### Video Transcript

Compute the left Riemann sum for π
of π₯ equals one over π₯ squared plus two on the closed interval negative three to
three, given that there are six subintervals of equal width. Approximate your answer to two
decimal places.

Remember when weβre writing a left
Riemann sum, we take values of π from zero to π minus one. And that gives us the value of π
at the left endpoint of each rectangle. The formula is the sum of π₯π₯
times π of π₯ π for values of π from zero to π minus one where π₯π₯ is π minus
π over π. Remember π and π are the lower
and upper limits of our interval, respectively, and π is the number of
subintervals. Then π₯ π is π plus π lots of
π₯π₯. Itβs only sensible to begin by
first calculating π₯π₯. We can see that our close interval
is from negative three to three. So weβre going to let π be equal
to negative three and π be equal to three.

Weβre interested in six
subintervals. So letβs let π be equal to
six. Then π₯π₯ is three minus negative
three over six, which is simply one. Next, weβll calculate what π₯ π
is. Itβs π, which we know to be
negative three, plus π₯π₯, which is one, lots of π. Thatβs of course negative three
plus π. Weβre looking to find what π of π₯
π is though. π of π₯ π must, therefore, be π
of negative three plus π. So letβs substitute π₯ equals
negative three plus π into our function. That gives us one over negative
three plus π squared plus two. And when we distribute the
parentheses, on the denominator we get π squared minus six π plus 11.

And now weβre ready to perform some
substitutions. Weβre finding the sum and weβre
taking values of π from zero to π minus one. Now, π is six. So π minus one is five. π₯π₯ is one and π of π₯ π is one
over π squared minus six π plus 11. But of course, we donβt really need
to write multiplied by one. So we need to evaluate this
sum. To do this, weβre going to
substitute values of π from zero through to five into this function and then find
their sum.

So when π equals zero, thatβs one
over zero squared minus zero plus 11. When π is one, thatβs one over one
squared minus six plus 11. When π is two, itβs one over two
squared minus 12 plus 11. And we repeat this process for π
equals three, π equals four, and π equals five. The very last thing to do is to
evaluate their sum. That gives us 1.5909 and so on,
which are correct to two decimal places is 1.59. Weβve computed the left Riemann sum
for our function over that closed interval, using six subintervals.