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.
WHAT YOU PROBABLY THINK
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. The lie dissolves on inspection: raise both curves by the same amount and the gap at every x is untouched, so the region can be shifted clear of the axis without changing its area. Top minus bottom works wherever the region sits.
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: the wrong way round, the answer arrives negative, and a negative area is the sign to swap the subtraction, not to abandon the method.
YOUR TURN
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.
The two curves are reflections in y = x, which is why the region is symmetric about that line; symmetry like this makes a good 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 matching the x-limits of the region. Converting to Cartesian form first is allowed whenever the conversion is clean, and doing the question both ways is the strongest check available.
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.
TRY IT UNSEEN
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.
The t-limits came straight from the question, but 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. Chop the region into strips of width δx, each of area about f(x) δx, and add. The sum is not the area, but it misses by less as the strips thin, and the definite integral is what the sums close in on,
which is why the integral sign is a stretched S for sum, and why dx sits where the strip width was.
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. No area needs computing; the mark is for reading the notation.
THE EXAM BIT
- 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 was the wrong way round; swap it and say so rather than starting again.
- For parametric areas, use ∫y (dx/dt) dt with t-limits, translating any given x-limits into t before integrating.
- A Cartesian conversion is a legal route and a free check when both routes are quick.
- For limit-of-a-sum questions, match Σ f(x) δx to ∫f(x) dx over the same interval; it is notation reading, not calculation.
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, 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.
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.
No animated video for this topic yet; these notes stand alone.