a constant and this
    condition then gives us a value
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    for C, so C must be equal to
    minus F of North.
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    And that value for C and go back
    in this result here. So we have
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    the final result that the area
    under the graph is given by big
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    F of X minus big F of note.
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    Let me just write that
    down again.
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    OK.
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    Now let's look at this problem.
    Supposing that I'm interested in
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    finding the area under the graph
    of Y is little F of X.
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    Up to the point.
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    Where X has the value be.
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    So I want the area from the Y
    axis, which is where we're
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    working from above the X axis up
    to the point.
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    Where X equals B.
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    We can use this boxed
    formula here.
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    When X is be will get a of B.
    That's the area up to be.
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    Will be F of B.
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    Minus F of note.
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    So this expression will give you
    the area up to be.
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    Suppose now I want.
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    Area up to A and let's
    Suppose A is about here.
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    So now I'm interested.
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    In this area in here, which is
    the area from the Y axis up to a
  • 20:13 - 20:17
    well, again using the same
    formula, the area up to a which
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    is a of A.
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    Is F evaluated at the X value
    which is a?
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    Subtract. Big F of note.
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    So I have two expressions on
    the page here, one for the
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    area up to be.
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    And one for the area up to A.
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    Ask yourself How do I find
    this area in here? That's
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    the area between A&B.
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    Well, the area between A&B we
    can think of as the area up to
  • 20:54 - 20:58
    be. Subtract the area up to A.
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    So if we find the difference of
    these two quantities, that will
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    give us the area between A&B.
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    So we get the area.
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    Under Wise F of X.
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    From X equals A to X equals
  • 21:19 - 21:23
    B. Is the area under
    this graph from X is A
  • 21:23 - 21:26
    to X is B is given by?
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    Once the area up to be, subtract
    the area up to A.
  • 21:31 - 21:33
    So if we just find the
    difference of these two
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    quantities will have F of B
    minus F of A.
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    Add minus F of not minus minus F
    of notes. These terms will
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    cancel out. So In other words
    that's the result we need. The
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    area under this graph between
    A&B is just.
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    Big F of B minus big FA.
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    Let me just write that
    down again.
  • 22:26 - 22:31
    The area and Y equals F of X
    between ex is an ex is B is
  • 22:31 - 22:36
    given by big F of B minus big F
    of a where big F remember is an
  • 22:36 - 22:37
    anti derivative of Little F.
  • 22:42 - 22:43
    Any answer derivative?
  • 22:54 - 22:58
    So this is a very important
    result because it means that if
  • 22:58 - 22:59
    I give you a function little F
  • 22:59 - 23:04
    of X. And you can calculate an
    anti derivative of its big F.
  • 23:05 - 23:08
    And you can find the area
    under the graph merely by
  • 23:08 - 23:11
    evaluating that anti
    derivative at B, evaluating
  • 23:11 - 23:14
    it A and finding the
    difference of the two
  • 23:14 - 23:14
    quantities.
  • 23:16 - 23:20
    Now in the previous video on
    integration by summation, the
  • 23:20 - 23:23
    area under under a graph was
    found in a slightly different
  • 23:23 - 23:25
    way. What was done there?
  • 23:25 - 23:31
    Was that the area under the
    graph between A&B was found by
  • 23:31 - 23:35
    dividing the area into lots of
    thin rectangular strips?
  • 23:36 - 23:40
    Finding the area of each of the
    rectangles and adding them all
  • 23:40 - 23:45
    up. And doing that we had this
    result that the area under the
  • 23:45 - 23:50
    graph between A&B was given by
    the limit as Delta X tends to
  • 23:50 - 23:55
    zero of the sum of all these
    rectangular areas, which was F
  • 23:55 - 24:01
    of X Delta X between X is A and
    X is be. If you're not familiar
  • 24:01 - 24:06
    with that, I would advise you go
    back and have a look at that
  • 24:06 - 24:09
    other video on integration by
    summation, but that formula.
  • 24:10 - 24:13
    For this area was derived
    in that video.
  • 24:14 - 24:18
    And this formula
    defines what we mean
  • 24:18 - 24:21
    by the definite
    integral from A to B
  • 24:21 - 24:24
    of F of X DX.
  • 24:25 - 24:28
    That's how we defined
    a definite integral.
  • 24:29 - 24:32
    And if we wanted to define a
    definite integral, if we wanted
  • 24:32 - 24:36
    to calculate a definite integral
    in that video, we had to do it
  • 24:36 - 24:38
    through the process of finding
    the limit of a sum.
  • 24:40 - 24:44
    Now the the process of finding
    the limit of a sum is quite
  • 24:44 - 24:47
    cumbersome and impractical. But
    what we've learned now is that
  • 24:47 - 24:51
    we don't have to use the limit
    of a sum to find the area. We
  • 24:51 - 24:53
    can use these anti derivatives.
  • 24:54 - 24:57
    Because we've got two
    expressions for the area, we've
  • 24:57 - 25:01
    got this expression. As a
    definite integral, and we've got
  • 25:01 - 25:03
    this expression in terms of
    antiderivatives, and if we put
  • 25:03 - 25:05
    all that together will end up
  • 25:05 - 25:09
    with this result. The definite
    integral from A to B.
