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Use the trapezoidal rule to estimate the definite integral between the limits of zero and one of the square root of π₯ plus one ππ₯ using four subintervals.
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Round your answer to three decimal places.
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Remember, the trapezoidal rule says that we can find an estimate for the definite integral evaluated between π and π of π of π₯ by splitting the area between the curve and the π₯-axis into π subintervals.
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The formula we will require is Ξπ₯ over two times π of π₯ naught plus π of π₯ π plus two lots of π of π₯ one all the way up to π of π₯ π minus one.
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And here Ξπ₯ is given by π minus π over π and π₯π is given by 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 question, weβre working with four subintervals.
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So π is equal to four.
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π is the lower limit of our integral.
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Itβs zero.
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And π is the upper limit.
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Itβs one.
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Ξπ₯ is therefore equal to one minus zero over four.
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We know that to be equal to a quarter or 0.25.
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The values for π of π₯ naught, π of π₯ one, and so on require little more work.
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But we can make this as simple as possible by adding a table.
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The π₯-values in our table run from π to π.
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Thatβs zero to one.
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And itβs worth noting that there will always be one more π₯ value than the value of π.
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So here Iβve included five columns.
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The other values of π₯ are found by repeatedly adding Ξπ₯ β thatβs 0.25 β to π, which is zero.
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So they are 0.25, 0.5, and 0.75.
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And this gives us our four strips of width 0.25 units.
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Weβre now going to substitute each of these π₯-values into our function.
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We begin with π of naught.
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Thatβs the square root of zero plus one, which is simply one.
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π of 0.25 is the square root of 0.25 plus one.
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This is root five over two.
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Now we could actually use a decimal value here.
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But for accuracy, it will be important to include at least six decimal places.
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We then have π of 0.5.
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Thatβs the square root of 0.5 plus one, which is root six over two.
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Repeating this process for 0.75 and one, we end up with root seven over two and root two, respectively.
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Letβs substitute everything we have into the formula for the trapezoidal rule.
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Ξπ₯ over two is 0.25 over two.
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π of π₯ naught and π of π₯ π are one and root two.
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We have two lots of everything else.
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Thatβs root five over two, root six over two, and root seven over two.
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And typing this into our calculator, we get a value of 1.21819 and so on.
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Correct to three decimal places, that gives us an estimate for the definite integral between zero and one of the square root of π₯ plus one as 1.218.
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And at this stage, we do have a couple of ways that we can check our solution.
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We could type the exact problem into our calculator.
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And when we do, we get a value of 1.21895 and so on.
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This is, of course, extremely close to the solution we got.
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Alternatively, we could use integration by substitution to evaluate the exact integral.