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NARRATOR: Exercise 5.

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Using the formula that
the sum of j from j = 1 to n

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is equivalent to
1 + 2 + etc, up to n,

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which is equivalent to
n times n + 1 all over 2,

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calculate the right endpoint estimate
for the area under the graph y = x

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between x = 0 and x = 1.

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Take a limit as n approaches infinity
to find the exact area.

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Confirm your answer
by elementary geometry.

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So first we'll consider
the right endpoint estimate

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under the function y = x
between x = 0 and x = 1.

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So this interval from 0 to 1
is divided into n rectangles.

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This means that each rectangle
has a width of 1 divided by n.

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This also means that each x value

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that marks a division
between the rectangles,

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which we'll denote xj,

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is equal to j times the width of
the rectangles, that is 1 over n.

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For example, x1 is at the position
1 times 1 over n,

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x2 is at the position
2 times 1 over n, which is 2 over n.

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xn is at the position
n times 1 over n,

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which is n over n,
which is indeed 1.

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So then we can establish
our right endpoint estimate.

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And we know that in
the right endpoint estimate

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the first rectangle has a height

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equivalent to the value
of the function at x1,

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which in this case,
given that the function is y = x,

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the height of the rectangle
is also just x1.

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And then we need to multiply it
by the width of the rectangle,

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which we've already established
to be 1 over n.

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The next rectangle has a height of x2
multiplied by 1 over n for the area.

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Right up to the final rectangle
which has a height of xn

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multiplied by the width 1 over n
for the area of that rectangle.

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Now, we've also established

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that x1, x2, x3, etc,

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in fact, in general xj,

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is equivalent to j over n.

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So that means that
x1 is equivalent to 1 over n,

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x2 is 2 over n, x3 is 3 over n, etc,

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right up to xn, which is n over n.

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And so we can modify

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our right endpoint estimate as shown.

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Factoring out the common factor
of 1 over n squared,

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we now end up with this expression,

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and so we can now use summation
notation to summarise that bracket.

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So we have a right endpoint estimate
equivalent to 1 over n squared

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multiplied by the sum of j
from j = 1 up to j = n.

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So now that we've established
our right endpoint estimate,

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we can now use the formula
provided in the question,

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which was that the sum of j
from j = 1 up to n

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is equivalent to
n times (n + 1) over 2.

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And so making that substitution
in our right endpoint estimate,

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we see that we can
simplify the expression

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to give the following.

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So we now have
a right endpoint estimate

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of half times (1 plus 1 over n).

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And to calculate the exact area
of that estimate,

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we would be looking to take the limit
as n approaches infinity.

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And what we're doing there

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is saying that we're going to divide
that region from x = 0 to x = 1

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into infinitely many rectangles,

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each with an infinitely small width,

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and hence that's going to
approach the exact area.

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So as n approaches infinity
in this expression,

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we'll have one over
a very, very large number,

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which is also approaching zero.

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And so we end up with
the exact area equal to one half.

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We can confirm that exact area
by using geometry.

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So we know that this exact area
forms a triangle

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and the area of a triangle is
half times the base times the height.

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We know that in this case
the base length of that triangle is 1

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and the height of
the triangle is also 1,

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so geometry also confirms that the
exact area under the curve y = x

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between x = 0 and x = 1

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is in fact one-half.

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And we see that this is equivalent
to the exact area that we obtained

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using the right endpoint estimate

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and then taking the limit
as n approaches infinity.