  • 25:10 - 25:12
    F of X DX.
  • 25:13 - 25:14
    Is FB.
  • 25:16 - 25:18
    Minus F of A.
  • 25:22 - 25:26
    In other words, the definite
    integral of little F between the
  • 25:26 - 25:30
    limits of A&B is found by
    evaluating this expression,
  • 25:30 - 25:34
    where big F is an anti
    derivative of Little F.
  • 25:40 - 25:42
    Let me write that formula
    down again.
  • 25:43 - 25:45
    The integral from A to B.
  • 25:47 - 25:54
    F of X DX is given by
    big FB minus big F of A.
  • 25:56 - 25:57
    Let me give you an example.
  • 26:02 - 26:06
    Suppose we're interested
    in the problem of finding
  • 26:06 - 26:11
    the area under the graph
    of Y equals X squared.
  • 26:15 - 26:17
    And let's suppose for the sake
    of argument we want the area
  • 26:17 - 26:22
    under the graph. Between X is
    not an ex is one, so we're
  • 26:22 - 26:23
    interested in this area.
  • 26:26 - 26:30
    In the previous unit on
    integration as a summation,
  • 26:30 - 26:34
    this was done by dividing
    this area into lots of thin
  • 26:34 - 26:35
    rectangular strips.
  • 26:36 - 26:40
    And finding the area of each of
    those rectangles separately.
  • 26:41 - 26:42
    And then adding them all up.
  • 26:43 - 26:48
    That gave rise to this formula
    that the area is the limit.
  • 26:49 - 26:51
    As Delta X tends to 0.
  • 26:52 - 26:57
    Of the sum from X equals not to
    X equals 1.
  • 26:58 - 27:00
    Of X squared Delta X.
  • 27:01 - 27:05
    So the area was expressed as the
    limit of a sum.
  • 27:06 - 27:12
    And in turn, that defines the
    definite integral. X is not to
  • 27:12 - 27:14
    one of X squared DX.
  • 27:16 - 27:20
    Now open till now if you
    wanted to work this area out
  • 27:20 - 27:25
    the way you would do it would
    be by finding the limit of
  • 27:25 - 27:27
    this some, but that's
    impractical and cumbersome.
  • 27:27 - 27:30
    Instead, we're going to use
    this result using
  • 27:30 - 27:30
    antiderivatives.
  • 27:32 - 27:35
    Our little F of X in this
    case is X squared.
  • 27:40 - 27:45
    And this formula at the top of
    the page tells us that we can
  • 27:45 - 27:47
    evaluate this definite
    integral by finding an
  • 27:47 - 27:50
    antiderivative capital F. So
    we want to do that first.
  • 27:50 - 27:53
    Well, if little F is X
    squared, big F of X, well we
  • 27:53 - 27:57
    want an anti derivative of X
    squared and if you don't know
  • 27:57 - 28:00
    what one is, you refer back to
    your table.
  • 28:01 - 28:05
    An anti derivative of X
    squared is X cubed over 3
  • 28:05 - 28:06
    plus a constant.
  • 28:14 - 28:19
    So in order to calculate this
    definite integral, what we need
  • 28:19 - 28:20
    to do is evaluate.
  • 28:21 - 28:25
    The Anti Derivative Capital F.
    At B. That's the upper limit,
  • 28:25 - 28:27
    which in this case is one.
  • 28:29 - 28:32
    And then at the lower limit,
    which in the in our case is
  • 28:32 - 28:34
    zero. Let's workout F of one.
  • 28:36 - 28:38
    Well, F of one is going to be 1
  • 28:38 - 28:41
    cubed. Over 3, which is
    just a third.
  • 28:42 - 28:43
    Plus C.
  • 28:45 - 28:47
    Let's workout big F of 0.
  • 28:48 - 28:50
    Big F of 0.
  • 28:51 - 28:54
    Is going to be 0 cubed over
    three, which is 0 plus. See, so
  • 28:54 - 28:58
    it's just see. And then we
    want to find the difference.
  • 28:59 - 29:01
    Half of 1 minus F of note.
  • 29:03 - 29:06
    Will be 1/3 plus C minus C, so
    the Seas will cancel and would
  • 29:06 - 29:08
    be just left with the third.
  • 29:09 - 29:12
    In other words, to calculate
    this integral here.
  • 29:13 - 29:16
    All we have to do is find the
    Anti Derivative Capital F.
  • 29:17 - 29:21
    Evaluate it at the upper limit
    evaluated at the lower limit.
  • 29:21 - 29:25
    Find the difference and the
    result is the third. That's the
  • 29:25 - 29:28
    area under this graph between
    North and one.
  • 29:30 - 29:32
    Now let me just show you how we
    would normally set this out.
  • 29:33 - 29:35
    We would normally set this
    out like this.
  • 29:39 - 29:44
    We find. An anti derivative of X
    squared, which we've seen is X
  • 29:44 - 29:46
    cubed over 3 plus C.
