IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~11 min read

Area Between Curve & y-axis

Rotate the perspective. Instead of vertical strips dx giving area ∫y dx (curve to x-axis), this sub-chapter integrates horizontal strips dy giving area ∫|x| dy (curve to y-axis). Two new wrinkles: you must rearrange y = f(x) into x = g(y) (the inverse step), and if the curve crosses the y-axis inside your y-range, the area splits — the part with negative x needs absolute-value treatment so it doesn’t cancel the positive part.

📘 What you need to know

Formula booklet: A = ∫_a^b |x| dy, where y runs from a to b. The modulus is essential — it stops areas on the LEFT of the y-axis from cancelling areas on the RIGHT.
Rearrange y = f(x) → x = g(y) before integrating. This is the inverse-function step: solve for x in terms of y. Square roots, exponentials, and inverse trig all appear naturally here.
Y-axis crossings split the integral: find y-values where x = 0 (set g(y) = 0). If any crossing lies in [a, b], split there and handle each piece separately.
Modulus mechanics: on a sub-interval where x > 0, integrate x directly. Where x < 0, integrate −x (i.e., flip the sign). Sum positive contributions only.
Limits a and b may not be given directly — they can come from the x-axis (y = 0) or from a curve root, just like with x-axis area problems. Use a GDC to sketch and find them.
Inverse functions are the rearrangement: y = e^x ⟹ x = ln y; y = arcsin x ⟹ x = sin y. The rearrangement is just the inverse function applied to y.
Don’t confuse with x-axis area: ∫y dx integrates y with respect to x (curve-to-x-axis); ∫x dy integrates x with respect to y (curve-to-y-axis). The differential tells you which is which.
Sketch first, integrate second. A quick GDC sketch identifies the bounded region, locates y-axis crossings, and confirms whether the area splits. Skipping the sketch is the #1 cause of errors here.

The formula and the split-area visual

Area between curve and y-axis — formula booklet A = ∫_a^b |x| dy where x = g(y), y ∈ [a, b]

When the curve doesn’t cross the y-axis in [a, b], the modulus is decorative — x doesn’t change sign. But when it does cross, the area splits into two pieces and the modulus becomes essential.

The curve x = 9 − y² (rearranged from y = √(9 − x)) is plotted against the y-axis. Between y = 0 and y = 3, x > 0 — the area lives to the RIGHT of the y-axis (green, contributes 18). Between y = 3 and y = 4, x < 0 — the area lives to the LEFT (orange, contributes 10/3). Dashed horizontal lines show the dy strips. The modulus |x| converts the orange strip widths into positive contributions, giving total 64/3.

Why rearrangement is the inverse-function step: solving y = e^x for x gives x = ln y — the inverse of e^x. Solving y = arcsin x gives x = sin y — the inverse of arcsin. So every “rearrange to x = g(y)” step is mechanically equivalent to writing down the inverse function, evaluated at y.

x-axis area vs y-axis area

Area between curve and x-axis

A = ∫_a^b |y| dx

Vertical strips of height |y| and thickness dx. Limits are x-values. Modulus splits the integral where the curve crosses the x-axis.

Area between curve and y-axis

A = ∫_a^b |x| dy

Horizontal strips of width |x| and thickness dy. Limits are y-values. Must rearrange to x = g(y) first. Modulus splits at y-axis crossings.

🧭 Recipe — area between curve and y-axis

Identify the limits a, b (in y-values) and sketch the curve on a GDC. The limits may be given directly or come from the x-axis (y = 0) or curve features. Confirm which bounded region the question is asking about.
Rearrange y = f(x) into x = g(y). This is the inverse-function step: solve for x in terms of y. Square both sides for roots, take ln for exponentials, etc.
Check for y-axis crossings in [a, b]: solve g(y) = 0. If a crossing y = c lies in the interval, you must split the integral at c.
Evaluate the integral with modulus handled: on each sub-interval, determine the sign of x. Where x > 0, integrate g(y) directly. Where x < 0, integrate −g(y) (negate to make it positive).
Sum the positive contributions from each sub-interval for the total area. For exact answers, keep fractions, π, e, etc. — don’t approximate unless asked.

Worked examples

WE 1

Basic — parabola in first quadrant

Find the area bounded by y = x², the y-axis, and the lines y = 0 and y = 4 (with x ≥ 0).

