Teacher page: definite integrals and areas between curves (AA 5.11)

The one thing to do with the figure

Slide the strip and ask what its height is.

Someone will say "the area". It is not: it is the height of one thin strip, top curve minus bottom curve, and the integral adds them up.

Watching the strip close to nothing at both ends is also the clearest possible argument for why the limits are 0 and 1 and must be found by solving.

Area and integral are different instructions

∫₀^2πsin x dx = 0 while the area is 4. The examiner means the word they wrote.

Make the habit: if the question says area, sketch it and look for crossings before writing anything.

The answers

∫₀²(x²+1)dx	14/3 ≈ 4.667.
Between y = x and y = x²	They meet at 0 and 1; area = ∫(x − x²) = 1/6 ≈ 0.1667.
sin over 0 to 2π	Integral 0, area 4.
1, 2, 3	4.667, 0.1667, 4.

Where the marks go

1 markFinding the limits by solving the curves equal.

1 markWriting the integral of the difference, correct way round.

1 markEvaluating, with brackets around the lower substitution.

1 markSplitting at a crossing when the question says area.

What each wrong answer tells you

They give	What it means
−1/6	Subtracted the wrong way round. An area cannot be negative, which is the check.
1/2 or 1/3	Integrated one curve only. They have not seen it as a single integral of a difference.
0 for the sine area	Gave the integral, which the question had already supplied.
Limits guessed	Reading them off a sketch rather than solving. Fine for a sanity check, not for the mark.

Other things they will say

"Which one is on top?" Test a point between the limits. Ten seconds, and it settles the sign permanently.

"What if part is below the x-axis?" Top minus bottom still works. That is the advantage of the difference form over two separate areas.

"Do I need the + C?" Not in a definite integral; it cancels. Show it cancelling once rather than just asserting it.

A possible order

	What is happening
1	The strip animation. What is the height of one strip?
2	Definite integrals by hand, with bracket discipline.
3	Signed against unsigned, with the sine example.
4	Between curves: find the limits, decide the order, integrate.
5	Questions 1 to 3.
6	A region where the top curve changes partway, as a preview of harder questions.

Two things not to say

Do not let them take the limits from a picture. Solving the two equations is the mark.

Do not teach between-curves as "area under the top minus area under the bottom". It is one integral of a difference, and the difference form is what survives when part of the region is below the axis.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

A logarithmEvaluate the integral of 1/x from 1 to e.
[ln x] from 1 to e = 1.
An exponentialEvaluate the integral of e^2x from 0 to 1.
(e² − 1)/2 = 3.194.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

Area between curvesFind the area between y = 2x and y = x².
They meet at x = 0 and x = 2. The integral of (2x − x²) is 4 − 8/3 = 4/3 = 1.333.
A trigonometric areaFind the area under y = sin x from 0 to π.
[−cos x] from 0 to π = 2.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

Find the limits firstExplain why the intersection points must be found before integrating between two curves.
They are the limits, and they also decide which curve is on top. Guessing them, or using the limits given for one curve alone, produces a number that is not the area. Here solving 2x = x² gives 0 and 2.
Signed against unsignedThe integral of sin x from 0 to 2π is zero. Explain why the area is not.
The second half lies below the axis and the integral counts it as negative, cancelling the first. The area is 4, found by splitting at π and taking the modulus of each piece.

Practicalities

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