  • 29:49 - 29:53
    We don't normally write the Plus
    C down, and the reason for that
  • 29:53 - 29:55
    is when we're finding definite
    integrals, the sea will always
  • 29:55 - 29:59
    cancel out as we saw here, the
    Seas cancelled out in here and
  • 29:59 - 30:00
    that will always be the case.
  • 30:00 - 30:04
    So we don't actually need to
    write a plus C down when we
  • 30:04 - 30:05
    write down an anti
    derivative of X squared.
  • 30:07 - 30:10
    It's conventional to write down
    the anti derivative in square
  • 30:10 - 30:15
    brackets. And to transfer the
    limits on the original integral,
  • 30:15 - 30:18
    the Norton one to the right hand
    side over there like so.
  • 30:20 - 30:24
    What we then want to do is
    evaluate the anti derivative at
  • 30:24 - 30:28
    the top limit that corresponded
    to the FB or the F of one. So we
  • 30:28 - 30:32
    work this out at the top limit
    which is 1 cubed over 3.
  • 30:35 - 30:39
    We work it out at the bottom
    limit. That's the F of a or
  • 30:39 - 30:40
    the F of Norte.
  • 30:41 - 30:45
    We work this out of the lower
    limits will just get zero and
  • 30:45 - 30:47
    then we want the difference
    between the two.
  • 30:48 - 30:52
    Which is just going to give
    us a third. So that's the
  • 30:52 - 30:54
    normal way we would set out a
    definite integral.
  • 31:00 - 31:01
    So as we've seen.
  • 31:03 - 31:07
    Definite integration is very
    closely associated with Anti
  • 31:07 - 31:08
    Differentiation.
  • 31:10 - 31:14
    Because of this, in general it's
    useful to think of Anti
  • 31:14 - 31:18
    Differentiation as integration
    and then we would often refer to
  • 31:18 - 31:22
    a table like this one of
    antiderivatives as simply a
  • 31:22 - 31:23
    table of integrals.
  • 31:24 - 31:27
    And there will be a table of
    integrals in the notes and in
  • 31:27 - 31:30
    later videos you'll be looking
    in much more detail at how to
  • 31:30 - 31:33
    use a table of integrals. So
    you think of the table of
  • 31:33 - 31:35
    integrals as your table of
    antiderivatives and vice versa.
  • 31:37 - 31:40
    Very early on in the video, we
    looked at this problem. We
  • 31:40 - 31:44
    started off with little F of X
    being equal to four X cubed.
  • 31:45 - 31:47
    Minus Seven X squared.
  • 31:48 - 31:51
    Plus 12X minus 4.
  • 31:52 - 31:54
    We differentiate it.
  • 31:57 - 32:02
    To give 12 X squared
    minus 14X plus 12.
  • 32:03 - 32:08
    And then we said that an anti
    derivative of 12 X squared minus
  • 32:08 - 32:09
    14X plus 12.
  • 32:10 - 32:13
    Was this function
    capital F plus C?
  • 32:14 - 32:17
    We've got a notation we
    can use now because we've
  • 32:17 - 32:20
    linked antiderivatives to
    integrals and the notation
  • 32:20 - 32:24
    we would use is that the
    integral of 12 X squared.
  • 32:25 - 32:31
    Minus 14X plus 12 DX
    is equal to.
  • 32:32 - 32:33
    4X cubed
  • 32:34 - 32:40
    minus Seven X squared plus 12X
    plus any arbitrary constant.
  • 32:41 - 32:44
    And we call this an
    indefinite integral.
  • 32:50 - 32:55
    And we say that the indefinite
    integral of 12 X squared minus
  • 32:55 - 33:00
    14X plus 12 with respect to X is
    4X cubed minus Seven X squared
  • 33:00 - 33:03
    plus 12X. Plus a constant
    of integration.
  • 33:07 - 33:11
    OK, In summary, what have
    we found? What we found
  • 33:11 - 33:12
    that a definite integral?
  • 33:14 - 33:20
    The integral from A to B of
    little F of X DX. We found that
  • 33:20 - 33:22
    this is a number.
  • 33:23 - 33:28
    And it's obtained from the
    formula F of B minus FA,
  • 33:28 - 33:33
    where big F is any anti
    derivative of Little F.
  • 33:34 - 33:38
    And we've also seen the
    indefinite integral of F
  • 33:38 - 33:40
    of X DX.
  • 33:41 - 33:46
    Is a function big F of X plus an
    arbitrary constant? Again, where
  • 33:46 - 33:51
    F is any anti derivative of
    Little F&C is an arbitrary
  • 33:51 - 33:56
    constant. So in this video we've
    learned how to do
  • 33:56 - 33:59
    differentiation in reverse.
    We've learned about
  • 33:59 - 34:00
    antiderivatives. Definite
  • 34:00 - 34:04
    integrals. And indefinite
    integrals in subsequent videos.
  • 34:04 - 34:08
    You learn a lot more teak
    techniques of integration.
Title:
www.mathcentre.ac.uk/.../9.2%20Integration%20as%20the%20reverse%20of%20differentiation.mp4
Video Language:
English
Duration:
34:14

English subtitles

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