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In this video, weβll learn how to use right, left, and midpoint Riemann sums to numerically approximate definite integrals.
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Up until this stage, youβve probably estimated the area between the curve and the π₯-axis by splitting into rectangles and finding their combined sum.
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Weβll now see how this process links to calculus and how it can help us to numerically approximate definite integrals.
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Suppose weβre looking to find the area between the curve π¦ equals π of π₯, the π₯-axis, and the vertical lines, denoted π₯ equals π and π₯ equals π.
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The region might look a little like this one shown, though it is possible that the function values could also be both positive and negative.
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The Riemann sums give us an approximation of this area by splitting it up into equal-sized rectangles.
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The heights of these rectangles will either be given by the function value at the left endpoint of each interval.
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Thatβs a left Riemann sum, the right endpoint for a right Riemann sum, or the midpoint of each interval.
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The formulae for the right and left Riemann sums are as shown.
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And remember, when weβre writing a right Riemann sum, we take values of π from one to π.
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And when weβre writing a left Riemann sum, we take values of π from zero to π minus one.
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And that gives us the value of π at the left endpoint of each rectangle.
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Now, here, Ξπ₯ is π minus π divided by π.
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π and π are the beginning and end points of the interval and π is the number of subintervals, in other words, the number of rectangles weβre splitting the region into.
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Then, π₯π looks a little bit confusing.
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But itβs π plus π lots of Ξπ₯.
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In other words, we start at the lower limit of our interval.
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And we repeatedly add Ξπ₯, the width of each rectangle.
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Now, letβs imagine weβre splitting the region into, letβs say, two rectangles.
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Here, Iβve chosen the height of each rectangle to be the value of the function of the left endpoint.
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Now, it follows that this is not going to give us a very good estimate for the area between the curve and the π₯-axis.
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But if we were to split further into, say, four rectangles, our estimate will be closer to the exact area.
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And splitting it further into eight rectangles, for example, and our estimate would be even closer.
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In fact, as the number of rectangles or subintervals, π, approaches infinity, the approximate area approaches the exact area between the curve and the π₯-axis.
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When evaluating right Riemann sums, we can say that the area π of the region that lies under the graph of a continuous function π is the limit as π approaches infinity of the sum of Ξπ₯ times π of π₯π for values of π from one to π.
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And when evaluating left Riemann sums, we say that the area of the region that lies under the graph of a continuous function π is the limit of the sum of Ξπ₯ times π of π₯π for values of π from zero to π minus one.
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And in fact, instead of using left or right endpoints, we could even take the height of the πth rectangle to be the value of π at any number π₯π star in the πth subinterval from π₯π minus one to π₯π.
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Recall π₯π star the sample point.
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In this case, we can generalize our formula, as shown.
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But then, we move on to yet another definition.
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And this is the definition of a definite integral.
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And you might have seen this before.
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If π is a function defined on the close interval π to π, we divide that interval into π subintervals of equal width.
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Thatβs Ξπ₯, where π₯ nought, π₯ one, π₯ two, and so on are endpoints of the subintervals.
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Then, π₯ one star, π₯ two star, up to π₯π sample points in the subintervals so that π₯π star lies in the π subinterval.
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Then, the definite integral of π from π to π is the limit as π approaches β of the sum of Ξπ₯ times π of π₯π star for values of π from one to π.
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That is, of course, providing this limit exists and gives us the same value for all possible choices of sample points.
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But hang on a minute!
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We just said that the area between the curve of π¦ equals π of π₯ and the π₯-axis between π₯ equals π and π₯ equals π is equal to this limit.
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So that must mean that the definite integral between the limits of π and π of our function is the exact area.
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And thatβs great because yes, there are some functions we can integrate easily.
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And therefore, we can evaluate the definite integral to find the exact area between the curve and the π₯-axis.
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But if we canβt, we now know that we can use Riemann sums to help approximate it.
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Letβs see what this might look like.
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The table shows the values of a function obtained from an experiment.
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Estimate the definite integral between five and 17 of π of π₯ with respect to π₯ using three equal subintervals with left endpoints.
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Remember, we can estimate a definite integral by using Riemann sums.
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In this case, weβre estimating the integral between five and 17 of π of π₯.
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Now, it doesnβt really matter that we donβt know what the function is.
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We have enough information in our table to perform the left Riemann sum.
