Maths › Integration › Areas, parametric curves and the limit of a sum

Areas, parametric curves and the limit of a sum

The region between two curves is one integral of top minus bottom, and the area under a curve given parametrically can be found without converting back to Cartesian form. Both rest on the definition of the definite integral as the limit of a sum of thinner and thinner strips.

Year 12-13EDEXCEL 9MA0 8.3, 8.4

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.

One integral of top minus bottom between the intersections gives the enclosed area wherever the region sits relative to the axis. Adding the same constant to both curves leaves every vertical difference unchanged, so moving the axis cannot change the result. State which curve is the upper one: a negative answer means the subtraction was taken the wrong way round.

Between two curves

The vertical distance between two curves is the difference of their y-values, top minus bottom, and integrating that difference between the points of intersection gives the enclosed area, whether or not the region lies above the x-axis. Adding the same constant to both curves leaves every difference unchanged, which is why the position of the axis does not affect the result.

Two parabolas enclosing a lens-shaped region shaded between their crossings at minus 2 and 2: integrating the gap between top and bottom curves in one pass gives the computed area of 64 over 3, with no need to treat the two curves separately.
FIG. 1Two parabolas, one lens: integrating the gap 8 − 2x² between the crossings at ±2 gives the area in a single integral, 64/3.

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.

State which curve is the upper one. Subtracting them the other way round gives the negative of the area, so a negative result means the subtraction should be reversed.

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.

The two curves are reflections of each other in y = x, so the region is symmetric about that line, which provides a check on the result.

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,

area=∫ydxdtdt\text{area} = \text{∫} y \, \frac{dx}{dt} \, dtNOT IN THE BOOKLET: LEARN IT

with t-limits corresponding to the x-limits of the region, and with x increasing over that range of t. Converting to Cartesian form first is also valid where the conversion is straightforward, and doing the question both ways checks the answer.

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 for y ≥ 0, and ∫04 2√x dx = 32/3, the same value by the other method.

The area under the parametric curve x equals t squared, y equals 2t from t 0 to 2: one strip of height y and width dx by dt times delta t, t-limits 0 and 2 becoming x-limits 0 and 4, area 32 over 3
Describe the drawing

The parametric curve x equals t squared, y equals 2t sweeping over a shaded region down to the x-axis, with one amber strip of height y and width dx by dt times delta t pulled out, dots at t equals 0, 1 and 2, and the t-limits translated beneath the axis into the x-limits 0 and 4: both integrals measure 32 over 3.

FIG. 2The region just measured, drawn: the curve x = t², y = 2t sweeping over shading that runs down to the x-axis, with one strip pulled out in amber. Its height is y and its width is (dx/dt) δt, which is where the extra factor in the integral comes from. Dots mark t = 0, 1 and 2, and beneath the axis the t-limits 0 and 2 are shown becoming the x-limits 0 and 4.

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.

The t-limits were given here. Where a question gives x-limits instead, convert them into values of t before integrating.

The limit of a sum

The definite integral is defined as a limit. Divide the region into strips of width δx, approximate each strip's area by f(x) δx, and add them. The sum approximates the area, with the error decreasing as δx decreases, and the definite integral is the limit of those sums as δx tends to zero.

limδx→0Σf(x)δx=∫abf(x)dx\text{lim}_{δx → 0} \, Σ \, f(x) \, δx = \text{∫}_{a}^{b} f(x) \, dx

The notation records that definition: the integral sign is an elongated S for sum, and dx occupies the position of the strip width δx.

Eight shaded rectangles tucked under a rising parabola, each of height f of x and width delta x: their computed total falls short of the true area 8 over 3, narrower strips close the gap, and the definite integral is the limit the sums are heading for.
FIG. 3Eight strips under y = x² total 2.1875 against a true 8/3 ≈ 2.667. A hundred strips reach 2.6268, ten thousand reach 2.6663: the sums close on the integral.

Questions on this definition ask for a translation rather than a calculation: given a sum of the form Σ f(x) δx over an interval, state the definite integral it tends to, including the limits.

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 was taken the wrong way round; reverse it and state which curve is on top.
  • For parametric areas use ∫y (dx/dt) dt with t-limits, converting any given x-limits into values of 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 give the ellipse area 2π, which agrees with πab for a = 2 and b = 1.

The area between two curves is one integral of top minus bottom between their intersections, wherever the region lies relative to the axis.

Parametric areas are found in t as ∫y (dx/dt) dt, and the definite integral is defined as the limit of a sum of strips as δx tends to zero.

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.

Question workbook

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Question workbook

PDF · 78 KB · screen edition · navy

Answer book

PDF · 100 KB · fully worked · navy
15 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the areas, parametric curves and the limit of a sum questions page.

USED LATER IN

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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.