IB Maths AA SL Topic 5 — Calculus Paper 1 & 2 ~8 min read

Area Between 2 Curves

When two curves enclose a region, the area is one clean integral: ∫(upper − lower) dx between their intersection points. The arithmetic is easy. The trap is identifying which curve is on top — and watching out for cases where they swap over and you need to split into multiple integrals.

📘 What you need to know

Area = ∫_a^b (y_upper − y_lower) dx, where a and b are intersection points.
Find intersections by setting f(x) = g(x) and solving.
If the curves cross more than twice, there are multiple regions — split the integral and watch for the upper/lower swap.
You don’t have to worry about the x-axis at all — (upper − lower) is always positive when set up correctly.
On Paper 2, your GDC handles all this with ∫|f(x) − g(x)| dx — modulus protects you from sign errors.

The big idea

Imagine the region between two curves as a stack of vertical strips. Each strip has a height equal to the top y-value minus the bottom y-value at that x. Adding up all those heights with an integral gives the area.

Area between two curves

A = ∫_a^b (y_upper − y_lower) dx

Stack of strips between two curves

Notice there’s no mention of the x-axis in this formula. That’s because (upper − lower) handles everything — even if both curves are below the axis, their difference is still the height of the strip.

Why you don’t need to worry about the x-axis

📍

Negatives cancel themselves out

If both curves are below the x-axis, both their y-values are negative. But (negative_upper) − (negative_lower) is still positive as long as you’ve correctly identified the upper curve. The formula self-corrects — you just need to get the order right.

When the curves cross more than twice

If the curves intersect at three or more points, the upper and lower curves switch between regions. You can’t just write one integral — you need one for each region, with the right curve on top each time.

Watch out — upper and lower may swap!

Region 1

∫ (f − g) dx

f is on top

Region 2

∫ (g − f) dx

g is on top — order flips!

Test a point in each region to know which curve is upper.

🧠

“Test a point in each region”

Pick any x-value inside each region and plug it into both curves. The bigger output is the upper curve for that region only. Don’t assume the same one stays on top everywhere.

Step-by-step method

Two-curves area recipe

Sketch both curves on the same axes (use your GDC).
Find all intersection points by solving f(x) = g(x). These are your limits.
Identify upper vs lower in each region by testing a point inside.
Write the integral(s) in the form ∫(upper − lower) dx for each region.
Integrate and add all the regions together.

The Paper 2 shortcut

If you have a calculator, you can skip the upper/lower analysis entirely. Just compute:

GDC modulus method

A = ∫_a^b |f(x) − g(x)| dx

The modulus flips any negative differences automatically, so it works whether f or g is on top — and it handles multi-region cases without any splitting.

Even if working is required, use the GDC at the end as a sanity check. If your manual answer doesn’t match the GDC’s number, something went wrong — usually a sign error or a missed region.

Worked examples

WE 1

Two parabolas — single region

Find the area enclosed by the curves y = −x² + 4 and y = x² − 4.

step 1 — find intersections −x² + 4 = x² − 4 → 2x² = 8 x² = 4 → x = ±2step 2 — which is upper? at x = 0: −x² + 4 = 4, x² − 4 = −4 −x² + 4 is upper everywhere on (−2, 2)step 3 — integrate A = ∫(−2 to 2) [(−x² + 4) − (x² − 4)] dx = ∫(−2 to 2) (8 − 2x²) dx = [8x − 2x³/3] from −2 to 2 = (16 − 16/3) − (−16 + 16/3) = 32/3 − (−32/3) = 64/3Area = 64/3 square units no need to worry about the x-axis — (upper − lower) handled everything!

WE 2

Two regions — upper and lower swap

Find the total area enclosed between the curves y = x³ and y = 2x.

step 1 — find intersections x³ = 2x → x³ − 2x = 0 x(x² − 2) = 0 → x = 0, ±√2 three intersections → two regions!step 2 — test each region on (−√2, 0): test x = −1 → x³ = −1, 2x = −2 x³ is upper here on (0, √2): test x = 1 → x³ = 1, 2x = 2 2x is upper here — they swapped!step 3 — region 1 integral A₁ = ∫(−√2 to 0) (x³ − 2x) dx = [x⁴/4 − x²] from −√2 to 0 = 0 − (1 − 2) = 1step 4 — region 2 integral A₂ = ∫(0 to √2) (2x − x³) dx = [x² − x⁴/4] from 0 to √2 = (2 − 1) − 0 = 1step 5 — sum Total area = A₁ + A₂ = 2 square units if you’d written one integral from −√2 to √2 with the same order, the two regions would cancel!

WE 3

Square root vs parabola

Find the area enclosed by the curves y = √x and y = x².

step 1 — find intersections √x = x² → square: x = x⁴ x⁴ − x = 0 → x(x³ − 1) = 0 x = 0 and x = 1step 2 — which is upper? at x = ½: √(½) ≈ 0.71, (½)² = 0.25 √x is upper on (0, 1)step 3 — integrate A = ∫(0 to 1) (√x − x²) dx = [⅔ x^(3/2) − x³/3] from 0 to 1 = ⅔ − ⅓ = 1/3Area = 1/3 square units always test a point — the answer to “which is upper” can flip mid-curve.

💡 Top tips

Sketch first, every time. Even a rough one tells you how many regions there are.
Set f(x) = g(x) to find limits — never use given x-values until you’ve checked they’re actually intersections.
Test a point in each region to identify the upper curve. Never guess.
Three or more intersections means multiple integrals. Don’t try to do it as one.
GDC modulus method on Paper 2 is bulletproof — no upper/lower analysis needed.
Squaring an equation (like √x = something) can introduce false roots — verify each one.

⚠ Common mistakes

Writing (lower − upper) instead of (upper − lower) — gives a negative number, then you forget it’s wrong.
Using one integral for a multi-region problem — the two halves can cancel and give a misleadingly small answer.
Assuming the same curve is always on top — at three or more intersections, the order changes.
Wrong intersections — careless algebra (or forgetting a negative root) leads to wrong limits.
Bringing in the x-axis when you don’t need to — area between curves doesn’t reference the x-axis at all.
Skipping the GDC check on Paper 2 — easy way to catch errors before submitting.

🎉 You’ve finished the entire Further Integration series! You can now integrate special functions, use reverse chain rule and substitution, evaluate definite integrals, handle negative integrals, and find areas between curves and lines or two curves. Calculus toolkit fully complete.

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