DSE Maths | Definite integration for area and volume lesson plan (for tutors)

DSE Maths | Definite integration for area and volume lesson plan (for tutors) | One curve, three questions

2026-11-03

Tutor lesson-prep guides · Secondary Maths

Students can integrate x squared but panic at the word area. This lesson uses a single curve to answer area, the area between curves and a volume of revolution, then explains why the trapezoidal estimate is high.

Objectives — Students can: (1) evaluate the definite integral of x squared from 0 to 2 as 2.667 and state that it measures signed area, with limits written correctly; (2) find the area between two curves by solving for intersections first and subtracting the lower from the upper; (3) compute a volume of revolution with pi times the integral of y squared (20.11 here); (4) estimate the same integral with four trapezia (2.75) and explain that the error of 0.083 (3.13 per cent) is an overestimate because the curve bends upwards.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Indefinite integral	What is the integral of x squared?	x cubed over three, plus C
Evaluating limits	How is F(2) – F(0) found?	Substitute the upper limit, subtract the lower
Area of a circle	Area with radius y?	Pi y squared

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Opener	Can rectangles from primary school add up to the area under a curve?	Estimate first, then check with integration
10-22 min	Integral means area	The integral of x squared from 0 to 2 is 2.667; write the antiderivative and limits	They compute 2.667
22-34 min	Between two curves	Solve for intersections, then upper minus lower	y = x and y = x squared enclose 0.167
34-46 min	Volume of revolution	Pi times the integral of y squared; square inside the integral	20.11
46-56 min	Trapezoidal rule and error	Four trapezia give 2.75, above the true 2.667	Explain why it overestimates
56-60 min	Exit check	Mark on the spot	Three correct

03 Teaching script (read it out)

Scene 1: A definite integral does two things: antiderivative, then substitute

The integral of x squared from 0 to 2 is x cubed over three evaluated at 2 minus at 0, which is 2.667. Three marks are lost here: forgetting dx, reversing the limits so the answer turns negative, and substituting 2 but forgetting to subtract the value at 0. Always shade the region on the sketch. Area is positive while an integral can be negative, so below the axis take the modulus or split the interval.

Ask them to: write the full three lines for the integral of 2x + 1 from 1 to 3.

Scene 2: Between curves: intersections first, upper minus lower second

y = x and y = x squared meet at x = 0 and x = 1 (solve x = x squared). Between them x is above x squared, so the area is the integral of x minus x squared, which is one half minus one third, 0.1667. Reversing the order gives a negative number, which is the signal that upper and lower are swapped. With three or more intersections the integral must be split and the order rechecked on each piece.

Follow up: integrating with respect to y uses x = y and x = y squared. Does the answer change?

Scene 3: Volume: pull pi out, square y

Rotating y = x squared about the x axis gives circular slices of radius y, so the volume is pi times the integral of y squared, that is pi times the integral of x to the fourth, which is 20.1062. The common slip is forgetting to square y, or multiplying by pi twice. Check with the trapezoidal rule: four strips of width 0.5 give 2.75, which is 0.083 above the true 2.667 (3.13 per cent), because the curve bends upwards and every chord lies above it.

Ask them: for a curve bending downwards, would the trapezoidal rule overestimate or underestimate?

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation	Area under a curve	Antiderivative, then substitute the limits	2.667
Calculation	Area between two curves	Intersections, upper minus lower, integrate	0.1667
Calculation	Volume about the x axis	Pi times the integral of y squared	20.11
Estimation	Trapezoidal value and error direction	Split, weight, compare	Overestimate by 3.13 per cent

Sanity check (teach this): Check: the exact value is 2.6667; four trapezia give 2.75, an error of 0.0833, about 3.13 per cent. The volume is pi times 32 over 5, 20.1062, in cubic units. Every area answer must be positive; a negative result means reversed limits or swapped curves, never a valid area. Write the integral, then the antiderivative, then the value, so method marks survive an arithmetic slip.

05 Independent practice (3 questions)

Question	Answer	What to watch
Integral of x squared from 0 to 2	2.667	Antiderivative x cubed over three
Area enclosed by y = x and y = x squared	0.1667	Intersections 0 and 1, upper minus lower
Volume for y = x squared about the x axis, 0 to 2	20.11	Pi times the integral of x to the fourth

06 Common mistakes and what to say

Mistake	Student writes	Say this
Forgetting to square	Using pi times the integral of y	“The slice is a circle of radius y, so its area is pi y squared.”
Reversing the limits	Copying a negative area	“Area is positive; a negative answer means the limits are reversed.”
Skipping the intersections	Integrating blindly from 0 to 1	“The intersections are the boundaries; without them you integrate the wrong piece.”
Treating the trapezoidal value as exact	Writing area equals 2.75	“2.75 is an estimate; the true value is 2.667, so state the error.”

07 Homework (paper, section and question numbers)

Maths chapter index — All chapters
Practice paper, Paper 1 Section B — Questions 12 to 14: area and volume

08 Exit check (last 5 minutes)

Question	Answer	Decision
Integral of x squared from 0 to 2	2.667	Right before probability
Area between the two curves	0.1667	One wrong: redo the intersections
Direction of the trapezoidal error	Overestimate	Two wrong: restart act three

09 Tutor FAQ

Must a sketch be drawn?

Not always, but it shows which curve is upper and where the interval ends, saving time.

Can the calculator integrate?

DSE calculators cannot integrate, so the antiderivative must be written by hand.

Will rotation about the y axis appear?

Yes; then integrate x squared with respect to y, using the same structure.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.