Area under a curve given parametricallyEdexcel International A Level Maths: Revision notes

For a curve with parametric equations $x=f(t)$ and $y=g(t)$, the area between the curve, the $x$-axis and the lines $x=a$ and $x=b$ is $\int_a^b y\,dx$. Writing $dx=\frac{dx}{dt}\,dt$ turns this into an integral in $t$ only: $A=\int_{t_1}^{t_2}y\frac{dx}{dt}\,dt,$ where $t_1$ and $t_2$ are the values of $t$ at $x=a$ and $x=b$. You do not need to sketch the curve, but you must know the curve lies above the $x$-axis over the range (or the area is negative).

The limits must be values of $t$, not $x$. Solve $f(t)=a$ and $f(t)=b$, and choose the values in the stated range of $t$. Example: $x=t^2$, $y=4t$, $t\ge0$, between $x=1$ and $x=9$. Then $t=1$ and $t=3$, and $A=\int_1^3 4t\cdot2t\,dt=\int_1^38t^2\,dt=\left[\frac{8t^3}{3}\right]_1^3=\frac{208}{3}$. If the region meets the $x$-axis, the limits come from $y=0$. For $x=t^2$, $y=4t-t^2$, $y=0$ at $t=0$ and $t=4$.

Example 1: $x=2t+1$, $y=t^2+2$, between $x=1$ and $x=7$. $t=0$ to $t=3$, $\frac{dx}{dt}=2$, so $A=\int_0^32(t^2+2)\,dt=2\left[\frac{t^3}{3}+2t\right]_0^3=30$. Example 2: $x=t^2$, $y=4t-t^2$, $0\le t\le4$. $\frac{dx}{dt}=2t$, so $A=\int_0^4(8t^2-2t^3)\,dt=\left[\frac{8t^3}{3}-\frac{t^4}{2}\right]_0^4=\frac{128}{3}$. Splitting at $x=4$ (that is $t=2$) gives $\frac{40}{3}$ and $\frac{88}{3}$.

With trigonometric equations you may need identities. For $x=2(t-\sin t)$, $y=2(1-\cos t)$, $\frac{dx}{dt}=2(1-\cos t)$ and the area under one arch is $\int_0^{2\pi}4(1-\cos t)^2\,dt$. Expand: $4(1-\cos t)^2=4-8\cos t+4\cos^2t$ and use $\cos^2t=\frac{1+\cos2t}{2}$ to get $6-8\cos t+2\cos2t$. Then $\left[6t-8\sin t+\sin2t\right]_0^{2\pi}=12\pi$.

Check the sign: an area should be positive. If $\frac{dx}{dt}<0$ over the range, $x$ decreases as $t$ increases, so the integral comes out negative and you should swap the limits (or take the modulus). Put units on contextual answers, for example m$^2$, and use the area for any further calculation: $\frac{40}{3}\times6=80$ plants.