# Ask Uncle Colin: An Area To Find

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: $\int_{\frac{1}{2}}^{4} \frac{4}{x} dx = \left[ 4 \ln(x) \right]_{\frac{1}{2}}^{4}$.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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##### Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.