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Use the trapezoidal rule to estimate the definite integral between zero and two of π₯ cubed with respect to π₯ using four subintervals.
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Remember, the trapezoidal rule says that we can find an approximation to the definite integral between the limits of π and π of π of π₯ by using the calculation Ξπ₯ over two times π of π₯ nought plus π of π₯ π plus two lots of π of π₯ one plus π of π₯ two all the way through to π of π₯ π minus one.
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Where Ξπ₯ is equal to π minus π over π.
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And π₯ subscript π is equal to π plus π times Ξπ₯.
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Letβs break this down and just begin by working out the value of Ξπ₯.
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Contextually, Ξπ₯ is the width of each of our subintervals.
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In this case, weβre working with four subintervals.
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So we could say that π is equal to four.
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π is the lower limit of our integral.
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So π is equal to zero, where π is the upper limit.
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And thatβs equal to two.
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Ξπ₯ is therefore two minus zero over four, which is one-half or 0.5.
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The values for π of π₯ nought and π of π₯ one and so on require a little more work.
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But we can make this as simple as possible by adding a table.
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Itβs useful to remember that there will always be π plus one columns required in the table.
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So here, thatβs four plus one, which is five.
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We have five columns in our table.
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The π₯-values run from π to π.
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Thatβs zero to two.
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And the ones in between are found by repeatedly adding Ξπ₯, thatβs 0.5, to π, which is zero.
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Thatβs 0.5, one, and 1.5.
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And that gives us our four strips of width 0.5 units.
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Weβre then simply going to substitute each π₯-value into our function.
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We begin with π of zero.
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Thatβs zero cubed, which is zero.
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Next, we have π of 0.5.
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Thatβs 0.5 cubed, which is 0.125.
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π of one is one cubed, which is still one.
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And we obtain the final two values in a similar manner.
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π of 1.5 is 3.375.
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And π of two is eight.
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And thatβs the tricky bit done with.
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All thatβs left is to substitute what we know into our formula for the trapezoidal rule.
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Itβs Ξπ₯ over two, which is 0.5 over two, times the first π of π₯ value plus the last π of π₯ value.
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Thatβs zero plus eight plus two lots of everything else.
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Thatβs two times 0.125 plus one plus 3.375.
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And that gives us a value of 17 over four.
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So using four subintervals, the trapezoidal rule gives us the estimate to the definite integral of π₯ cubed between zero and two to be 17 over four.
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Now where possible, this can be checked in a number of ways.
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You could evaluate a Riemann or midpoint sum or here simply perform the integration.
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When we integrate π₯ cubed, we get π₯ to the fourth power divided by four.
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Evaluating this between the limits of zero and two gives us two to the fourth power divided by four minus zero to the fourth power divided by four, which is 16 over four.
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And thatβs really close to the answer we got, suggesting weβve probably performed our calculations correctly.