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Find the lower Riemann sum approximation for π of π₯ equals five minus π₯ squared on the close interval one to two, given that π equals four subintervals.
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The lower Riemann sum approximation is going to be the smallest estimate for the area between the curve of π of π₯ and the π₯-axis and bounded by the lines π₯ equals one on π₯ equals two that weβll get.
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So we recall if π of π₯ is increasing, that is, π prime of π₯ is greater than zero over some interval, then a left Riemann sum will yield an underestimate for the area between the curve on the π₯-axis in that interval.
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And a right Riemann sum will yield an overestimate.
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For a decreasing function, the reverse is true.
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So letβs begin by finding the derivative of our function π of π₯ and working out whether itβs greater than or less than zero over the closed interval one to two.
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The first derivative of our function is negative two π₯.
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Now for values of π₯ that are greater than zero, in other words, positive values of π₯, negative two π₯ is going to be less than zero.
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This means that our function must be a decreasing function in the closed interval one to two.
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So we need to evaluate a right Riemann sum.
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To find a right Riemann sum, 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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And in our case, π is equal to one and π is equal to two.
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And of course, π is the number of subintervals.
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And π₯π is π plus π lots of Ξπ₯.
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Letβs begin by working out the value of Ξπ₯.
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Itβs π minus π, thatβs two minus one, all divided by π, which is four.
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Two minus one divided by four is a quarter or 0.25.
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And now, we have enough information to be able to evaluate π₯π.
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Itβs π which is one plus Ξπ₯ times π; thatβs 0.25π.
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Remember, to evaluate the summation, weβre going to need to find π of π₯π.
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Thatβs, of course, π of one plus 0.25π.
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So letβs swap π₯ in our function for one plus 0.25π.
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When we do, we obtain π of π₯π to be five minus one plus 0.25π all squared.
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Now whilst we could distribute these parentheses and simplify, thereβs really no need.
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We have enough to substitute into the summation formula.
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Weβre interested in values of π from one to π, so thatβs one to four.
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Then itβs Ξπ₯ which is 0.25 times five minus one plus 0.25π squared.
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Now, itβs useful to know that 0.25 is independent of π.
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So we can take this out as a common factor.
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And we need to work out the sum of five minus one plus 0.25π squared between π equals one and four and then multiply that by 0.25.
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To do this, we simply substitute values of π from one to four into this formula and then add them all together.
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When π is one, itβs five minus one plus 0.25 squared.
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When π is two, we get five minus one plus 0.5 squared.
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When π is three, itβs five minus one plus 0.75 squared.
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And when π is four, its five minus one plus one all squared.
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Remember, weβre gonna add these together and then multiply it all by 0.25.
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When we do, we obtain an answer of 2.28125 which correct to two decimal places is 2.28.
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The lower Riemann sum approximation for our function is 2.28.