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    "Fundamental"genre: Noise
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    Calculus is fundamental.

    Lyrics
    the theorem:

    if

    little f of t is continuous on the open interval I
    and
    a is a constant on I
    and
    x and t are in I
    and
    big F of t is any antiderivative of little f of t,

    then

    the integral of little f of t
    with respect to t
    from a to x
    is equal to
    big F of x

    the proof:

    define big A of x to be
    the integral of little f of t
    with respect to t
    from a to x

    thus d big A dx
    is equal to
    the limit as h goes to naught of
    big A of quantity x plus h less big A of x
    all over h

    for any h greater than naught
    such that x plus h is in I
    big A of quantity x plus h
    is equal to
    the integral of little f of t
    with respect to t
    from a to quantity x plus h

    thus big A of quantity x plus h less big A of x
    is equal to
    the integral of little f of t
    with respect to t
    from a to quantity x plus h
    less the integral of little f of t
    with respect to t
    from a to x

    which is equal to
    the integral of little f of t
    with respect to t
    from x to a
    plus the integral of little f of t
    with respect to t
    from a to quantity x plus h

    [note that an integral
    from r to s
    is equal to
    the same integral
    from s to r
    multiplied by negative one]

    which is equal to
    the integral of little f of t
    with respect to t
    from x to quantity x plus h

    [note that an integral
    from b to c
    added to the same integral
    from c to d
    is equal to
    the same integral
    from b to d]

    let little m be the minimum value
    of little f of t
    on the interval
    from x to quantity x plus h

    let big M be the maximum value
    of little f of t
    on the interval
    from x to quantity x plus h

    little m multiplied by h
    must be less than (or equal to)
    the integral of little f of t
    with respect to t
    from x to quantity x plus h
    which must be less than (or equal to)
    big M multiplied by h

    if we divide the inequality through by h
    we see that
    little m
    is less than (or equal to)
    the integral of little f of t
    with respect to t
    from x to quantity x plus h
    all over h
    which is less than (or equal to)
    big M

    this inequality is the same as
    little m
    is less than (or equal to)
    big A of quantity x plus h
    less big A of x
    all over h
    which is less than (or equal to)
    big M

    take the limit as h goes to naught

    little m
    goes to little f of x

    big A of quantity x plus h
    less big A of x
    all over h
    goes to d big A dx

    big M
    goes to little f of x

    [remember that
    little m
    and big M
    were defined as
    the minimum
    and maximum
    of little f of t
    on the interval
    from x to quantity x plus h
    respectively

    as h goes to naught
    the minimum
    and maximum
    go towards little f of x]

    thus little f of x
    is less than (or equal to)
    d big A dx
    which is less than (or equal to)
    little f of x

    or
    little f of x
    equals
    d big A dx

    or
    the antiderivative of little f of x
    is big A of x

    or
    the integral of little f of t
    with respect to t
    from a to x
    is equal to
    big F of x

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