Video Transcript
Compute the left Riemann sum for 𝑓
of 𝑥 equals one over 𝑥 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 we’re writing a left
Riemann sum, we take values of 𝑖 from zero to 𝑛 minus one. And that gives us the value of 𝑓
at the left endpoint of each rectangle. The formula is the sum of 𝛥𝑥
times 𝑓 of 𝑥 𝑖 for values of 𝑖 from zero to 𝑛 minus one where 𝛥𝑥 is 𝑏 minus
𝑎 over 𝑛. Remember 𝑎 and 𝑏 are the lower
and upper limits of our interval, respectively, and 𝑛 is the number of
subintervals. Then 𝑥 𝑖 is 𝑎 plus 𝑖 lots of
𝛥𝑥. It’s only sensible to begin by
first calculating 𝛥𝑥. We can see that our close interval
is from negative three to three. So we’re going to let 𝑎 be equal
to negative three and 𝑏 be equal to three.
We’re interested in six
subintervals. So let’s let 𝑛 be equal to
six. Then 𝛥𝑥 is three minus negative
three over six, which is simply one. Next, we’ll calculate what 𝑥 𝑖
is. It’s 𝑎, which we know to be
negative three, plus 𝛥𝑥, which is one, lots of 𝑖. That’s of course negative three
plus 𝑖. We’re looking to find what 𝑓 of 𝑥
𝑖 is though. 𝑓 of 𝑥 𝑖 must, therefore, be 𝑓
of negative three plus 𝑖. So let’s substitute 𝑥 equals
negative three plus 𝑖 into our function. That gives us one over negative
three plus 𝑖 squared plus two. And when we distribute the
parentheses, on the denominator we get 𝑖 squared minus six 𝑖 plus 11.
And now we’re ready to perform some
substitutions. We’re finding the sum and we’re
taking values of 𝑖 from zero to 𝑛 minus one. Now, 𝑛 is six. So 𝑛 minus one is five. 𝛥𝑥 is one and 𝑓 of 𝑥 𝑖 is one
over 𝑖 squared minus six 𝑖 plus 11. But of course, we don’t really need
to write multiplied by one. So we need to evaluate this
sum. To do this, we’re going to
substitute values of 𝑖 from zero through to five into this function and then find
their sum.
So when 𝑖 equals zero, that’s one
over zero squared minus zero plus 11. When 𝑖 is one, that’s one over one
squared minus six plus 11. When 𝑖 is two, it’s one over two
squared minus 12 plus 11. And we repeat this process for 𝑖
equals three, 𝑖 equals four, and 𝑖 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. We’ve computed the left Riemann sum
for our function over that closed interval, using six subintervals.
