NARRATOR: Exercise 5.
Using the formula that the sum of j from j = 1 to n
is equivalent to 1 + 2 + etc, up to n,
which is equivalent to n times n + 1 all over 2,
calculate the right endpoint estimate for the area under the graph y = x
between x = 0 and x = 1.
Take a limit as n approaches infinity to find the exact area.
Confirm your answer by elementary geometry.
So first we'll consider the right endpoint estimate
under the function y = x between x = 0 and x = 1.
So this interval from 0 to 1 is divided into n rectangles.
This means that each rectangle has a width of 1 divided by n.
This also means that each x value
that marks a division between the rectangles,
which we'll denote xj,
is equal to j times the width of the rectangles, that is 1 over n.
For example, x1 is at the position 1 times 1 over n,
x2 is at the position 2 times 1 over n, which is 2 over n.
xn is at the position n times 1 over n,
which is n over n, which is indeed 1.
So then we can establish our right endpoint estimate.
And we know that in the right endpoint estimate
the first rectangle has a height
equivalent to the value of the function at x1,
which in this case, given that the function is y = x,
the height of the rectangle is also just x1.
And then we need to multiply it by the width of the rectangle,
which we've already established to be 1 over n.
The next rectangle has a height of x2 multiplied by 1 over n for the area.
Right up to the final rectangle which has a height of xn
multiplied by the width 1 over n for the area of that rectangle.
Now, we've also established
that x1, x2, x3, etc,
in fact, in general xj,
is equivalent to j over n.
So that means that x1 is equivalent to 1 over n,
x2 is 2 over n, x3 is 3 over n, etc,
right up to xn, which is n over n.
And so we can modify
our right endpoint estimate as shown.
Factoring out the common factor of 1 over n squared,
we now end up with this expression,
and so we can now use summation notation to summarise that bracket.
So we have a right endpoint estimate equivalent to 1 over n squared
multiplied by the sum of j from j = 1 up to j = n.
So now that we've established our right endpoint estimate,
we can now use the formula provided in the question,
which was that the sum of j from j = 1 up to n
is equivalent to n times (n + 1) over 2.
And so making that substitution in our right endpoint estimate,
we see that we can simplify the expression
to give the following.
So we now have a right endpoint estimate
of half times (1 plus 1 over n).
And to calculate the exact area of that estimate,
we would be looking to take the limit as n approaches infinity.
And what we're doing there
is saying that we're going to divide that region from x = 0 to x = 1
into infinitely many rectangles,
each with an infinitely small width,
and hence that's going to approach the exact area.
So as n approaches infinity in this expression,
we'll have one over a very, very large number,
which is also approaching zero.
And so we end up with the exact area equal to one half.
We can confirm that exact area by using geometry.
So we know that this exact area forms a triangle
and the area of a triangle is half times the base times the height.
We know that in this case the base length of that triangle is 1
and the height of the triangle is also 1,
so geometry also confirms that the exact area under the curve y = x
between x = 0 and x = 1
is in fact one-half.
And we see that this is equivalent to the exact area that we obtained
using the right endpoint estimate
and then taking the limit as n approaches infinity.