Area Between Curves | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

The strip height is top minus bottom

At one fixed $x$ , subtract the lower y-coordinate from the upper y-coordinate. This gives the vertical gap, even when both coordinates are negative: from −5 to −2 the gap is $-2-(-5)=3$ .

The graphs meet at x = −1 and x = 3. At x = 1, their vertical gap is 5 − 1 = 4.

A=\int_a^b\big(y_{\text{top}}-y_{\text{bottom}}\big)\,dx

The limits must describe the requested region. If it is enclosed just by the two graphs, set their equations equal to find the intersections. Test one input between consecutive intersections to decide which graph is higher.

Find intersections, then keep the subtraction order

Worked example

Find the area between y = 2x + 3 and y = x².

Solve $x^2-2x-3=0$ . A polynomial solver gives −1 and 3; the convenient roots give the written factorisation $(x+1)(x-3)=0$ . At $x=0$ , the line is higher.

\begin{gathered}\int_{-1}^3(3+2x-x^2)\,dx\\=\left[3x+x^2-\frac{x^3}{3}\right]_{-1}^3\\=9-\left(-\frac53\right)\\=\frac{32}{3}\end{gathered}

For awkward decimal roots, use exact algebra or the quadratic formula when exact limits are required; do not round the boundaries early.

If the graphs exchange order at another intersection inside the requested interval, split there and reverse which function is subtracted. Crossing the x-axis alone does not require a split when the gap remains positive.

Your turn9709/11 May/Jun 2025 Q4 [5 marks]

The curve $y=5x^{3/2}-20x$ and line $y=x-16$ meet at $x=1$ and 16. Find the enclosed area.

Show worked answer

At $x=4$ , the line gives −12 and the curve gives −40. The line is above the curve.

\begin{aligned}&A=\int_1^{16}(21x-16)\,dx\\&\quad-\int_1^{16}5x^{3/2}\,dx\\&=\left[\frac{21x^2}{2}-16x\right]_1^{16}\\&\quad-\left[2x^{5/2}\right]_1^{16}\\&=384-\left(-\frac{15}{2}\right)\end{aligned}

Answer

\frac{783}{2}\text{ square units}