Take a breath — this chapter is genuinely one of the friendliest in Class 12 Maths, and I mean that. After the long grind of indefinite integrals, substitutions and partial fractions, Application of Integrals asks you to do something almost restful: look at a picture, shade a region, and find how much space it covers. That is it. The integration you already learned does the heavy lifting; your new job is mostly about seeing the region clearly and setting up the right limits.
I want you to think of a definite integral here as a very patient area-counter. Imagine you have a curved flower bed in your garden and you want to know how much soil to buy. You cannot use a simple rectangle formula, because one edge is curved. So instead you slice the bed into hundreds of paper-thin vertical strips. Each strip is so narrow that it is basically a rectangle — and you can find the area of a rectangle. Add up all the strips, and you have the area of the whole bed. That adding-up is exactly what the integral sign ∫ does.
We will go slowly. We will start with the plain idea of a strip, move to areas under lines and simple curves, learn what to do when the region dips below the x-axis, and then work carefully through the curves the syllabus actually names: lines, circles, parabolas and ellipses in standard form. Every single number on this page has been checked symbolically and numerically before it was written down, so you can trust the answers while you build your own confidence.
Meet Your Tutor
Application of Integrals is less about difficult integration and more about describing a region correctly. I will help you sketch the boundaries, choose vertical or horizontal strips, split at every sign or curve change, and check the final area geometrically. Draw the picture before the integral every single time; that habit prevents most errors.
What You’ll Learn
- Area As A Definite Integral: The Thin-Strip Idea
- Area Under A Curve With Respect To The X-Axis
- Area With Respect To The Y-Axis
- Regions Below The Axis And Why You Take The Modulus
- Area Of A Circle And Its Parts
- Area Under A Parabola
- Area Of An Ellipse And Its Quadrants
- Areas Bounded By A Line And A Standard Curve
- Symmetry Shortcuts That Save You Real Time
- Standard Area Results At A Glance
- Area Between Two Curves (Good To Know)
- How To Set Up Any Area Question In Five Steps
- Common Mistakes And How To Avoid Them
Your Game Plan
Please do not read this chapter like a novel. Work it like a gym session — a little effort in each set, with a rest in between. Here is the order I would sit beside you and insist on.
- Read The Thin-Strip Idea until the picture makes sense in your head. Do not move on until you can explain “why does an integral give area?” to a friend in two sentences.
- Master area under a curve w.r.t. the x-axis with straight lines first. Lines are easy to check because you can also find the area by school geometry — do both and watch them agree. That agreement is what builds trust.
- Do the y-axis version next while the x-axis version is still fresh. They are the same idea with the letters swapped, and learning them close together stops you confusing them later.
- Spend real time on regions below the axis. This is where most marks quietly leak away in the board exam.
- Now do the three named curves one at a time: circle, then parabola, then ellipse. One sitting each is plenty.
- Learn the symmetry shortcuts only after you can do the long way. Shortcuts you do not understand are traps.
- Finish with the Practice Worksheet. Attempt every question with pen on paper and a rough sketch before you open the answer.
Study Notes
Area As A Definite Integral: The Thin-Strip Idea
Look at the figure above. We want the area of the shaded region — the patch trapped between the blue curve, the x-axis, and the two vertical dashed lines at x = a and x = b. There is no formula in mensuration for that shape, because the top edge is curved.
So here is the trick, and it is genuinely the whole chapter in one idea. Chop the region into a large number of very thin vertical strips, like the yellow ones in the picture. Take any one strip. Its width is a tiny amount we call dx. Its height is simply the y-value of the curve at that point, that is, y = f(x). Because the strip is so thin, the curve across the top of it is practically a flat line, so the strip is practically a rectangle:
Now add up every strip from the left edge x = a to the right edge x = b. “Add up infinitely many infinitely thin things” is precisely what the integral sign was invented to mean. So:
That is the master formula. Everything else in this chapter is this one line, applied to different curves and different boundaries. When you meet a hard-looking board question, come back and whisper to yourself: what is my strip, and where does it start and stop?
