6412
/www/nrich/html/content/id/6412/
1
2011-02-01T00:00:01
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The graphs of six functions and the graphs of their integrals have been mixed up below. Can you match them together?<br></br>
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You might also like to use accurate charts suitable for reading off measurements (you will need to print these in landscape mode)<br></br>
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<a href="/content/id/6412/accurateGraphs1.jpg">First 6 charts</a><br></br>
<a href="/content/id/6412/accurateGraphs2.jpg">Second 6 charts</a><br></br>
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Can you use the charts to estimate the integrals between various points on the graphs of the differentials? Do your estimates match the readings off the graphs of the integrals?<br></br>
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<em>Extension: Can you suggest sensible algebraic forms for the graphs (which are all built from 'standard' functions)</em><br></br></mdoxml>
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<h3>Why do this problem?</h3>
<div><a href="http://nrich.maths.org/public/viewer.php?obj_id=6412&part=">This problem</a> offers students insights into differentiation, integration and the relationships between the two without needing to get involved with technical manipulations. It would be well suited to use as an introduction or summary to differentiation or integration. It is very good for giving intuitive meaning to
the procedures and features of integration and differentiation, so would suit students with a range of technical skills.</div>
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<div>The accurate charts can also be used as a problem involving fitting curves to equations. They can also be used to practise numerical integration.</div>
<h3>Possible approach</h3>
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You can print off the graphs <a class="pdflink" href="/content/id/6412/Integration%20Matcher.pdf">here</a>, or as accurate charts with scales <a class="pdflink" href="/content/id/6412/IntegrationMatcher2.pdf">here</a>.<br></br>
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<div>(<span style="font-weight: bold;">One approach</span>) Give the graphs to the class and say that they come in integral/differential pairs. For classes who have not yet met integration: "These graphs come in pairs. For each function $f(x)$ there is a graph of the function $A(x)$ where $A(x)$ is the area under the curve $y=f(t)$ between $t=0$ and $t=x$. Can you match each graph of a function
$f$ with its corresponding graph of the area function $A$?"<br></br>
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Students could work in pairs for this activity. It is important that they give evidence for why they believe that each pair goes together. Students who have met the idea of integration as anti-differentiation could use this to check their pairings by looking for turning points mapping to a zero on the derivative graph. Students with a more sophisticated understanding should be expected to give
the most rigorous answers.</div>
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<div>(<span style="font-weight: bold;">Another approach</span>) Give the graphs out, but don't say that they come in integral differential pairs. Working in teams, ask how the graphs can be grouped. Ask the teams to share their ideas of groupings. Do any emerge as the best groupings? The notion of integrals/differentials is likely to emerge, but you might give a hint in this direction if groups
are struggling. With this approach, be prepared to go with the flow of interesting pairing suggestions. Does the class as a whole agree that there is a 'best' overall grouping?</div>
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<div>As always with matching questions, an emphasis should be placed on clear explanation of results. Guesses or poorly justified reasoning should be challenged most strongly!</div>
<h3>Key questions</h3>
<div>What are the key points on each graph?</div>
<div>Are there any statements you could make about the sign of the integral of each curve?</div>
<div>What evidence can you give for your statement? Does this convince you? Does it convince the rest of the class?</div>
<div>What algebraic forms might match these curves?</div>
<h3>Possible extension</h3>
<div>The extension in the question is challenging. Students might suggest the algebraic forms and give evidence to support their suggestion (after all -- the charts are only pictures. It is logically possible that the values come from any number of functions)</div>
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<div>If the students were to invent a similar problem, which curves would they choose? Why?</div>
<h3>Possible support</h3>
<div>Encourage students to start with the 'easiest' curves. They could make sketches of what the area function would look like for each curve, and look to see if their sketch matches any of the cards.</div></mdoxml>
