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

“About 8.316.”

“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?)”


Hope that helps!

- Uncle Colin