IB Maths AA HL Topic 5 — Calculus Paper 2 ~8 min read

Finding Areas Using a GDC

A definite integral ∫ₐᵇ f(x) dx computes the area between the curve y = f(x) and the x-axis from x = a to x = b. The “+ c” cancels in the subtraction F(b) − F(a). On Paper 2 your GDC can evaluate any definite integral in seconds — but you still need to set up the right integral first by identifying the correct limits.

📘 What you need to know

Definite integral: ∫ₐᵇ f(x) dx = F(b) − F(a), where F is any antiderivative of f.
“+ c” cancels: F(b) − F(a) eliminates the constant, so don’t bother writing it in definite integrals.
Area under a curve: when f(x) ≥ 0 on [a, b], the integral equals the area between the curve and the x-axis.
Limits a and b correspond to vertical boundary lines x = a and x = b. Sometimes a or b is a root of f (where the curve meets the x-axis).
Find limits from a sketch: vertical boundary lines give the limits directly; intersections with the x-axis need solving f(x) = 0.
GDC workflow: enter the integrand, the lower limit, and the upper limit; the GDC returns the value (often as a decimal that you then convert to an exact fraction).
For exact answers: the GDC may give 6.5333… — your job is to recognise this as 98/15, or work it out by hand.
Beware of negative areas: if the curve dips below the x-axis between limits, the integral subtracts that region. This is fine — but if the question asks for “area enclosed”, you may need to integrate in pieces and take absolute values (covered in the next chapter).

What is a definite integral?

Fundamental Theorem of Calculus ∫_a^b f(x) dx = F(b) − F(a)

The integration limits a and b replace “+ c“. You evaluate the antiderivative at x = b, then subtract its value at x = a. The result is a number — the area under the curve when f(x) ≥ 0 on [a, b].

The shaded region is bounded by the curve, the x-axis, and the vertical lines x = a (lower limit) and x = b (upper limit). The definite integral evaluates this area when the curve stays above the x-axis.

Setting up the integral — finding the limits

What the boundary is	How to find the limit
Vertical line x = a (given)	limit is just a — read it from the question/diagram
The y-axis	limit is x = 0
A root of f (where curve meets x-axis)	solve f(x) = 0; that root value is the limit
Two curves intersect	solve f(x) = g(x) for x — limits between the two intersections (covered in next chapter)

Using your GDC

GDC — without graphing

∫_□^□□ dx

enter integrand, lower limit, upper limit — GDC returns the value

GDC — with graphing

graph y = f(x) → “area”

use the “area under curve” feature; specify limits a, b

Decimal → exact fraction: if your GDC shows 6.533333…, that’s 98/15. For repeating decimals, try multiplying by a small denominator (2, 3, 4, 5, …) until you get a whole or near-whole number. Many GDCs have a “fraction” or “exact” mode that converts directly.

🧭 Recipe — find an area using a definite integral

Sketch the situation if there isn’t already a diagram — identify the curve, the x-axis, and the vertical boundaries.
Find the limits a and b: from given vertical lines, or by solving f(x) = 0 for roots.
Check the curve is above the x-axis on [a, b] — test a value in between. If yes, the integral equals the area directly.
Write the integral: Area = ∫ₐᵇ f(x) dx.
Evaluate by hand (for Paper 1) or with GDC (for Paper 2) — convert any decimal to exact form if required.

Worked examples

WE 1

Basic definite integral — by hand

Evaluate ∫₀² (3x² + 4x) dx.

Find the antiderivative F(x) = ∫(3x² + 4x) dx = x³ + 2x² (no need for + c — it cancels) Apply F(b) − F(a) F(2) = 8 + 8 = 16 F(0) = 0 + 0 = 0 ∫₀² (3x² + 4x) dx = 16 − 0 = 16 ∫₀² (3x² + 4x) dx = 16 GDC check: enter integrand, lower 0, upper 2 → 16

WE 2

Area under a curve — given limits

Find the area bounded by the curve y = 6 − x², the x-axis, and the vertical lines x = −2 and x = 2.

Check the curve is above the x-axis on [−2, 2] y(−2) = 6 − 4 = 2 > 0; y(0) = 6 > 0; y(2) = 2 > 0 ✓ Set up the integral A = ∫₋₂² (6 − x²) dx Find the antiderivative and evaluate F(x) = 6x − x³/3 F(2) = 12 − 8/3 = 36/3 − 8/3 = 28/3 F(−2) = −12 + 8/3 = −36/3 + 8/3 = −28/3 A = 28/3 − (−28/3) = 56/3 Area = 56/3 square units curve is symmetric about y-axis; the two negatives in F(−2) became positive on subtraction

WE 3

Find limits from roots — area enclosed

Find the exact area enclosed between the curve y = 3x − x² and the x-axis.