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Differentiation relates to gradients and integration relates to
areas.<br></br>
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The two are inverse procedures.<br></br>
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The graphs are<br></br>
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<table border="1">
<tbody>
<tr>
<td>K</td>
<td>F</td>
<td>cos(x)</td>
<td>sin(x)</td>
<td> </td>
</tr>
<tr>
<td>G</td>
<td>E</td>
<td>sin^2(x)</td>
<td>0.5x-0.25 sin(2x)</td>
<td> </td>
</tr>
<tr>
<td>C</td>
<td>L</td>
<td>0.25ln(x/4)</td>
<td>0.25xln(x/4)-x/4</td>
<td> </td>
</tr>
<tr>
<td>D</td>
<td>B</td>
<td>0.25x^3-3x^2+8x+1</td>
<td>x^4/16-x^3+4x^2+x</td>
<td> </td>
</tr>
<tr>
<td>I</td>
<td>H</td>
<td>4/(1+x^2)</td>
<td>4tan^{-1}(x)</td>
<td> </td>
</tr>
<tr>
<td>J</td>
<td>A</td>
<td>1</td>
<td>x</td>
<td> </td>
</tr>
<tr>
<td> </td>
<td> </td>
<td>function</td>
<td>integral</td>
<td> </td>
</tr>
</tbody>
</table>
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<span class="editorial">I really liked this excellent solution sent in by Alex from Stoke-on-Trent Sixth Form College. The logical deductions were very clearly presented, with a good sense of where assumptions needed to be made.The solution made really good use of the fact that integration is a summation process. An alternative approach would be to look at differentiation properties of the
functions in question.</span><br></br>
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The philosopher Wittgenstein once commented it is impossible to induct the next term in a series from the terms known because there always exists a formula which would be true for the known terms and set the next term to any desired value. Likewise, it is impossible to know the true algebraic forms of the graphs, or how they would behave for $x &gt; 10$ or $x &lt; 0$.<br></br>
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Chart J's function looks like y=1 because it is a straight horizontal line. The integral of $1$ is $x$, and Chart A looks like $y=x$. So a sensible pairing is:<br></br>
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$$A = \int J$$<br></br>
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Now, in general we have<br></br>
$$<br></br>
f(x) \geq 0\mbox{ for }0 \leq x \leq 10 \Rightarrow \int f(x) \geq 0 \mbox{ for }0 \leq x \leq 10<br></br>
$$<br></br>
This means that graphs which go below the x-axis at points cannot be the integrals of graphs which do not.<br></br>
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Graphs C, D, F, K and L are negative at some points, so cannot be the integrals of the purely positive graphs B, E, G, H, I. If any of these positive graphs are not integrals of another graph, they must have another of these graphs as their integral. There are an odd number of these graphs, so they cannot all be paired off in this way. As the remaining graph could not have a non-positive graph as
its integral, it must be the integral of one of the non-positive graphs.<br></br>
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Of the graphs which go below the x-axis at some points, C and L are initially negative, so their integral graph must be initially negative also. This leaves D, F and K as candidates for pairing with one of the positive graphs.<br></br>
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Chart D is positive for $0 &lt; x &lt; 4.2$ then negative for $4.2 &lt; x &lt; 7.9$. This means its integral should rise during the $0 &lt; x &lt; 4.2$ and fall during the $4.2 &lt; x &lt; 7.9$. Chart B matches this behaviour, and none of the other graphs does, so must be the integral of D.<br></br>
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$$B = \int D $$<br></br>
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Of the sinusoidal-shaped graphs, only G is positive for all $x$. Its integral should be expected to vary the same period as G, but also be one of the positive graphs. The period between the "turning points" is the same of the period of G $(3.1)$.<br></br>
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$$E = \int G$$<br></br>
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F and K are the remaining graphs which look like a sine wave. K crosses the x-axis at $0$, while F crosses it at $1$. This is the same behaviour as the sine and cos graphs. $\int \cos x = \sin x$, so<br></br>
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$$F = \int K$$<br></br>
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Of the remaining graphs C, H, I and L; L is totally negative. It would therefore have a totally negative integral graph, but as none of the other graphs are, it must be an integral of one other other functions. C is the other non-positive graph, therefore:<br></br>
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$$L = \int C$$<br></br>
The integral of both H and I would be expected to grow slowly, however only H does this, so:<br></br>
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$$H = \int I $$<br></br></mdoxml>
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