IB Maths AA SLTopic 5 — CalculusPaper 1 & 2~8 min read
Area Between a Curve and a Line
When the boundary of a region is part curve, part straight line, you split the calculation into two pieces. The curve part needs an integral. The line part is just a triangle or trapezium — basic geometry, no integration needed. Then you either add the two areas, or subtract one from the other, depending on how the region sits.
📘 What you need to know
The region’s boundary is made of two pieces: a piece of curve and a piece of straight line.
The line part can be computed as a triangle (½bh) or trapezium (½h(a+b)) — no integration needed.
You must decide whether the answer is the sum or the difference of those two areas — a sketch tells you which.
Always find the intersection of the curve and the line first — it gives you the limits to integrate between.
On Paper 2, you can shortcut the entire problem: integrate ∫ab|f(x) − g(x)| dx on a GDC.
Two cases: sum or difference
The shape of the region tells you whether to add or subtract.
Sum case vs difference case
SUM case
Total = ∫(curve) + ½bh
curve and line meet at top — neither is “inside” the other
DIFFERENCE case
R = ∫(curve) − ½bh
line cuts across — region is between them
If you can’t tell whether to add or subtract, just sketch it. The picture always reveals which pieces belong to your region.
The line part — three options
For the area under a straight line, you don’t need to integrate. Pick the easiest geometric shape:
Three ways to get the line area
▲ triangle
A = ½ · b · h
▱ trapezium
A = ½ · h · (a + b)
∫ integral
A = ∫ yline dx
📍
Pick whatever’s fastest
Triangle if the line meets the x-axis. Trapezium if the line is between two non-zero y-values. Integral if the limits are awkward — but on a calculator paper, the GDC handles it just as fast.
Step-by-step method
Curve + line area recipe
Sketch first. If no diagram is given, plot both on your GDC.
Find the key x-values — roots of the curve, root of the line, and where curve meets line. These give your limits.
Decide: sum or difference? Look at how the region sits — is the line cutting through, or sitting alongside?
Compute each piece — curve area with an integral, line area with a triangle or trapezium.
Combine — add them (sum case) or subtract (difference case).
🧠
“Sketch · split · solve · combine”
Four short words. The sketch sets you up; the split separates curve from line; the solve handles each piece in its easiest form; the combine gives the final answer.
The Paper 2 shortcut
On a calculator paper, you don’t need to split anything. Just integrate the difference of the two functions between their intersection points:
Direct GDC method (Paper 2)
A = ∫ab |f(x) − g(x)| dx
a, b = where curve and line meet
The modulus handles the order automatically — even if you can’t tell which is “upper” and which is “lower”, the GDC gives the correct positive area.
Worked examples
WE 1
Sum case — curve plus triangle
The region R is bounded by the curve y = x² (from x = 0 to x = 2), the line connecting (2, 4) to (4, 0), and the x-axis. Find the area of R.
step 1 — area under curveI = ∫(0 to 2) x² dx = [x³/3] from 0 to 2= 8/3step 2 — area under line (triangle)vertices (2, 0), (4, 0), (2, 4)A = ½ × 2 × 4 = 4step 3 — sumR = 8/3 + 4 = 8/3 + 12/3 = 20/3Area of R = 20/3 square unitscurve and line don’t overlap — they meet at one point and bound separate chunks!
WE 2
Difference case — curve minus triangle
The region R is bounded by the curve y = −x² + 6x − 5 and the line y = x − 1. R lies entirely in the first quadrant. Find the area of R.
step 1 — find intersections−x² + 6x − 5 = x − 1x² − 5x + 4 = 0 → (x−1)(x−4) = 0x = 1 (both = 0) and x = 4 (both = 3)step 2 — area under curve from 1 to 4∫(1 to 4) (−x² + 6x − 5) dx= [−x³/3 + 3x² − 5x] from 1 to 4at 4: −64/3 + 48 − 20 = 20/3at 1: −1/3 + 3 − 5 = −7/3= 20/3 − (−7/3) = 27/3 = 9step 3 — area under line (triangle)line from (1, 0) to (4, 3): triangleA = ½ × 3 × 3 = 9/2step 4 — subtractR = 9 − 9/2 = 9/2Area of R = 9/2 square unitsline cuts under the curve — region is the lens-shape, so subtract!
WE 3
Curve and line meet at two points
Find the area enclosed by the curve y = √x and the line y = x/2.
step 1 — find intersections√x = x/2 → square: x = x²/4x² − 4x = 0 → x(x − 4) = 0x = 0 and x = 4step 2 — which is on top?at x = 1: √1 = 1, line = 0.5curve is above line on (0, 4)step 3 — integrate (upper − lower)A = ∫(0 to 4) (√x − x/2) dx= [⅔ x^(3/2) − x²/4] from 0 to 4= (⅔ × 8 − 4) − 0 = 16/3 − 4 = 4/3Area = 4/3 square unitswhen both ends are intersections, just integrate (upper − lower) directly — no triangle needed!
💡 Top tips
Sketch before anything. Your GDC plots both functions in two clicks — use it.
Find all the key x-values: roots of the curve, root of the line, and intersections of curve and line. These set your limits.
Pick the easiest line shape — usually a triangle (line meets x-axis) or a trapezium.
For two intersections, the cleanest method is ∫(upper − lower) dx between them.
Paper 2: GDC modulus shortcut. ∫|f − g| dx between intersections gives the answer directly.
Mark up the diagram with intercepts, intersections, and a clear shading of the region.
⚠ Common mistakes
Adding when you should subtract (or vice versa). A sketch prevents this every time.
Wrong limits — using a y-value, or forgetting to find the curve-line intersection.
Computing area under the wrong line — make sure your triangle vertices are correct, especially when the line doesn’t go through the origin.
Squaring sloppily when solving √x = … — always check for extraneous roots after squaring.
Forgetting to subtract the “lower” curve in the (upper − lower) method.
Not using the GDC on Paper 2 — manual integration is fine, but the GDC is faster and a check.
Next up: Area Between 2 Curves — same idea but with two curves instead of curve-and-line. The line was easy because triangles and trapeziums have known formulas; with two curves you’ll need an integral on both pieces.
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