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The left Riemann sum involves taking the heights of our rectangles as the function value at the left endpoint of the subinterval.
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We want to use three equally sized subintervals.
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So letβs recall the formula that allows us to work out the size of each subinterval, in other words, the widths of the rectangle.
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Itβs Ξπ₯ equals π minus π over π, where π and π are the endpoints of our interval and π is the number of subintervals.
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In our case, weβre looking to evaluate the definite integral between five and 17.
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So we let π be equal to five and π be equal to 17.
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And we want three equal subintervals.
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So weβll let π be equal to three.
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Ξπ₯ is then 17 minus five all divided by three, which is simply four.
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Then when writing a left Riemann sum, we take values of π from zero to π minus one.
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Itβs the sum of Ξπ₯ times π of π₯π for values of π from zero to π minus one.
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π₯π is π plus π lots of Ξπ₯.
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In this case, we know that π is equal to five, and Ξ π₯ is equal to four.
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So our π₯π value is given by five plus four π.
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Well, since weβre using the left Riemann sum, we begin by letting π be equal to zero.
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We need to work out π₯ zero.
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Itβs five plus four times zero, which is simply five.
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We can find π of π₯ nought in our table.
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Itβs negative three.
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Next, we let π be equal to one.
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And we get π₯ one to be five plus four times one, which is nine.
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We look up the value π₯ equals nine in our table.
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And we see that π of nine is negative 0.6.
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Next, we let π be equal to two.
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And remember, weβre looking for values of π up to π minus one.
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Well, three minus one is two.
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So this is the last value of π weβre interested in.
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This time, thatβs five plus four times two which is 13.
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We look up π₯ equals 13 in our table.
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And we get that π of 13 and π of π₯ two is 1.8.
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Then, according to our summation formula, we find the sum of the products of Ξπ₯ and these values of π of π₯π.
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And so, an estimate for our definite integral is four times negative three plus four times negative 0.6 plus four times 1.8, which is negative 7.2.
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An estimate for the definite integral between five and 17 of π of π₯ with respect to π₯ using three equal subintervals is negative 7.2.
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Now, we donβt need to worry here that our answer is negative.
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Remember, when weβre working with Riemann sums, weβre looking at areas.
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But when the function values are negative, the rectangle sits below the π₯-axis.
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And so, its area is subtracted.
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Letβs now have a look at how we can use the formulae with right endpoints.
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Approximate the definite integral between negative two and two of three π₯ squared minus five π₯ with respect to π₯ using a Riemann sum with right endpoints.
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Take π to be eight.
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Remember, we can estimate the solution to a definite integral by using Riemann sums, in this case is the definite integral of three π₯ squared minus five π₯ between the limits of negative two and two.
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And weβre going to be using right endpoints.
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When writing Riemann sums with right endpoints, we take values of π from one to π.
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Itβs the sum of Ξπ₯ times π of π₯π for these values of π, where Ξπ₯ is π minus π divided by π.
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Bear in mind here that π is the number of subintervals and π₯π is equal to π plus π lots of Ξπ₯.
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So letβs look at what we actually have.
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Our limits are from negative two to two.
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So weβll let π be equal to negative two and π be equal to two.
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Weβre told that we need to take π to be eight.
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Geometrically, this tells us the number of rectangles we have.
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And now, we can work out Ξπ₯.
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Thatβs the width of each rectangle.
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According to our formula, Ξπ₯ is π minus π over π.
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Thatβs two minus negative two over eight, which is one-half.
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And once we have Ξπ₯, we can then work out π₯π.
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Itβs π which we said is negative two plus Ξπ₯.
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Thatβs half times π.
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In other words, π₯π is negative two plus π over two.
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Now, for our Riemann sum, we need to work out π of π₯π.
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Thatβs clearly π of negative two plus π over two.
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We achieve this by substituting negative two plus π over two into our formula three π₯ squared minus five π₯.
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And if we distribute our parentheses and simplify, we see that π of π₯π is three-quarters π squared minus 17 over two π plus 22.
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Letβs now substitute Ξπ₯ and π of π₯π into our summation formula.
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We now see that an approximation to our definite interval is the sum of a half times three-quarters π squared minus 17 over two π plus 22 for values of π from one to eight.
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Now, actually, this constant factor a half is independent of π.
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So we could actually take it outside of the sum.