One small point of language. The dx at the end is not decoration — it is telling you the strips are vertical and their thickness is measured along the x-direction. Later, when we use dy, the strips will be horizontal instead. Keeping that picture in mind will stop a lot of confusion.
Solution. The strip is vertical, so we use dx. Its height is the curve value y = x²/4 + 1. It sweeps from x = 1 to x = 4. Therefore
Area = ∫14 (x²/4 + 1) dx.
That is all the “setting up” ever asks of you: height, then limits.
Solution. ∫14 (x²/4 + 1) dx = [ x³/12 + x ]14
= ( 64/12 + 4 ) − ( 1/12 + 1 )
= ( 16/3 + 4 ) − ( 1/12 + 1 )
= 28/3 − 13/12 = 112/12 − 13/12 = 99/12 = 33/4 = 8.25 square units.
Notice the answer carries the words “square units”. Area always does. Do not drop them.
Solution. Area = ∫14 (3x + 2) dx = [ 3x²/2 + 2x ]14 = (24 + 8) − (3/2 + 2) = 32 − 7/2 = 57/2 = 28.5 square units.
Check it without calculus. The region is a trapezium standing on the x-axis. Its parallel sides are the heights at x = 1 and x = 4, i.e. 5 and 14, and the distance between them is 3. Trapezium area = ½ × (5 + 14) × 3 = ½ × 19 × 3 = 57/2. The two methods agree exactly. Do this kind of cross-check a few times — it turns “I hope this is right” into “I know this is right”.
Do not move on until that thin-strip picture feels comfortable. If you can look at any shaded region and immediately say “my strip is vertical, its height is this, it runs from here to here”, the rest of this chapter is just arithmetic.
Area Under A Curve With Respect To The X-Axis
This is the standard case and the one you will meet most often. The region sits between the curve and the x-axis, and it is fenced in on the left and right by two vertical lines.
Area = ∫ab y dx.
Three things to check every single time, in this order:
- Is y written as a function of x? If the equation is given as y² = 4x, you must first make y the subject: y = 2√x (taking the upper branch). You cannot integrate an equation, only a function.
- Where are the left and right boundaries? Sometimes they are handed to you (“between x = 1 and x = 4”). Sometimes you must find them, usually by seeing where the curve meets the x-axis (put y = 0 and solve) or where it meets a given line.
- Is the curve above the axis throughout? If any part dips below, stop — that needs the special treatment two sections from now.
Solution. On 0 ≤ x ≤ 2 the cube x³ is never negative, so the region sits above the axis and we can integrate directly.
Area = ∫02 x³ dx = [ x⁴/4 ]02 = 16/4 − 0 = 4 square units.
Solution. No limits are given, so the curve itself must supply them. The region is closed off where the curve meets the axis, so set y = 0:
4x − x² = 0 ⇒ x(4 − x) = 0 ⇒ x = 0 or x = 4.
Between 0 and 4 the parabola opens downwards and lies above the axis, so
Area = ∫04 (4x − x²) dx = [ 2x² − x³/3 ]04 = ( 32 − 64/3 ) − 0 = 96/3 − 64/3 = 32/3 square units.
Solution. The parabola y = 2x² passes through the origin, so the left boundary is x = 0 and the right boundary is x = 3.
Area = ∫03 2x² dx = [ 2x³/3 ]03 = 2(27)/3 = 18 square units.
Area With Respect To The Y-Axis
Sometimes the region is squeezed against the y-axis instead: bounded on the left by the y-axis, on the right by a curve, and capped by two horizontal lines y = c and y = d. Turning your head sideways would fix it, and that is essentially what we do algebraically.
Area = ∫cd x dy, where x has been written as a function of y.
Everything mirrors what you already know. Where you used to write y = f(x) and integrate dx, you now write x = g(y) and integrate dy. The only genuinely new habit is rearranging the equation to make x the subject.
Solution. x is already the subject — lovely. Horizontal strips of length x = 2y + 1 and thickness dy, running from y = 0 up to y = 3:
Area = ∫03 (2y + 1) dy = [ y² + y ]03 = (9 + 3) − 0 = 12 square units.
