Dear Uncle Colin,

I need to find the area between the curves $y=16x$, $y= \frac{4}{x}$ and $y=\frac{1}{4}x$, as shown. How would you go about that?

Awkward Regions, Exhibit A

Hi, AREA, and thanks for your message!

As usual, there are several possible approaches here, but I’m going to write up the most obvious one.

It involves:

• Finding the area between the x-axis, the steeper line and the curve
• Then subtracting the area beneath the shallower line

### Where do they cross?

Before anything else, we need the points of intersection!

Obviously, the two lines cross at the origin.

The steeper line crosses the curve when $16x = \frac{4}{x}$, which gives $x = \pm \frac{1}{2}$ - from the picture, we clearly want the positive value, so the upper crossing is at $\left( \frac{1}{2}, 8\right)$.

The shallower line crosses the curve when $\frac{1}{4}x=\frac{4}{x}$, which gives $x= \pm 4$; again, we want the positive value, so the crossing point is $(4,1)$.

### Now let’s integrate!

The area between the steeper line, the x-axis and the line $x=\frac{1}{2}$ forms a triangle, so we don’t even need to integrate. The area is $\frac{1}{2} \times \frac{1}{2} \times 8 = 2$ square units.

Under the curve, we do need to integrate: $\int_{\frac{1}{2}}^{4} \frac{4}{x} dx = \left[ 4 \ln(x) \right]_{\frac{1}{2}}^{4}$.

That works out to $4\ln(4) - 4\ln\br{\frac{1}{2}} = 12\ln(2)$.

So the area under the steep line and the curve totals $2 + 12\ln(2)$ - but we still have the lower triangle to take away.

### The lower triangle

The lower triangle has a base of 4, a height of 1 and an area of 2 - which leaves a total area for the region of just $12\ln(2)$.

#### Whoosh

“Thank you, sensei. (I reckon they’ve done $12 \times \frac{7}{10}$ and dropped it by 1%. It’s actually 8.318, but what’s the third decimal place between friends?)”