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Tawjihi (Grade 12) · Math · Integration

Applications of integration: a paper in the national format

An integration paper for Grade 12, sat in 90 minutes

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A 6-question paper on the area between two curves and the volume of a solid of revolution, laid out in the national exam's order and carrying its mark allocation. A student who finishes it has met every shape of question the paper is likely to ask.

Published: June 9, 2026Updated: August 30, 2026Duration: 90 minutesTaught in: Jordan

90 minutes

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What the student does

1
Read the whole paper first
The questions run from shortest to longest with the mark printed beside each one, so the student divides the clock before picking up a pen.
2
Sketch before integrating
Every area question opens with a rough sketch of the region. A region that was never drawn tends to get integrated between the wrong limits, or between two of the three it needed.
3
Find the intersection points himself
In four of the six questions the limits are not given. They come out of an equation that has to be worked before any integrating starts.
4
Write the steps, not only the result
The mark split is printed on the paper: one for the right integral, one for the right limits, one for the result. The student sees where a mark goes missing before it goes missing.
5
Check the sign
An area is never negative. Any negative result on an area question points at a reversed order, or at an interval that should have been split.

What's inside

6 questions in the national paper's order, with its mark allocation
A model answer with the marks spread step by step
A sketch of the region attached to every area question
Two volume questions, about two different axes

Outcomes

Finds the area between two curves after working out the intersection points himself.

Splits the interval at the point where the two functions swap places, and adds the pieces.

Sets up the volume integral about the x-axis and about the y-axis, and knows which one calls for integrating with respect to y.

What the teacher focused on

The student knows how to integrate. What he loses in the exam is the two limits: from where to where. So I built the paper on that alone.

Teaching strategies

The sketch as a precondition, not a decoration
Limits left out of most of the questions
The mark split shown to the student, not hidden from him
Questions ordered from shortest to longest

He left the limits out of four of the six questions, and turned down the common format that hands them to you in the wording. A question that gives you both limits tests integration on its own, and integration is not where his students lose marks. The loss happens earlier, when the region is being pinned down.

Questions from inside it

1Find the area of the region enclosed between the curve y = x² and the line y = x + 2.
2The region between the curve y = √x and the x-axis from x = 0 to x = 4 is turned one full revolution about the x-axis. Find the volume of the solid formed.
3Find the area of the region enclosed between the curve y = x³ − x and the x-axis.

Curriculum

Grade: Tawjihi (Grade 12)
Subject: Math
Unit: Integration
Taught in: Jordan

Where this lesson sits in the curriculum

The same lesson appears in these countries' curricula. One page, five paths to it.

When to use it

The last period before the revision week

One paper shows a student how much he knows and how much is left, while he can still put his hand up and ask you.

A mock sat under a real clock

90 minutes without a break. A student who spends 25 minutes on a question he knows loses the last question in full.

One student, after school

Work question three in front of him and nothing else. Its missing limits are the whole unit in a single question.

The explanation

The area between two curves is the integral of their difference — the upper function minus the lower one — across the interval between the two intersection points. Those points come from solving the equation that sets the functions equal, which is why the work starts with an equation and not with an integral. If the two functions swap places inside the interval, one integral subtracts one area from the other and returns a number smaller than the real area; the fix is to split the interval at the swap and add the pieces. The volume of a solid of revolution about the x-axis is the integral of π y² with respect to x, not the integral of π y. The square is not a detail in the formula. It is the area of the circular cross-section being swept: radius y, area π y².

Common mistakes it catches

Writing the volume as π ∫ y dx instead of π ∫ y² dx

What that produces is an area multiplied by π, not a volume. Ask him for the units: ∫ y dx comes out in square units and a volume is cubic. Then tie y² back to πr², the area of the circle the rotation is sweeping.

Integrating x³ − x from −1 to 1, getting 0, and writing that the area is zero

The curve crosses the axis at 0, so the piece below it integrates to a negative that cancels the piece above. The area is there; it is the integral that vanished. Split at 0 and add the absolute values. The working rule: an area of zero on an exam paper is not an answer, it is a root nobody looked for.

Substituting u = x² + 1 in a definite integral and leaving the limits at 0 and 2

The limits are written in x, so when the variable changes they change with it: 0 becomes 1 and 2 becomes 5. The alternative is to go back to x before putting the limits in — longer, but safer under time pressure.

More from this teacher

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