Integration Matcher
The graphs are
| K |
F |
cos(x) |
sin(x) |
|
| G |
E |
sin^2(x) |
0.5x-0.25 sin(2x) |
|
| C |
L |
0.25ln(x/4) |
0.25xln(x/4)-x/4 |
|
| D |
B |
0.25x^3-3x^2+8x+1 |
x^4/16-x^3+4x^2+x |
|
| I |
H |
4/(1+x^2) |
4tan^{-1}(x) |
|
| J |
A |
1 |
x |
|
| |
|
function |
integral |
|
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.
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 $x > 10$ or $x < 0$.
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:
$$A = \int J$$
Now, in general we have
$$
f(x) \geq 0\mbox{ for }0 \leq x \leq 10 \Rightarrow \int f(x) \geq 0 \mbox{ for }0 \leq x \leq 10
$$
This means that graphs which go below the x-axis at points cannot be the integrals of graphs which do not.
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.
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.
Chart D is positive for $0 < x < 4.2$ then negative for $4.2 < x < 7.9$. This means its integral should rise during the $0 < x < 4.2$ and fall during the $4.2 < x < 7.9$. Chart B matches this behaviour, and none of the other graphs does, so must be the integral of D.
$$B = \int D $$
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 $(3.1)$.
$$E = \int G$$
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
$$F = \int K$$
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:
$$L = \int C$$
The integral of both H and I would be expected to grow slowly, however only H does this, so:
$$H = \int I $$