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And now, whilst this step isnβt entirely necessary, it canβt simplify things on occasion.
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Weβre now going to substitute values of π from one through to eight into three-quarters π squared minus 17 over two π plus 22 and find their sum.
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When π is one, we get 0.75 minus 8.5 plus 22, which is 14.25.
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When π is two, we get eight.
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When π is three, we get 3.25.
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When π is four, we get zero.
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When π is five, itβs negative 1.75.
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When π is six, we get negative two.
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When itβs seven, we get negative 0.75.
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And when π is eight, we get two.
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Finding the sum of these values and then multiplying it by one-half and we get 23 over two.
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And so, an approximation to our definite integral weβre using a right Riemann sum with eight subintervals is 23 over two.
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So far, weβve considered how to estimate integrals with left and right Riemann sums.
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Letβs now look at how we might work using midpoints.
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Using the midpoint rule with π equals five, round the definite integral from two to five of two π₯ over three π₯ plus two with respect to π₯ to four decimal places.
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Remember, we can estimate a definite integral by using Riemann sums.
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We split the region into subintervals and create a rectangle in each.
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The total area of the rectangles gives us an estimate to the integral.
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In a midpoint Riemann sum, the height of each rectangle is equal to the value of the function at the midpoint of its base.
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Now, working with midpoints isnβt quite as nice as using the left or right Riemann sum.
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In this case, thereβs nothing to stop us evaluating each of the areas in term.
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And letβs begin by working out the width of each subinterval.
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Geometrically, it tells us the width of the rectangles.
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And itβs given by Ξπ₯ equals π minus π over π, where π and π are the endpoints of the interval and π is the number of subintervals.
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In our case, our lower limit is two and our upper limit is five.
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We, therefore, let π be equal to two and π be equal to five.
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And weβre told that π is equal to five.
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Ξπ₯ is five minus two over five, which is three-fifths or 0.6.
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And a table can make the next step a little easier.
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In our table, weβre going to begin by working out each of the subintervals.
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We know the lower limit of our definite integral to be two.
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To find the right endpoint of our first rectangle on our first subinterval, we add 0.6 to two to get 2.6.
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This means our next rectangle starts at π₯ equals 2.6.
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This time, we add 0.6 again.
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And we find the right endpoint to be 3.2.
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The left endpoint of our next rectangle must, therefore, be 3.2.
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And the right endpoint is 3.2 plus Ξπ₯.
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Thatβs 3.8.
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Our next rectangle begins at 3.8.
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And adding 0.6, we find that it ends at 4.4.
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And our fifth and final rectangle β Remember, we wanted π to be equal to five β begins at 4.4 and then ends at 4.4 plus 0.6, which is five.
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And thatβs a really good start because we know the upper limit of our interval is indeed five.
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Next, weβre going to work out the midpoint of each of these subintervals.
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Now, we can probably do this in our head.
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But if weβre struggling, we add the two values and divide by two.
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And when we do, we obtain the midpoints to be 2.3, 2.9, 3.5, 4.1, and 4.7, respectively.
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To work out the height of each rectangle, we need to work out the value of the function at these points.
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So, for example, in this first row, weβll start by working out π of 2.3.
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To do that, we substitute 2.3 into our function two π₯ over three π₯ plus two.
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And we get 46 over 89.
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Then, we repeat this process for π₯ equals 2.9.
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When we substitute 3.5 into our function, we get 14 over 25.
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And the height of our final two rectangles are 82 over 143 and 94 over 161 units, respectively.
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And our very final step is to calculate the area of each rectangle by multiplying its width by its height.
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Of course, the width of each rectangle is Ξπ₯ at 0.6.
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So we multiply each of these function values by 0.6 and then find their sum.
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Now, we could do this in turn or we could find their sum and multiply by 0.6.
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Weβll get the same answer.
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When we do find the total of all the values in our column titled Ξπ₯ times π of π₯π, we get 1.66571 and so on.
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And if we run this correct to four decimal places, we find that an estimate that the definite integral between two and five of two π₯ over three π₯ plus two with respect to π₯ is roughly 1.6657.
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In this video, we saw that the definite integral of some function between the limits of π and π can be approximated using left or right Riemann sums or the midpoint rule.
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And we used the summation formulae for the left and right Riemann sums.
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And we also saw that the midpoint rule can be a little more complicated, but that itβs absolutely fine to use a table to estimate the solution.