Geometry check. Again a trapezium: parallel sides x = 1 (at y = 0) and x = 7 (at y = 3), distance 3 apart. Area = ½ × (1 + 7) × 3 = 12. Perfect match.
Solution. Horizontal strips of length x = y²:
Area = ∫13 y² dy = [ y³/3 ]13 = 27/3 − 1/3 = 9 − 1/3 = 26/3 square units.
Solution. Make x the subject: x = √(25 − y²), taking the positive root because we are in the first quadrant.
Area = ∫05 √(25 − y²) dy.
Using the standard result ∫√(a² − t²) dt = (t/2)√(a² − t²) + (a²/2) sin−1(t/a):
= [ (y/2)√(25 − y²) + (25/2) sin−1(y/5) ]05
= ( 0 + (25/2)(π/2) ) − ( 0 + 0 ) = 25π/4 square units.
Sanity check: a full circle of radius 5 has area 25π, and a quarter of that is 25π/4. It agrees.
Regions Below The Axis And Why You Take The Modulus
Here is the single most common place students lose marks in this chapter, so let us be very careful and very honest about what is happening.
The definite integral does not actually compute area. It computes signed area. When the curve is above the x-axis, y is positive, each strip contributes a positive amount, and the integral comes out positive — that is the area. But when the curve dips below the axis, y is negative, so every strip contributes a negative amount and the integral comes out negative.
Area, of course, can never be negative. A region does not stop existing because it happens to sit below a line you drew. So we have to intervene.
Area = | ∫ab y dx |.
Equivalently, and often tidier, just swap the order and integrate ∫(0 − y) dx, which is positive from the start.
The nastier case is when the curve crosses the axis somewhere in the middle, as in the figure above. Then the negative part quietly cancels some of the positive part, and the single integral gives you a number that is neither the area nor anything useful.
Area = | ∫ac y dx | + | ∫cb y dx |.
Solution. On 0 ≤ x ≤ 2 we have x² ≤ 4, so y is negative throughout — the whole region hangs below the axis.
∫02 (x² − 4) dx = [ x³/3 − 4x ]02 = ( 8/3 − 8 ) − 0 = −16/3.
Taking the modulus, Area = 16/3 square units.
Solution. First, where does the curve cross the axis? x² − 1 = 0 gives x = ±1, and x = 1 lies inside our interval. So we must split at x = 1.
Piece 1 (0 to 1, below the axis): ∫01 (x² − 1) dx = [ x³/3 − x ]01 = 1/3 − 1 = −2/3. Modulus: 2/3.
Piece 2 (1 to 2, above the axis): ∫12 (x² − 1) dx = [ x³/3 − x ]12 = (8/3 − 2) − (1/3 − 1) = 2/3 − (−2/3) = 4/3. Already positive.
Total area = 2/3 + 4/3 = 2 square units.
Why the split mattered. If you had lazily integrated straight through from 0 to 2 you would have got −2/3 + 4/3 = 2/3 — a perfectly real number, and completely wrong. That single missing step is worth several marks.
Solution. Tempting shortcut: ∫−22 x³ dx = [ x⁴/4 ]−22 = 4 − 4 = 0. Area zero? Obviously not — there is clearly a region there.
What happened is that the piece from −2 to 0 lies below the axis (area 4) and exactly cancels the piece from 0 to 2 above the axis (area 4). Split at the crossing point x = 0:
Area = | ∫−20 x³ dx | + | ∫02 x³ dx | = 4 + 4 = 8 square units.
Area Of A Circle And Its Parts
The circle in standard form is x² + y² = a², centred at the origin with radius a. To use it in an area integral we make y the subject:
The + sign gives the upper semicircle, the − sign the lower one. For a region in the first quadrant you always want the + branch.
Almost every circle question reduces to one integral, so it is worth knowing this standard result by heart. It comes from the substitution x = a sin θ, and you have met it in the Integrals chapter:
Now watch how one integral gives you the whole family of circle results. Integrate the upper branch from 0 to a and you get a quadrant. Multiply by 4 and you get the whole circle. Integrate from −a to a and you get the semicircle. Integrate from some x = k out to a and you get a segment.