Step 1 — limits in y y from 0 to 4 (given directly) Step 2 — rearrange to x = g(y) y = x² ⟹ x = √y (positive root since x ≥ 0) Step 3 — y-axis crossings? x = √y = 0 only at y = 0 — boundary point, no internal crossing x ≥ 0 throughout [0, 4] — no split needed Step 4 — integrate A = ∫ from 0 to 4 of √y dy = ∫ y^(1/2) dy = [(2/3) y^(3/2)] from 0 to 4 = (2/3)(4)^(3/2) – 0 = (2/3)(8) = 16/3 A = 163 square units classic warm-up. Note that integrating y dx between x = 0 and x = 2 would give a different region (the area UNDER y = x², not between y = x² and the y-axis). The shapes complement each other to form a 2 × 4 rectangle: 16/3 + 8/3 = 24/3 = 8 ✓.

WE 2

Logarithmic curve — rearranging y = ln x to x = e^y

Find the area bounded by y = ln x, the y-axis, and the lines y = 1 and y = 2.

Step 1 — limits in y y from 1 to 2 (given directly) Step 2 — rearrange (inverse of ln) y = ln x ⟹ x = e^y Step 3 — y-axis crossings? x = e^y > 0 for all real y — no crossings, no split Step 4 — integrate A = ∫ from 1 to 2 of e^y dy = [e^y] from 1 to 2 = e² – e A = e² − e ≈ 4.67 square units the rearrangement step is just “apply the inverse function”. Same idea would work for y = arctan x → x = tan y, etc. Note that e^y > 0 always — that’s why no modulus and no split is needed.

WE 3

Split across y-axis — sideways parabola

Find the area bounded by y = √(9 − x), the y-axis, and the lines y = 0 and y = 4.

Step 1 — limits in y y from 0 to 4 (given) Step 2 — rearrange (square both sides) y = √(9-x) ⟹ y² = 9-x ⟹ x = 9 – y² Step 3 — y-axis crossings? (set x = 0) 9 – y² = 0 ⟹ y = 3 (within [0, 4] — must split!) For y ∈ [0, 3]: x = 9 – y² ≥ 0 (right of y-axis) For y ∈ [3, 4]: x = 9 – y² ≤ 0 (left of y-axis) Step 4 — split the integral A = ∫ from 0 to 3 of (9 – y²) dy + ∫ from 3 to 4 of -(9 – y²) dy = ∫ from 0 to 3 of (9 – y²) dy + ∫ from 3 to 4 of (y² – 9) dy Step 5 — evaluate each piece Right: [9y – y³/3] from 0 to 3 = (27 – 9) – 0 = 18 Left: [y³/3 – 9y] from 3 to 4 = (64/3 – 36) – (9 – 27) = 64/3 – 36 + 18 = 64/3 – 18 = 10/3 Total: 18 + 10/3 = 54/3 + 10/3 = 64/3 A = 643 square units this is exactly the SVG above. Notice how the LEFT integral ∫(y² – 9) dy comes out positive (10/3), not negative — that’s the modulus at work. If you integrated the same expression WITHOUT flipping the sign on the left part, you’d get 18 – 10/3 = 44/3 ≈ 14.67, the “signed” answer, which UNDERSTATES the actual area.

WE 4

Inverse function with integration by parts — clean answer

Find the area bounded by y = e^x, the y-axis, and the lines y = 1 and y = e.

Step 1 — limits in y y from 1 to e (given) Step 2 — rearrange y = e^x ⟹ x = ln y Step 3 — y-axis crossings? (set x = 0) ln y = 0 ⟹ y = 1 (it’s the lower limit — boundary, not internal) For y ∈ [1, e]: ln y ≥ 0 — no split Step 4 — integrate (parts, since ∫ ln y dy needs parts) ∫ ln y dy = y ln y – y + c (from parts: u = ln y, dv = dy) A = ∫ from 1 to e of ln y dy = [y ln y – y] from 1 to e Step 5 — evaluate At y = e: e · 1 – e = 0 At y = 1: 1 · 0 – 1 = -1 A = 0 – (-1) = 1 A = 1 square unit (exactly) the integral ∫ ln y dy uses integration by parts (the “1·ln y” single-function trick from the previous note). The clean integer answer 1 comes from the limit y = e — that’s the trademark sign that e was chosen deliberately.

WE 5

Inverse trig — rearrange y = arcsin x

Find the area bounded by y = arcsin x, the y-axis, and the lines y = 0 and y = π/2.

