### Area L

Draw the graph of a continuous increasing function in the first quadrant and horizontal and vertical lines through two points. The areas in your sketch lead to a useful formula for finding integrals.

### Integral Equation

Solve this integral equation.

### Integral Sandwich

Generalise this inequality involving integrals.

# Integration Matcher

##### Stage: 5 Challenge Level:
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!

 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 $\cos x$