Solution. Here a = 4, and in the first quadrant y = √(16 − x²). The region runs from x = 0 to x = 4.
Area = ∫04 √(16 − x²) dx
= [ (x/2)√(16 − x²) + 8 sin−1(x/4) ]04
= ( 0 + 8 × (π/2) ) − ( 0 + 0 ) = 4π square units.
Check: the whole circle has area π(4)² = 16π, and a quarter of that is 4π. It agrees, exactly as it must.
Solution. a = 3, so y = √(9 − x²), and the upper half stretches from x = −3 to x = 3.
Area = ∫−33 √(9 − x²) dx = [ (x/2)√(9 − x²) + (9/2) sin−1(x/3) ]−33
= ( 0 + (9/2)(π/2) ) − ( 0 + (9/2)(−π/2) ) = 9π/4 + 9π/4 = 9π/2 square units.
Again it checks out: half of π(3)² = 9π is 9π/2.
Solution. The line x = 1 slices the circle vertically; the smaller piece is the bit to the right of the line, running from x = 1 out to the rim at x = 2. That piece is symmetric about the x-axis, so find the top half and double it.
Area = 2 × ∫12 √(4 − x²) dx
= 2 [ (x/2)√(4 − x²) + 2 sin−1(x/2) ]12
= 2 [ ( 0 + 2 × (π/2) ) − ( (1/2)√3 + 2 × (π/6) ) ]
= 2 [ π − √3/2 − π/3 ]
= 2 [ 2π/3 − √3/2 ] = 4π/3 − √3 square units (about 2.457).
Notice how the sketch did the hard work: it told us which piece was “smaller” and that we could halve the labour using symmetry.
Area Under A Parabola
Two standard forms turn up constantly, and it pays to recognise them instantly.
- y² = 4ax — opens to the right, vertex at the origin, symmetric about the x-axis. Its focus is at (a, 0) and its latus rectum is the vertical line x = a.
- y = ax² — opens upwards, vertex at the origin, symmetric about the y-axis.
For y² = 4ax you must decide which branch you are on. Solving gives y = ±2√(ax), so y = 2√(ax) is the upper half and y = −2√(ax) the lower half. Because the curve is symmetric about the x-axis, the neat move is nearly always: find the area of the upper half and double it.
Area = 2 × ∫0a 2√(ax) dx = 8a²/3 square units.
Derive it rather than memorise it — the derivation takes four lines and never lets you down.
Solution. Compare with y² = 4ax: 4a = 8, so a = 2 and the latus rectum is the line x = 2.
Upper branch: y = √(8x) = 2√2 × √x.
Upper half area = ∫02 √(8x) dx = 2√2 [ (2/3) x3/2 ]02 = 2√2 × (2/3) × 2√2 = (4/3) × 2 × 2 = 16/3.
By symmetry about the x-axis, total Area = 2 × 16/3 = 32/3 square units.
Cross-check with the general result 8a²/3 = 8(4)/3 = 32/3. It matches.
Solution. Upper branch y = 2√x. The region runs from the vertex at x = 0 to the line x = 4, and is symmetric about the x-axis.
Upper half = ∫04 2√x dx = 2 [ (2/3) x3/2 ]04 = (4/3)(8) = 32/3.
Total Area = 2 × 32/3 = 64/3 square units.
Solution. Upper half = ∫01 2√x dx = 2 [ (2/3) x3/2 ]01 = 4/3.
Total Area = 2 × 4/3 = 8/3 square units.
Compare with the previous example: widening the cut from x = 1 to x = 4 multiplied the area by 8. That is the x3/2 growth showing itself — a nice reminder that area does not grow in proportion to the width.
Solution. This is the y = ax² form with a = 1, so no branch worries at all — y is a genuine function of x.
Area = ∫02 x² dx = [ x³/3 ]02 = 8/3 square units.
A pleasing observation: the rectangle enclosing this region (width 2, height 4) has area 8, and our answer is exactly one third of it. That “one third” relationship holds for every region under y = ax² from the origin, and makes a lovely mental check.