Find where the curve crosses the x-axis 3x − x² = 0 → x(3 − x) = 0 → x = 0 or x = 3 Check the curve is above the x-axis on (0, 3) at x = 1.5: y = 4.5 − 2.25 = 2.25 > 0 ✓ Set up the integral with limits 0 and 3 A = ∫₀³ (3x − x²) dx Find the antiderivative and evaluate F(x) = (3/2)x² − x³/3 F(3) = 27/2 − 9 = 27/2 − 18/2 = 9/2 F(0) = 0 A = 9/2 − 0 = 9/2 Area = 9/2 square units “area enclosed by the curve and x-axis” means the region trapped between them — the roots automatically become the limits

WE 4

Definite integral with a root function

Evaluate ∫₁⁴ (√x + 3x) dx.

Rewrite and integrate ∫(x^(1/2) + 3x) dx = (2/3) x^(3/2) + (3/2) x² Evaluate at the upper limit x = 4 F(4) = (2/3)(8) + (3/2)(16) = 16/3 + 24 = 16/3 + 72/3 = 88/3 Evaluate at the lower limit x = 1 F(1) = (2/3)(1) + (3/2)(1) = 2/3 + 3/2 = 4/6 + 9/6 = 13/6 Subtract ∫₁⁴ = 88/3 − 13/6 = 176/6 − 13/6 = 163/6 ∫₁⁴ (√x + 3x) dx = 163/6 4^(3/2) = (√4)³ = 2³ = 8 — knowing power-of-4 values pays off here

WE 5

Irrational limits — exact surd answer

Find the exact area enclosed between the curve y = 8 − x² and the x-axis.

Find x-intercepts 8 − x² = 0 → x² = 8 → x = ±√8 = ±2√2 Curve is above x-axis on (−2√2, 2√2): at x = 0, y = 8 > 0 ✓ Set up integral with limits ±2√2 A = ∫₋₂√₂^(2√2) (8 − x²) dx Antiderivative F(x) = 8x − x³/3 Evaluate at x = 2√2 F(2√2) = 16√2 − (2√2)³/3 (2√2)³ = 2³·(√2)³ = 8·2√2 = 16√2 F(2√2) = 16√2 − 16√2/3 = 48√2/3 − 16√2/3 = 32√2/3 By symmetry F(−2√2) = −32√2/3 A = 32√2/3 − (−32√2/3) = 64√2/3 Area = 64√2/3 square units when limits are irrational, EXACT calculation by hand is often cleaner than a GDC, which gives 30.17… — you’d have to recognise this as 64√2/3

WE 6

GDC workflow — give an exact p/q answer from a decimal

Use your GDC to evaluate ∫₀² (4x⁵ − 3x² + 7) dx. Give your answer in the form p/q where p and q are integers.

Enter the integrand, lower limit 0, upper limit 2 into GDC GDC display: 48.6666666… Convert decimal to fraction 0.666… = 2/3, so 48.666… = 48 + 2/3 = 146/3 Verify by hand ∫(4x⁵ − 3x² + 7) dx = (2/3)x⁶ − x³ + 7x F(2) = (2/3)(64) − 8 + 14 = 128/3 + 6 = 128/3 + 18/3 = 146/3 F(0) = 0 → ∫₀² = 146/3 ✓ ∫₀² (4x⁵ − 3x² + 7) dx = 146/3 GDCs return decimals — your job is to spot the exact fraction. 0.666… → 2/3 is the most common conversion to memorise

💡 Top tips

Sketch first if no diagram is given — even a quick freehand sketch confirms the limits and helps avoid sign errors.
Use the GDC’s exact/fraction mode if available — it converts decimals like 6.5333… to 98/15 automatically.
Common decimal-to-fraction conversions: 0.5 = 1/2, 0.333… = 1/3, 0.25 = 1/4, 0.666… = 2/3, 0.2 = 1/5, 0.1666… = 1/6.
“+ c” is unnecessary in definite integrals — it cancels in F(b) − F(a).
Check curve sign on the interval — if the curve dips below the x-axis, the integral gives a NEGATIVE value (which equals minus the area there). For “area enclosed” you may need pieces.

⚠ Common mistakes

Including “+ c” in a definite integral — it’s wrong notation. Definite integrals give a NUMBER, not an expression with c.
Mixing up the limits — F(b) − F(a), upper minus lower. Reversing them flips the sign.
Sign error with F(−a) — for an odd-power term like x³ at x = −2, you get −8, not 8. Watch parentheses on negative x-values.
Assuming positive area when curve dips below x-axis — for full enclosed area you may need to split into intervals and take absolute values.
Accepting a decimal as the final answer when an exact value is asked for — always convert to exact fraction or surd if the question says “exact”.

That closes the Integration chapter! You can now: integrate powers of x, find the constant of integration using initial conditions, and evaluate definite integrals to find areas under curves. Up next comes Techniques & Applications of Integration — integrating trig, exponential, and reciprocal (1/x) functions, the reverse chain rule, substitution, areas between curves, and signed vs unsigned areas. All built on the foundation you’ve laid here.

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