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Ben sent us a lovely detailed solution.

Consider A, which is obviously a simple $y=x$ graph. We can easily compute

the integral and derivative of this: the integral is $y=x^2/2 + c$, and

the derivative is $y=1$. We can easily see that graph J is $y=1$, so A=$\int$J.

Considering B, we can see that the graph has a stationary point near to

$x=4$, and also one near $x=8$. Its derivative would therefore have roots at

those points. The only one which matches this is D, so B=$\int$D.

Consider C, which looks like a shifted logarithmic graph. The gradient of

this graph is always positive, so the derivative would also always be

positive. However, the rate at which $y$ increases as $x$ increases slows down

as $x$ tends to infinity. Therefore the derivative would start off with $x$

very high, and as $x$ tends to infinity, it would tend towards zero. This

matches I. Unfortunately, I could also be the derivative of H, so we must

look closer. It can be seen that C crosses the axis at x=4, so if C was

actually the derivative of some graph, then it would have a stationary

point at $x=4$, with gradient negative below that, and positive above that. L

seems to match that, so L=$\int$C but we've also found that H=$\int$I.

Consider E. This function does indeed look rather strange, but looking

closer, it seems to be periodic in some way. Indeed, its points of

inflection seem to occur at multiples of $\pi$, which almost guarantees that

a trigonometric function is involved. It is also worth noting that the

gradient is always greater than or equal to zero, so its derivative would

always be greater or equal to zero. Looking at G, its roots occur at

intervals of pi, and it is always greater than or equal to zero, so it

turns out that E=$\int$G. (It can in fact be seen that the

equation of G is $y=(\sin x)^2$, so the equation of E is

$y=\frac{1}{2}(x-\sin(x)\cos(x))$.

Consider F. It is fairly obvious that this is in fact the graph of

$y=\sin x$, the derivative of which is $y=\cos x$, which is graph K, so F=$\int$K.

Excellent work. The answers are in the table below to make checking easier. Well done to everyone who found a possible function for each graph, and to Oliver who got most of them!

Consider A, which is obviously a simple $y=x$ graph. We can easily compute

the integral and derivative of this: the integral is $y=x^2/2 + c$, and

the derivative is $y=1$. We can easily see that graph J is $y=1$, so A=$\int$J.

Considering B, we can see that the graph has a stationary point near to

$x=4$, and also one near $x=8$. Its derivative would therefore have roots at

those points. The only one which matches this is D, so B=$\int$D.

Consider C, which looks like a shifted logarithmic graph. The gradient of

this graph is always positive, so the derivative would also always be

positive. However, the rate at which $y$ increases as $x$ increases slows down

as $x$ tends to infinity. Therefore the derivative would start off with $x$

very high, and as $x$ tends to infinity, it would tend towards zero. This

matches I. Unfortunately, I could also be the derivative of H, so we must

look closer. It can be seen that C crosses the axis at x=4, so if C was

actually the derivative of some graph, then it would have a stationary

point at $x=4$, with gradient negative below that, and positive above that. L

seems to match that, so L=$\int$C but we've also found that H=$\int$I.

Consider E. This function does indeed look rather strange, but looking

closer, it seems to be periodic in some way. Indeed, its points of

inflection seem to occur at multiples of $\pi$, which almost guarantees that

a trigonometric function is involved. It is also worth noting that the

gradient is always greater than or equal to zero, so its derivative would

always be greater or equal to zero. Looking at G, its roots occur at

intervals of pi, and it is always greater than or equal to zero, so it

turns out that E=$\int$G. (It can in fact be seen that the

equation of G is $y=(\sin x)^2$, so the equation of E is

$y=\frac{1}{2}(x-\sin(x)\cos(x))$.

Consider F. It is fairly obvious that this is in fact the graph of

$y=\sin x$, the derivative of which is $y=\cos x$, which is graph K, so F=$\int$K.

Excellent work. The answers are in the table below to make checking easier. Well done to everyone who found a possible function for each graph, and to Oliver who got most of them!

Chart | Function | Chart | Function | |

A | $y=x$ | is the integral of | J | $y=1$ |

B | $\frac{x^4}{16}-x^3+4x^2+x$ | is the integral of | D | $\frac{x^3}{4}-3x^2+8x+1$ |

L | $\frac{x}{4}\ln\left(\frac{x}{4}\right)-\frac{x}{4}$ | is the integral of | C | $\frac{1}{4}\ln\left(\frac{x}{4}\right)$ |

H | $4\tan^{-1}(x)$ | is the integral of | I | $\frac{4}{(1+x^2)}$ |

E | $\frac{x}{2}-\frac{1}{4}\sin(2x)$ | is the integral of | G | $\sin^2 x$ |

F | $\sin x$ | is the integral of | K | $\cos x$ |