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Represent the area under the curve of the function π of π₯ equals π₯ squared minus one on the close interval zero to three in sigma notation using a right Riemann sum with π subintervals.
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Remember when weβre finding a right Riemann sum, we find the sum of π₯π₯ times π of π₯ π for values of π from one to π.
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π₯π₯ is π minus π over π where π and π are the lower and upper limits of our interval, respectively, and π is the number of subintervals.
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π₯ π is π plus π lots of π₯π₯.
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We always begin by working out what π₯π₯ is.
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In our case, π is equal to zero, π is equal to three, and, well, π is just π.
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This means π₯π₯ is three minus zero over π or just three over π.
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Next, weβre going to work out what π₯ π is.
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Itβs π, which we know to be zero, plus π₯π₯, which is three over π, times π.
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Weβll write this as three π over π.
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Of course, we want to know what π of π₯ π is.
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So it follows that to find π of π₯ π, we find π of three π over π.
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Letβs substitute three π over π into our formula.
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Thatβs three π over π all squared minus one which is nine π squared over π squared minus one.
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Weβre now ready to use the summation formula.
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Weβre evaluating our sum for values of π from one to π.
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Its π₯π₯, which is three over π, multiplied by nine π squared over π squared minus one.
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We distribute our parentheses and then weβre going to look to create a common denominator.
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We can do that by multiplying both the numerator and denominator of our second fraction by π squared.
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That gives us three π squared over π cubed, leaving us just to simply combine the numerators.
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We have 27 π squared minus three π squared over π cubed.
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Now, we can simplify this somewhat.
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The numerators share a factor of 27 and three.
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And of course they have a common denominator of π cubed.
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Both three and π cubed are indepnent of π.
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This means we can take three over π cubed outside of the sigma symbol, and that means weβre done.
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Weβve represented the area under the curve of the function in sigma notation with a right Riemann sum.
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Itβs three over π cubed times the sum of nine π squared minus π squared for values of π from one π.