Area Under a Curve Given Parametrically

The area under $x=4\cos t$, $y=3\sin t$ ($0\le t\le\frac{\pi}{2}$) between $x=0$ and $x=4$ — the case where matching the limits means reversing the natural order of $t$

1. 01

   Differentiate $x$ with respect to $t$: $x=4\cos t \Rightarrow \frac{dx}{dt}=-4\sin t$.

   Earns: M1 — correct derivative of $x(t)$. Already negative for every $t$ in $\left(0,\frac{\pi}{2}\right)$ — worth noticing now, because it's the reason the limits need care in stage 2.

2. 02

   Find the t-values matching the two given x-limits. $x=0$: $4\cos t=0\Rightarrow\cos t=0\Rightarrow t=\frac{\pi}{2}$ (the only solution in $0\le t\le\frac{\pi}{2}$). $x=4$: $4\cos t=4\Rightarrow\cos t=1\Rightarrow t=0$. So $x=0$ corresponds to $t=\frac{\pi}{2}$, and $x=4$ corresponds to $t=0$ — the LARGER x-limit belongs to the SMALLER t-value.

   Earns: M1 — correctly solves for both t-values, restricted to the given domain of t. This is the step spec 6.5's own guidance is entirely about: converting x-limits into t-limits.

3. 03

   Write the integral with the t-limits in the order that matches the x-limits: $x=0$ (the lower x-limit) is the LOWER limit of the t-integral, so its t-value, $\frac{\pi}{2}$, goes on the bottom; $x=4$ (the upper x-limit) is the UPPER limit, so its t-value, $0$, goes on top: $A=\int_0^4 y\,dx = \int_{\pi/2}^{0} y\,\frac{dx}{dt}\,dt = \int_{\pi/2}^{0} (3\sin t)(-4\sin t)\,dt$.

   Earns: A1 — correctly ordered t-limits, matched to their corresponding x-limits rather than left in the order they were found in stage 2. Reversing this order (writing $\int_0^{\pi/2}$ instead) is the single most direct way to lose this exact mark, and produces a negative final value.

4. 04

   Simplify the integrand before integrating: $(3\sin t)(-4\sin t) = -12\sin^2 t$.

   Earns: M1 — correct product, with the sign carried through explicitly rather than dropped.

5. 05

   Integrate, using $\sin^2 t = \frac{1-\cos 2t}{2}$: $-12\sin^2 t = -6(1-\cos 2t) = -6+6\cos 2t$, so $\int(-12\sin^2 t)\,dt = -6t+3\sin 2t$.

   Earns: M1 — attempts the double-angle identity and integrates term by term.

6. 06

   Evaluate at the limits found in stage 3: $\left[-6t+3\sin 2t\right]_{\pi/2}^{0} = (0+0) - \left(-6\cdot\frac{\pi}{2}+3\sin\pi\right) = 0-(-3\pi+0)=3\pi$.

   Earns: A1 — correct exact final value, and positive, as an area has to be. If the t-limits from stage 2 had been used in the wrong order ($0$ to $\frac{\pi}{2}$ instead of $\frac{\pi}{2}$ to $0$), this same working would evaluate to $-3\pi$ — same magnitude, wrong sign, from a limits-ordering slip rather than any error in the calculus itself.

Complete the chain — the area under $x=4\sin t$, $y=3\cos t$ ($0\le t\le\frac{\pi}{2}$) between $x=0$ and $x=4$ — the SAME arc as the worked example above, traced the other way

1. 01

   Differentiate $x$ with respect to $t$: $x=4\sin t \Rightarrow \frac{dx}{dt}=4\cos t$ — positive for every $t$ in $\left(0,\frac{\pi}{2}\right)$, unlike the worked example above.

2. 02

   Find the t-values matching the two x-limits. $x=0$: $4\sin t=0\Rightarrow t=0$. $x=4$: $4\sin t=4\Rightarrow\sin t=1\Rightarrow t=\frac{\pi}{2}$. This time the SMALLER x-limit ($x=0$) matches the SMALLER t-value ($t=0$), and the LARGER x-limit matches the LARGER t-value — the natural order of $t$ already matches the order of the x-limits, so no reversal is needed here: $A=\int_0^4 y\,dx = \int_0^{\pi/2} y\,\frac{dx}{dt}\,dt$.

