Maths › Integration › Areas, parametric curves and the limit of a sum
Areas, parametric curves and the limit of a sum
Three upgrades to area-finding: the gap between two curves surrenders to a single integral of top minus bottom, curves given parametrically integrate without ever leaving the parameter, and underneath it all sits the definition, an integral as the limit of a sum of ever-thinner strips.
Builds on Definite integrals and areas and Parametric equations.
IN THIS TOPIC
- Find the area between two curves with one integral of top minus bottom.
- Find areas under parametric curves using ∫y (dx/dt) dt, with a Cartesian check.
- Read a definite integral as the limit of a sum of strips of width δx.
COMMON MISCONCEPTION
The area between two curves only makes sense when both sit above the x-axis.
Between two curves
The gap between two curves is a height difference, top minus bottom, and integrating that difference between the crossing points measures the enclosed region in one pass, wherever the axis sits. Raise both curves by the same amount and the gap at every x is untouched, so you can always shift the region clear of the axis without changing its area. Top minus bottom works wherever the region happens to sit.
WORKED EXAMPLE
One integral, both boundaries
Find the area enclosed between y = 8 − x2 and y = x2.
Crossings first: x2 = 8 − x2 gives x = ±2.
Integrate the gap: ∫−22 (8 − 2x2) dx = [8x − 2x3/3].
Evaluating, (16 − 16/3) − (−16 + 16/3) = 64/3.
Which curve is on top matters. Get it the wrong way round and the answer arrives negative, which is a signal to swap the subtraction, not to abandon the method.
GUIDED PRACTICE
A tighter lens
Find the area enclosed between y = √x and y = x2, before opening the working.
Show the working
The curves cross where √x = x2, at x = 0 and x = 1, with √x on top between them.
∫01 (x1/2 − x2) dx = [(2/3)x3/2 − x3/3] = 2/3 − 1/3 = 1/3.
These two curves are reflections of each other in y = x, so the region is symmetric about that line. Spotting a symmetry like that makes a decent sanity check on the answer.
Areas from parametric curves
A parametric curve gives x and y through a parameter t, and the area integral converts to run entirely in t. Since dx = (dx/dt) dt,
with t-limits that match the x-limits of the region. The booklet does not carry this one, so quote it yourself. Converting to Cartesian form first is allowed whenever the conversion is clean, and doing the question both ways is the strongest check you have.
WORKED EXAMPLE
Staying in the parameter
The curve x = t2, y = 2t is traced for t ≥ 0. Find the area under it from t = 0 to t = 2.
dx/dt = 2t, so the area is ∫02 2t × 2t dt = ∫02 4t2 dt.
That is [4t3/3] from 0 to 2 = 32/3.
Cartesian check: the curve is y2 = 4x, so y = 2√x, and ∫04 2√x dx = 32/3 as well. Same region, two routes, one answer.
INDEPENDENT PRACTICE
Another trace
Find the area under the curve x = t3, y = t2 from t = 0 to t = 2.
Show the working
dx/dt = 3t2, so the area is ∫02 t2 × 3t2 dt = ∫02 3t4 dt.
That is [3t5/5] from 0 to 2 = 96/5.
Here the t-limits came straight from the question. When a question gives x-limits instead, translating them into t-values is the step that separates the marks.
The limit of a sum
Underneath the square-bracket routine sits the definition of the thing itself. Chop the region into strips of width δx, take each strip's area as roughly f(x) δx, and add them up. The sum is not the area. It misses by less as the strips thin, and the definite integral is the number those sums close in on.
The notation is a fossil of that idea. The integral sign is a stretched S for sum, and dx sits exactly where the strip width used to be.
Exam questions on this point are recognition questions. Given a sum of the form Σ f(x) δx over an interval, name the definite integral it tends to, limits included. Nothing gets computed. The mark is for reading the notation.
ASSESSMENT FOCUS
- Between two curves, find the crossings, integrate top minus bottom between them, and state which curve is on top.
- A negative answer means the subtraction went the wrong way round. Swap it and say so; do not start again.
- For parametric areas use ∫y (dx/dt) dt with t-limits, translating any given x-limits into t before integrating.
CHECK YOURSELF
The curve x = 2 cos t, y = sin t for 0 ≤ t ≤ π/2 is a quarter of an ellipse. Show that the area under it is π/2.
Show a hint
As t runs π/2 → 0, x runs 0 → 2; then a sin² rewrite.
Show the answer
dx/dt = −2 sin t, and x increases as t falls, so the area is ∫ from π/2 to 0 of sin t × (−2 sin t) dt = ∫0π/2 2 sin2 t dt.
The double angle rewrite gives ∫0π/2 (1 − cos 2t) dt = [t − ½ sin 2t] = π/2.
Four such quarters make the full ellipse area 2π, matching the πab formula with a = 2 and b = 1, so the answer survives its own audit.
The area between curves is one integral of top minus bottom, wherever the region sits.
Parametric areas run in t via ∫y (dx/dt) dt, and every integral is secretly a limit of strip sums.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the areas, parametric curves and the limit of a sum questions page.
CHECK YOUR PROGRESS
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- Find the area between two curves with one integral of top minus bottom.
- Find areas under parametric curves using ∫y (dx/dt) dt, with a Cartesian check.
- Read a definite integral as the limit of a sum of strips of width δx.
Open the full revision checklist to see every objective in the course in one place.