Area Under a Curve | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Area does not cancel across the axis

An integral adds signed contributions. Geometric area measures size, so every separate region contributes positively. First identify the curve, the axis and the other boundaries. The y-axis is $x=0$ ; an x-axis intersection satisfies $y=0$ .

The two geometric areas are equal. Their signed contributions cancel, but their sizes add.

If the curve stays above the x-axis, integrate directly with increasing limits. If it stays below, negate that integral. If it crosses the axis, split at the crossings before adding the positive sizes.

Find the zeros inside the region

Worked example

Find the total area between y = x(x − 2) and the x-axis, from x = 0 to x = 3.

The roots are 0 and 2. At $x=1$ , the product is negative; at $x=5/2$ , it is positive. The change at 2 needs separate integrals.

\begin{gathered}F(x)=\frac{x^3}{3}-x^2\\\int_0^2(x^2-2x)\,dx\\=F(2)-F(0)=-\frac43\\\int_2^3(x^2-2x)\,dx\\=F(3)-F(2)=\frac43\end{gathered}

Answer

\begin{gathered}A=\frac43+\frac43\\=\frac83\text{ square units}\end{gathered}

The signed integral over the whole interval is zero, despite the non-zero area.

Common mistake

Taking the absolute value of one integral after positive and negative parts have cancelled cannot recover the area. Split first. A calculator integral is signed too.

Your turnPractice

Find the total area between $y=x^2-4$ and the x-axis for $0\le x\le3$ .

Show worked answer

Only the root $x=2$ is inside this interval. The curve is negative before it and positive after it.

\begin{aligned}&A=-\int_0^2(x^2-4)\,dx\\&\qquad+\int_2^3(x^2-4)\,dx\\&=\frac{16}{3}+\frac73=\frac{23}{3}\end{aligned}

The other root, −2, is outside the requested region.