Find the exact area under the curve given parametrically by $x=t^2$, $y=t$, for $t\ge0$, between $x=0$ and $x=4$. (VERIDIAN-original question — this course's research pass found no primary Pearson document evidencing spec 6.5 specifically, so this question, and every mark allocation attached to it, is modelled on the general M/A/B conventions verified in WMA14-verified-facts.md §4, not transcribed from a real mark scheme, which for this spec point this research pass did not find one of.)

2 valid methods · every one reaches $\frac{16}{3}$ · 2 marks available

A curve $C$ has parametric equations $x=t^2+1$, $y=4t-t^2$, for $t\ge0$. The region $R$ is bounded by $C$, the x-axis, and the lines $x=1$ and $x=5$. (a) Find $\frac{dx}{dt}$. (1) (b) Show that the x-limits $x=1$ and $x=5$ correspond to $t=0$ and $t=2$ respectively. (2) (c) Hence find the exact area of $R$. (3) — VERIDIAN-original question. This course's research pass found no primary Pearson document evidencing spec 6.5, so this question and every mark code and criterion attached to it below are modelled on the general M/A/B conventions verified in WMA14-verified-facts.md §4 (method marks for 'knowing a method and attempting to apply it'; accuracy marks dependent on the method mark; the general Pure-Mathematics principle that an integration method mark is earned once the power of at least one term is increased by 1), not transcribed from a real mark scheme, which for this spec point does not exist in the material this course has reviewed.

6 marks available

A curve $C$ has parametric equations $x=8\cos^3\theta$, $y=6\sin^2\theta$, $0\le\theta\le\frac{\pi}{2}$. The point $P$ lies on $C$ and has parameter $\theta=\frac{\pi}{3}$. (a) Find the coordinates of $P$. (2) The line $l$ is the normal to $C$ at $P$. (b) Find an equation for $l$ in the form $y=mx+c$, where $m$ and $c$ are constants. (5) The finite region $S$ is bounded by the curve $C$, the line $l$, the $y$-axis and the $x$-axis. (c) Show that the area of $S$ is given by $4+144\displaystyle\int_0^{\pi/3}(\sin\theta\cos^2\theta-\sin\theta\cos^4\theta)\,d\theta$. (6) (d) Hence, using algebraic integration, find the exact area of $S$. (3) — REAL Pearson Edexcel International A-Level question, WMA14 Paper 01A, January 2026, Q7 (Publications Code WMA14_01_2601_MS, Question Paper Log Number P87595A, sat Thursday 22 January 2026). Independently re-fetched and re-verified against the primary-source question-paper and mark-scheme PDFs (pdftotext -layout and -raw cross-checked, and the curve's own exponent placement and the mark scheme's final printed fraction confirmed by direct visual inspection of the rendered PDF page, not from pdftotext's layout output alone) and cross-checked by computer algebra (sympy) before being written in here — this is the first genuinely real, cited past-paper question this lesson has, closing the header's own honestly-stated gap that no real Pearson question or examiner commentary for spec 6.5 was ever found. It also teaches a structure absent from every other item in this lesson: region $S$ is bounded PARTLY by a straight line (the normal $l$, found using spec 5.1's parametric-differentiation technique) and PARTLY by the parametric curve $C$ itself, so its area is found as a trapezium-under-the-line component plus a parametric-integral component — not a region bounded purely by the curve and vertical/horizontal cut lines, the only shape every other item in this lesson uses. One real limitation, stated rather than smoothed over: no examiner report for this series was found publicly available at the time of writing, so — like the first marked-solution block above — this item can verify the real mark-scheme MECHANICS but does not yet contribute a documented candidate MISCONCEPTION; it was deliberately not added to the trap-taxonomy below for that reason.

16 marks available