Step 1 — limits in y y from 0 to π/2 (given) Step 2 — rearrange y = arcsin x ⟹ x = sin y Step 3 — y-axis crossings? (set x = 0) sin y = 0 at y = 0, π, 2π, … In [0, π/2], only the boundary y = 0 — no internal crossing For y ∈ (0, π/2]: sin y > 0 — no split Step 4 — integrate A = ∫ from 0 to π/2 of sin y dy = [-cos y] from 0 to π/2 = -cos(π/2) – (-cos(0)) = -0 + 1 = 1 A = 1 square unit (exactly) another clean integer answer. The “rearrange to inverse” step is mechanical: arcsin’s inverse is sin, so x = sin y immediately. Don’t get confused by which function is “outer” — once you’ve identified y = (something) · x, just apply the inverse function on both sides.

WE 6

Split area — full SME archetype with different numbers

Find the area bounded by y = 3 − √(x + 1), the y-axis, and the lines y = 0 and y = 3.

Step 1 — limits in y y from 0 to 3 (given) Step 2 — rearrange (isolate the square root, then square) y = 3 – √(x+1) y – 3 = -√(x+1) 3 – y = √(x+1) (3 – y)² = x + 1 x = (3 – y)² – 1 Step 3 — y-axis crossings? (set x = 0) (3 – y)² – 1 = 0 ⟹ (3 – y)² = 1 ⟹ 3 – y = ±1 y = 2 or y = 4 y = 2 is in [0, 3] — MUST split at y = 2 Step 4 — sign analysis For y ∈ [0, 2]: (3-y) > 1, so (3-y)² > 1, so x > 0 (right of y-axis) For y ∈ [2, 3]: (3-y) < 1, so (3-y)² < 1, so x < 0 (left of y-axis) Step 5 — split and integrate (use x = y² – 6y + 8 expanded form) Right: ∫ from 0 to 2 of (y² – 6y + 8) dy = [y³/3 – 3y² + 8y] from 0 to 2 = (8/3 – 12 + 16) – 0 = 8/3 + 4 = 20/3 Left: ∫ from 2 to 3 of -(y² – 6y + 8) dy = ∫ (-y² + 6y – 8) dy = [-y³/3 + 3y² – 8y] from 2 to 3 = (-9 + 27 – 24) – (-8/3 + 12 – 16) = -6 – (-8/3 – 4) = -6 + 8/3 + 4 = -2 + 8/3 = 2/3 Total: 20/3 + 2/3 = 22/3 A = 223 square units two sub-intervals; left contribution (2/3) is much smaller than right (20/3) because the curve barely dips into the left half-plane (only for y between 2 and 3 with x between -1 and 0). Always check the sign of x on EACH sub-interval — sign-tracking errors are the #1 cause of wrong answers on split-area problems.

💡 Top tips

Sketch first on your GDC. Identify the bounded region, the limits in y, and any crossings of the y-axis. Trying to work blind on this topic invites errors.
Always rearrange first: y = f(x) → x = g(y). Common patterns: square roots become squares, exp becomes ln, trig becomes inverse trig.
Find y-axis crossings by setting g(y) = 0. If any crossing y = c lies inside [a, b], you MUST split the integral at c.
On sub-intervals where x < 0, flip the sign of the integrand to make the contribution positive. The modulus is non-negotiable — without it, areas cancel.
GDC integrates the modulus directly: if your calculator supports it, ∫|g(y)| dy over the full interval gives the right answer in one step — no need to manually split.

⚠ Common mistakes

Forgetting to rearrange: writing A = ∫y dy (nonsensical) instead of ∫x dy after substituting x = g(y).
Ignoring negative x: integrating g(y) directly without the modulus gives a SIGNED area where negative-x regions cancel positive-x ones. SME’s worked example would give 23/3 instead of 37/3 if you skip the modulus.
Missing y-axis crossings: not checking g(y) = 0 and assuming a single integral handles the whole region. Always check for crossings inside the integration interval.
Confusing this with x-axis area: integrating ∫y dx (curve-to-x-axis) when the question asks for ∫x dy (curve-to-y-axis). The differential and the limits tell you which is which.
Wrong-branch rearrangement: for y² = x, both x = y² (correct here, since y² is single-valued in y) and the other branches are issues. For y = √(x), the inverse is only the positive branch of x = y² (with y ≥ 0).

Up next: Volumes of Revolution. Rotate a region around an axis and you get a 3D solid; integrating π·(radius)² gives its volume. Two flavours: rotation around the x-axis gives V = π∫y² dx; rotation around the y-axis gives V = π∫x² dy. The y-axis version uses exactly the rearrangement skill from this note — solve y = f(x) for x = g(y), then integrate g(y)² with respect to y. Squaring is often pedagogically helpful: a function with √ inside simplifies dramatically when you square it.

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