Area Of An Ellipse And Its Quadrants
The ellipse in standard form is x²/a² + y²/b² = 1. Picture a circle that someone has gently stretched: it reaches out to ±a along the x-axis and ±b along the y-axis. When a = b it is a circle, which is a useful thing to remember when checking answers.
⇒ y = ± (b/a) √(a² − x²).
That factor b/a in front is the whole difference between an ellipse and a circle. Everything after it is the circle integral you already know.
Because of that, an ellipse question is really a circle question wearing a coat. Pull the constant b/a outside the integral, do the familiar root integral, and multiply back in at the end.
Whole ellipse = 4 × (quadrant) = πab square units.
Notice that setting b = a recovers πa² for the circle — a comforting consistency check.
Solution. Here a² = 16 and b² = 9, so a = 4 and b = 3. In the first quadrant y = (3/4)√(16 − x²), and x runs from 0 to 4.
Area = ∫04 (3/4)√(16 − x²) dx
= (3/4) [ (x/2)√(16 − x²) + 8 sin−1(x/4) ]04
= (3/4) [ ( 0 + 8 × (π/2) ) − 0 ] = (3/4)(4π) = 3π square units.
Check against the formula: πab/4 = π(4)(3)/4 = 3π. It agrees.
Solution. a = 3, b = 2. The upper half is y = (2/3)√(9 − x²), stretching from x = −3 to x = 3.
Upper half = ∫−33 (2/3)√(9 − x²) dx = (2/3) [ (x/2)√(9 − x²) + (9/2) sin−1(x/3) ]−33
= (2/3) [ (9/2)(π/2) − (9/2)(−π/2) ] = (2/3)(9π/2) = 3π.
Doubling for the lower half, Area = 6π square units.
Formula check: πab = π(3)(2) = 6π. Agreed.
Areas Bounded By A Line And A Standard Curve
This is the natural next step, and it is squarely inside the syllabus wording because a line is one of the named curves. The region is fenced by a straight line on one side and a circle, parabola or ellipse on the other.
The method has one extra step compared with everything so far, and then it is business as usual.
- Sketch both, roughly but honestly. You must be able to see which region is being asked for — a line usually cuts a curve into two pieces, and the question will say “smaller” or “larger”, or the shading will tell you.
- Find where they meet by solving the two equations simultaneously. These intersection points give you the limits of integration.
- Decide what the strip height is in the region you want. Sometimes the strip runs from the axis up to the line; sometimes from the axis up to the curve; sometimes you need two integrals because the top boundary changes partway across.
- Integrate and add.
Solution. The line meets the axes at (3, 0) and (0, 2), which are exactly the ends of the ellipse quadrant. So the line is a chord joining those two points, and the region we want is the thin sliver between the curved arc and the straight chord.
Ellipse quadrant: ∫03 (2/3)√(9 − x²) dx = (2/3)[ (x/2)√(9 − x²) + (9/2) sin−1(x/3) ]03 = (2/3)(9π/4) = 3π/2.
Triangle under the chord: the line is y = 2(1 − x/3), so ∫03 2(1 − x/3) dx = [ 2x − x²/3 ]03 = 6 − 3 = 3. (Or just: ½ × 3 × 2 = 3.)
Sliver = arc region − triangle = 3π/2 − 3 = (3π − 6)/2 square units (about 1.712).
The idea worth carrying away: when a line cuts a standard curve, you can very often get the answer as bigger region minus smaller region rather than by setting up something complicated.
Solution. Substituting x = 1 into the circle gives 1 + y² = 4, so y = ±√3 — the line cuts the circle at (1, √3) and (1, −√3). The region we want runs from x = 1 to the rim x = 2 and is symmetric about the x-axis.
Area = 2 × ∫12 √(4 − x²) dx = 2 [ (x/2)√(4 − x²) + 2 sin−1(x/2) ]12
= 2 [ π − (√3/2 + π/3) ] = 2 [ 2π/3 − √3/2 ] = 4π/3 − √3 square units (about 2.457).
Reassurance check: the whole circle has area 4π ≈ 12.57, and our sliver is about 2.46 — comfortably less than a quarter of the circle, which matches the picture.
