Video: Finding the Area under a Curve Using Sigma Notation with Riemann Sums

Compute the left Riemann sum for 𝑓(π‘₯) = 1/(π‘₯Β² + 2) on [βˆ’3, 3], given that there are six subintervals of equal width. Approximate your answer to two decimal places.

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Video Transcript

Compute the left Riemann sum for 𝑓 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 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 π›₯π‘₯ times 𝑓 of π‘₯ 𝑖 for values of 𝑖 from zero to 𝑛 minus one where π›₯π‘₯ is 𝑏 minus π‘Ž over 𝑛. Remember π‘Ž and 𝑏 are the lower and upper limits of our interval, respectively, and 𝑛 is the number of subintervals. Then π‘₯ 𝑖 is π‘Ž plus 𝑖 lots of π›₯π‘₯. It’s only sensible to begin by first calculating π›₯π‘₯. We can see that our close interval is from negative three to three. So we’re going to let π‘Ž 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 π›₯π‘₯ is three minus negative three over six, which is simply one. Next, we’ll calculate what π‘₯ 𝑖 is. It’s π‘Ž, which we know to be negative three, plus π›₯π‘₯, which is one, lots of 𝑖. That’s of course negative three plus 𝑖. We’re looking to find what 𝑓 of π‘₯ 𝑖 is though. 𝑓 of π‘₯ 𝑖 must, therefore, be 𝑓 of negative three plus 𝑖. So let’s substitute π‘₯ 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. π›₯π‘₯ is one and 𝑓 of π‘₯ 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.

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