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

Compute the left Riemann sum for π(π₯) = 1/(π₯Β² + 2) on [β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 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 weβre 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₯. Itβs only sensible to begin by first calculating π₯π₯. We can see that our close interval is from negative three to three. So weβre going to let π be equal to negative three and π be equal to three.

Weβre interested in six subintervals. So letβs let π be equal to six. Then π₯π₯ is three minus negative three over six, which is simply one. Next, weβll calculate what π₯ π is. Itβs π, which we know to be negative three, plus π₯π₯, which is one, lots of π. Thatβs of course negative three plus π. Weβre looking to find what π of π₯ π is though. π of π₯ π must, therefore, be π of negative three plus π. So letβs 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 weβre ready to perform some substitutions. Weβre finding the sum and weβre 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 donβt really need to write multiplied by one. So we need to evaluate this sum. To do this, weβre going to substitute values of π from zero through to five into this function and then find their sum.

So when π equals zero, thatβs one over zero squared minus zero plus 11. When π is one, thatβs one over one squared minus six plus 11. When π is two, itβs 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. Weβve computed the left Riemann sum for our function over that closed interval, using six subintervals.