Symmetry Shortcuts That Save You Real Time
The four curves in this syllabus are all beautifully symmetric, and using that symmetry can halve or quarter your work. But — and this matters — only use a shortcut once you can do the question the long way. A shortcut applied to the wrong region is worse than no shortcut at all.
Ellipse x²/a² + y²/b² = 1 has exactly the same symmetries. Find one quadrant, multiply by 4.
Parabola y² = 4ax is symmetric about the x-axis only. Find the upper half, multiply by 2.
Parabola y = ax² is symmetric about the y-axis only. If your region spans x = −k to x = k, find 0 to k and multiply by 2.
There is one condition you must check before using any of these, and it is easy to state: the region itself must share the symmetry, not just the curve. If a question asks for the part of a circle in the first quadrant only, multiplying by 4 is simply wrong — the curve is symmetric but the region is not.
Solution. The ellipse is symmetric about both axes, and the region we want (the whole interior) shares that symmetry. So the four quadrant-pieces are identical.
Quadrant = ∫04 (3/4)√(16 − x²) dx = 3π (worked out in the ellipse section).
Total = 4 × 3π = 12π square units.
Formula check: πab = π(4)(3) = 12π. Agreed.
Solution. The parabola is symmetric about the x-axis, yes — but the question explicitly restricts the region to above the x-axis. The region is not symmetric, only the curve is. The correct answer is 36 square units; the doubling is unjustified.
Read the region description, not just the curve. “In the first quadrant”, “above the x-axis” and “the smaller region” are all instructions that switch symmetry shortcuts off.
Standard Area Results At A Glance
Here is a summary of every standard result in this chapter, all of which we derived above. Use this table to check your answers, not to replace the integration — board questions that say “using integration” want to see the integral.
| Curve (standard form) | Region | Set-up | Area |
|---|---|---|---|
| Line y = mx + c | Under the line, x = a to b | ∫ab (mx + c) dx | trapezium value |
| Circle x² + y² = a² | One quadrant | ∫0a √(a² − x²) dx | πa²/4 |
| Circle x² + y² = a² | Upper semicircle | ∫−aa √(a² − x²) dx | πa²/2 |
| Circle x² + y² = a² | Whole circle | 4 × (quadrant) | πa² |
| Circle x² + y² = a² | To the right of x = k | 2 × ∫ka √(a² − x²) dx | segment value |
| Parabola y² = 4ax | Up to the latus rectum x = a | 2 × ∫0a 2√(ax) dx | 8a²/3 |
| Parabola y² = 4ax | Up to a line x = k | 2 × ∫0k 2√(ax) dx | (8/3)√(a) k3/2 |
| Parabola y = ax² | x = 0 to k, above the x-axis | ∫0k ax² dx | ak³/3 |
| Ellipse x²/a² + y²/b² = 1 | One quadrant | ∫0a (b/a)√(a² − x²) dx | πab/4 |
| Ellipse x²/a² + y²/b² = 1 | Whole ellipse | 4 × (quadrant) | πab |
| Any curve x = g(y) | Between the curve and the y-axis, y = c to d | ∫cd x dy | as evaluated |
Area Between Two Curves (Good To Know)
Read that carefully: it names area under simple curves. The phrase “area between two curves” does not appear in the 2026-27 curriculum text, and we are not going to pretend otherwise. So treat this section as an extension, not as guaranteed examinable content. It is genuinely useful — it deepens your understanding of the strip idea, it appears in many older papers and reference books, and it costs you very little once the rest of the chapter is solid. But before you spend a whole revision session on it, check the current syllabus circular and your school’s exam pattern with your teacher. If you are short of time, the sections above are the priority.
The idea is a small and rather elegant extension of everything you already know. If two curves overlap and you want the area of the region trapped between them, slice it into vertical strips again. Each strip now runs from the lower curve up to the upper curve, so its height is the difference of the two y-values.
Area = ∫ab [ f(x) − g(x) ] dx.
The limits a and b are the x-coordinates where the two curves intersect, found by solving the equations simultaneously.
