What's the area between $f_k(x) = x(x-k)(x-2)$ and the $x$-axis?

What's the area between $f_k(x) = x(x-k)(x-2)$ and the $x$-axis?

Question

Let $0< k <2$. Below is sketched a graph of $y=f_k(x)$ where $f_k(x) = x(x-k)(x-2)$. Let $A(k)$ denote the area of the shaded region.

1. Without evaluating them, write down an expression for $A(k)$ in terms of two integrals.

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1. Explain why $A(k)$ is a polynomial in $k$ of degree $4$ or less. [You are not required to calculate $A(k)$ explicitly.]

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1. Verify that $f_k (1+t) = -f_{2-k}(1-t)$ for any $t$.

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1. How can the graph of $y=f_k(x)$ be transformed to the graph of $y = f_{2-k}(x)$? Deduce that $A(k)=A(2-k)$.

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1. Explain why there are constants $a, b, c$ such that $A(k) = a(k-1)^4 + b(k-1)^2 + c.$ [You are not required to calculate $a,b, c$ explicitly.]

Suggestion

How is the value of an integral affected if a function gives negative values (is below the $x$-axis)?

What is the highest of power of $k$ we can get by integrating $f$, with the limits given?

What happens if we use part (iii), replacing $1+t$ with $x$?

Can we compare $f_{2-k}(x)$ to $f_k (x)$? What do the transformations from part iv) do to the area $A(k)$?

1. Explain why there are constants $a, b, c$ such that $A(k) = a(k-1)^4 + b(k-1)^2 + c.$ [You are not required to calculate $a,b,c$ explicitly.]

Can we show $A(1+k) = A(1-k)$? What implications for symmetry does this have?

Solution

We need to consider the two shaded regions separately, because one is above the $x$-axis (giving a positive area), and one is below the $x$-axis (giving a negative area), so we can’t just take the integral $\displaystyle\int_0^2 \! f_k(x) \, \mathrm{d}x$.

Instead, we have

$$A(k) = \displaystyle\int_0^k \! f_k(x) \, \mathrm{d}x - \displaystyle\int_k^2 \! f_k(x) \, \mathrm{d}x,$$

as the negative sign in front of the second integral cancels out the negative sign of the area between $x=k$ and $x=2$.

Therefore,

$$A(k) = \displaystyle\int_0^k \! x(x-k)(x-2) \, \mathrm{d}x - \displaystyle\int_k^2 \! x(x-k)(x-2) \, \mathrm{d}x.$$

$A(k)$ is certainly a polynomial in $k$. Could it include a $k^5$ term?

Only if you have an $x^5$ term or a $kx^4$ term after integrating, which means having an $x^4$ term or a $kx^3$ term before integrating.

Neither is possible, by looking at $x(x-k)(x-2)$, and the same argument works for powers of $k$ higher than $5$.

Hence $A(k)$ is a polynomial in $k$ of degree at most $4$.

Let’s start by thinking about what $f_k(1+t)$ and $f_{2-k}(1-t)$ are, given that $f_{k}(x)=x(x-k)(x-2)$.

To find $f_{k}(1+t)$, we need to substitute $1+t$ for $x$ in the formula for $f_{k}(x).$ This gives us $f_k(1+t) = (1+t)(1+t-k)(1+t-2).$

To find $f_{2-k}(1-t)$, we need to substitute $1-t$ for $x$, but we also need to substitute $2-k$ for $k.$ Therefore $f_{2-k}(1-t) = (1-t)(1-t-(2-k))(1-t-2).$

Now we can manipulate

$-f_{2-k}(1-t)$

to try to obtain

$f_k(1+t).$

$$\begin{align*} -f_{2-k}(1-t) \hspace{2mm} & =-(1-t)(1-t-(2-k))(1-t-2) \\ & = -(1-t)(-1-t+k)(-1-t) \\ & = (t-1)(1+t-k)(1+t) \quad \text{by taking a factor of }-1\text{ from each bracket} \\ & = (1+t)(1+t-k)(t-1) \quad \text{by rearranging the order of the brackets.} \\ \end{align*}$$

If we think about what we are aiming for, we can see that the first two brackets involve $1+t$, which is a good sign. It would be helpful to rewrite the last bracket in a way that also involves $1+t.$ We can write $t-1$ as $1+t-2$ which means that

$$\begin{align*} -f_{2-k}(1-t) \hspace{2mm} & =(1+t)(1+t-k)(1+t-2) \\ & = f_{k}(1+t) \quad \text{as required.}\\ \end{align*}$$

We started from the RHS because it looks messier and we wanted to get it to a simpler expression. We could have started with the LHS.

From part (iii), we have that

$$f_k(1+t) = -f_{2-k}(1-t).$$

If we substitute $t$ with $1-x$, the above equation becomes equivalent to

$$-f_k(2-x) =f_{2-k}(x).$$

Now, $f_k(2-x)$ is the reflection of $f_k(x)$ in the line $x=1$.

A second transformation is now required because of the minus sign in front of $f_{k}$. This is a reflection in the $x$-axis.

Together these transformations are the same as a rotation $180^{\circ}$ around the point $(1,0)$.

In fact, combining two reflections always gives you a rotation about the point where the two mirror lines intersect.

As we can see in the sketch above, since we’re rotating $180^\circ$ around a point on the x axis, the area between the graph and the $x$-axis remains the same for $f_k$ and $f_{2-k}$, i.e. $A(k) = A(2-k)$.

Or else we could say that since both of the reflections preserve the shaded area, combining them does too.

We know that $A(k) = A(2-k)$, so $A(1-k) = A(2-(1-k)) = A(1+k)$, and so $y = A(k)$ is symmetrical about the line $k = 1$.

So if we translate $y = A(k)$ by $1$ to the left, then it is even, so if we express $A(k)$ as powers of $(x-1)$, only even powers are possible.

Therefore we have $A(k) = a(k-1)^4 + b(k-1)^2 + c$ for some constants $a, b, c$.