Areas with Changing Boundaries | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Trace the whole boundary before integrating

A region bounded by a curve, a line and an axis may not have the same upper and lower boundaries across its whole width. Move an imaginary vertical strip from left to right. Split where its top or bottom changes from one boundary to another.

Left of x = 1 the lower boundary is the axis. Right of x = 1 it is the tangent y = 4x − 4.

A triangle can replace one integral

Worked example

Find the area bounded by y = x², its tangent at (2, 4), and the x-axis.

The tangent gradient is $2(2)=4$ . The point-gradient equation gives $y-4=4(x-2)$ , or $y=4x-4$ . It meets the x-axis at $x=1$ .

From 0 to 1, the region lies between the curve and the axis. From 1 to 2, it lies between the curve and the tangent.

\begin{aligned}&A=\int_0^1x^2\,dx\\&\;+\int_1^2(x^2-4x+4)\,dx\end{aligned}

More quickly, take all the area under the curve from 0 to 2 and subtract the triangle under the tangent. Its base is $2-1=1$ and its height is 4.

\begin{gathered}A=\left[\frac{x^3}{3}\right]_0^2-2\\=\frac83-2=\frac23\end{gathered}

Do not integrate curve minus tangent from 0: the tangent is not the lower boundary on the first part.

Your turnRead the boundaries

The normal to $y=\sqrt{2x+1}$ at $(4,3)$ is $y=-3x+15$ . Find the region bounded by the curve, this normal and the x-axis.

Show worked answer

The curve reaches the axis at −1/2; the normal reaches it at 5. The region is under the curve up to 4, then under the normal from 4 to 5.

\begin{gathered}\int_{-1/2}^{4}\sqrt{2x+1}\,dx\\=\left[\frac{(2x+1)^{3/2}}3\right]_{-1/2}^{4}\\=9\\A=9+\frac32=\frac{21}{2}\end{gathered}

This triangle is added, not subtracted: it is part of the requested region beyond the curve-normal meeting point.