Note one pleasant feature: because you are subtracting, you do not need to worry about whether the region is above or below the x-axis. As long as f is genuinely above g, the difference f − g is positive and the integral comes out positive automatically.
Solution. Intersections: x = x² ⇒ x(1 − x) = 0 ⇒ x = 0 and x = 1.
Between 0 and 1, which is on top? Test x = ½: the line gives 0.5 and the parabola gives 0.25. So the line is above.
Area = ∫01 (x − x²) dx = [ x²/2 − x³/3 ]01 = 1/2 − 1/3 = 1/6 square units.
Solution. Intersections: substitute y² = 4x into the circle, giving x² + 4x = 32, so x² + 4x − 32 = 0, i.e. (x + 8)(x − 4) = 0. Only x = 4 is admissible (x = −8 would make y² negative). Then y² = 16, so y = ±4. The curves meet at (4, 4) and (4, −4). The circle has radius 4√2 ≈ 5.657.
The region is symmetric about the x-axis, so work with the upper half and double.
Going left to right, the top boundary changes at x = 4: first it is the parabola, then it is the circle. So the upper half needs two integrals.
Part A (0 to 4, under the parabola): ∫04 2√x dx = 2[(2/3)x3/2]04 = (4/3)(8) = 32/3.
Part B (4 to 4√2, under the circle): ∫44√2 √(32 − x²) dx = [ (x/2)√(32 − x²) + 16 sin−1(x/(4√2)) ]44√2
= ( 0 + 16 × (π/2) ) − ( 2 × 4 + 16 × (π/4) ) = 8π − 8 − 4π = 4π − 8.
Upper half = 32/3 + 4π − 8 = 8/3 + 4π.
Total area = 2(8/3 + 4π) = 16/3 + 8π square units (about 30.47).
A reality check: the whole circle has area π(4√2)² = 32π ≈ 100.5, and our region is a bit under a third of it. The picture agrees.
How To Set Up Any Area Question In Five Steps
When you are under exam pressure and the question looks unfamiliar, do not try to remember which “type” it is. Run this checklist instead. It works on every question in this chapter, without exception.
- Sketch. Draw the axes, sketch each curve given, and mark the intercepts. Rough is fine; labelled is essential.
- Shade. Shade the exact region the question describes. Read the words “smaller”, “larger”, “first quadrant”, “above the x-axis” very carefully — they change the answer completely.
- Choose your strip. Vertical (dx) if the fences are vertical lines or the x-axis. Horizontal (dy) if the fences are horizontal lines or the y-axis. Pick whichever needs less rearranging.
- Find the limits. Either they are given, or you find them by setting y = 0, or by solving two equations simultaneously to get intersection points.
- Check for sign trouble, then integrate. Does the curve cross the axis inside your limits? If yes, split and take moduli. Then integrate, and finish with “square units”.
Step 1 — sketch. An upward parabola through the origin, and a horizontal line at height 4.
Step 2 — shade. The region sits to the left of the parabola, right of the y-axis, below y = 4. (Not the region under the parabola — read again.)
Step 3 — strip. The fences are the y-axis and the horizontal line y = 4. Horizontal strips, so dy.
Step 4 — limits. y runs from 0 to 4. Rearranging, x = √y (positive root, first quadrant).
Step 5 — integrate. Area = ∫04 √y dy = [ (2/3) y3/2 ]04 = (2/3)(8) = 16/3 square units.
Check. The enclosing rectangle (width 2, height 4) has area 8. The region under the parabola is 8/3 (we found this earlier), so the region beside it must be 8 − 8/3 = 16/3. The two agree.
Common Mistakes And How To Avoid Them
I have gathered the errors I see most often. Read this list once now, and again the night before your exam — it is short and it is worth real marks.
Practice Worksheet
Ten questions, arranged roughly from gentle to board-level. Please attempt each one on paper — with a rough sketch — before you open the answer. Reading a solution feels like learning; it is not. Writing one is.
Q1. Find the area bounded by the line y = 2x + 3, the x-axis, and the lines x = 0 and x = 4.
