Food for thought

## Solution

In the Warm-up we scaled (or stretched) a function $f(x)$ so that $\int_{-1}^{2} a\,f(x) \, dx = 1.$

• Which of the following functions could we scale to satisfy this condition and what is the value of $a$ in each case?

• Which ones lead to problems and why?

• How would your answers be different if we changed the condition to $\textit{an area of }\,1 \textit{ is enclosed by the curve }y=a\,f(x) \textit{, the }x\textit{-axis and the lines }x=-1\textit{ and }x=2\text{ ?}$

Can you think of some other functions which cannot be scaled to satisfy one or both of our conditions?

Any function which is undefined over part of the interval will be a problem for the integral condition.

Any function with a singularity (vertical asymptote) in the interval will be a problem, except in a few special cases.

When a function has rotational symmetry about the mid point of the interval, i.e. $x=\frac{1}{2}$, then the area above and below the axis will be equal. The integral will evaluate to $0$ which cannot be scaled to $1$ however hard we try! So the integral condition cannot be satisfied, but often the area condition can.

There are other cases with zero integral which are not symmetrical in this way. For instance, starting with a simple function like $f(x)=x^2-c$ we can find $\int_{-1}^2 x^2-c\,dx= \left[\frac{x^3}{3}-cx\right]_{-1}^2$ and choose a value of $c$ which gives $\int_{-1}^2f(x)\,dx=0$. Can you extend that to find other functions?

Are there any functions without singularities for which an area of $1$ cannot be achieved?