Area = ∫04 (2x + 3) dx = [ x² + 3x ]04 = (16 + 12) − 0 = 28 square units.
Check by geometry: a trapezium with parallel sides 3 and 11, distance 4 apart. ½ × (3 + 11) × 4 = 28. Agreed.
Q2. Find the area bounded by the curve y = x² + 1, the x-axis, and the lines x = 1 and x = 3.
Area = ∫13 (x² + 1) dx = [ x³/3 + x ]13 = (9 + 3) − (1/3 + 1) = 12 − 4/3 = 32/3 square units.
Q3. Find the area enclosed between the curve y = 4x − x² and the x-axis.
Area = ∫04 (4x − x²) dx = [ 2x² − x³/3 ]04 = 32 − 64/3 = 32/3 square units.
Q4. Find the area bounded by y = x² − 9, the x-axis, and the lines x = 0 and x = 3.
∫03 (x² − 9) dx = [ x³/3 − 9x ]03 = 9 − 27 = −18.
Taking the modulus, Area = 18 square units.
Q5. Find the area bounded by y = x² − 4, the x-axis, and the lines x = 0 and x = 4.
Piece 1 (0 to 2): [ x³/3 − 4x ]02 = 8/3 − 8 = −16/3. Modulus: 16/3.
Piece 2 (2 to 4): [ x³/3 − 4x ]24 = (64/3 − 16) − (8/3 − 8) = 16/3 − (−16/3) = 32/3.
Total = 16/3 + 32/3 = 48/3 = 16 square units.
Warning: integrating straight through gives 16/3, which is not the area.
Q6. Find the area of the region in the first quadrant enclosed by the circle x² + y² = 36 and the coordinate axes.
Area = ∫06 √(36 − x²) dx = [ (x/2)√(36 − x²) + 18 sin−1(x/6) ]06
= ( 0 + 18 × (π/2) ) − 0 = 9π square units.
Check: the whole circle is π(6)² = 36π, and a quarter is 9π. Agreed.
Q7. Find the area of the region bounded by the parabola y² = 12x and its latus rectum.
Upper branch y = √(12x) = 2√3 × √x.
Upper half = ∫03 √(12x) dx = 2√3 [ (2/3)x3/2 ]03 = 2√3 × (2/3) × 3√3 = (4/3) × 3 × 3 = 12.
By symmetry about the x-axis, Area = 2 × 12 = 24 square units.
Check with the general result: 8a²/3 = 8(9)/3 = 24. Agreed.
Q8. Using integration, find the area enclosed by the ellipse x²/25 + y²/16 = 1.
Upper half = ∫−55 (4/5)√(25 − x²) dx = (4/5)[ (x/2)√(25 − x²) + (25/2) sin−1(x/5) ]−55
= (4/5)[ (25/2)(π/2) − (25/2)(−π/2) ] = (4/5)(25π/2) = 10π.
Doubling, Area = 20π square units.
Check: πab = π(5)(4) = 20π. Agreed.
Q9. Find the area bounded by the curve x = y² + 1, the y-axis, and the lines y = 0 and y = 2.
Area = ∫02 (y² + 1) dy = [ y³/3 + y ]02 = 8/3 + 2 = 14/3 square units.
Q10. Find the area of the region in the first quadrant bounded by the circle x² + y² = 16 and the y-axis, between y = 0 and y = 4, by integrating with respect to y.
Area = ∫04 √(16 − y²) dy = [ (y/2)√(16 − y²) + 8 sin−1(y/4) ]04
= ( 0 + 8 × (π/2) ) − 0 = 4π square units.
Check: this is a quarter of the circle x² + y² = 16, whose total area is 16π. A quarter is 4π. Agreed — and notice you get the same answer whether you slice vertically or horizontally, which is a nice confirmation that both methods are really the same idea.
Kaizen closing thought. You do not need to be brilliant at this tomorrow. You need to be slightly better than you were today — aim for one more correct question than yesterday, every single day. Ten days of that is a hundred questions of progress, and it compounds quietly until one morning the whole chapter simply feels easy. Be patient with yourself, keep the sketches honest, and keep going.