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Study GuideYear 12Level 3 · ChallengeHSCVCE

Free Year 12 Integration and Area Under Curves

An HSC/VCE-style study guide to definite integrals and area: the standard integrals, the fundamental theorem of calculus, areas below the x-axis, areas between two curves and the trapezoidal rule. Each worked example shows the antiderivative in square brackets, substitutes the limits and interprets the sign, with a six-question self-check and answers.

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Year
Year 12
Subject
Mathematics
Topic
Calculus
Difficulty
Level 3 · Challenge
Estimated time
30 minutes
Curriculum
Australian Curriculum
Answers
Not applicable
Format
PDF (A4) + print

Students will practise

  • finding antiderivatives of polynomial, linear-power, exponential, reciprocal and trigonometric functions
  • evaluating definite integrals using the fundamental theorem of calculus
  • finding the area between a curve and the x-axis, including regions below the axis
  • finding the area between two curves and approximating area with the trapezoidal rule

Curriculum: Australian Curriculum. We show specific outcome codes only where they have been verified against the official curriculum document.

What's next?

Completed: Integration and Area Under Curves

  1. 1HSC/VCE-style Calculus Practice with Worked Solutions
  2. 2Year 12 Differentiation (Power Rule) Worksheet — Level 3
  3. 3Year 12 Differentiation (Power Rule) Worksheet — Level 3 (Set 2)
  4. 4Year 12 Integration (Polynomials) Worksheet — Level 2
Study guide

How to use this study guide

  1. Read it together first, pausing at each worked example to try the step before reading the answer.
  2. Attempt the "Check yourself" questions at the end without looking back.
  3. Then practise with a worksheet from the pathway above and finish with the topic test.

Common questions

Who is this study guide for?

Year 12 students (typically ages 17–18) working on calculus. It is pitched at level 3 · challenge.

Are the answers included?

This is a study guide, so there is no separate answer sheet; the 'Check yourself' questions include answers.

How long does it take?

About 30 minutes. Short, regular sessions work best: two or three a week beats one long one.

Do I need to sign up to download?

No. Click Download Free PDF and it opens immediately. It is free for personal, classroom and homeschool use.

What should we do next?

Try HSC/VCE-style Calculus Practice with Worked Solutions.

Preview

Year 12 · Mathematics · Calculus

Integration and Area Under Curves

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What you need to know

Integration reverses differentiation. An indefinite integral ∫ f(x) dx is a family of functions F(x) + C whose derivative is f(x). A definite integral has limits: by the fundamental theorem of calculus, ∫ from a to b of f(x) dx = F(b) − F(a), written [F(x)] with the limits b and a. When f(x) ≥ 0 on the interval, this number is the area between the curve and the x-axis from x = a to x = b.

Function f(x)Antiderivative F(x)Note
xn (n ≠ −1) + Craise the power by 1, divide by the new power
(ax + b)n (n ≠ −1) + Calso divide by the coefficient of x
eaxeax + Cex is its own integral
ln |x| + Cthe n = −1 case
cos xsin x + Cx in radians
sin x−cos x + Cnote the sign

Signed area. Where the curve is below the x-axis the integral is negative. To find a physical area, integrate each part separately and add the absolute values. Area between two curves: find where they intersect, then integrate (top curve − bottom curve) between those x-values — this works even if part of the region is below the axis.

Evaluate ∫ from 0 to 2 of (3x2 + 2x) dx

  1. Antiderivative: [x3 + x2] from 0 to 2
  2. = (23 + 22) − (0 + 0) = 8 + 4 = 12

Find ∫ (2x + 1)4 dx and ∫ from 0 to 1 of e2x dx

  1. ∫ (2x + 1)4 dx = + C = [frac (2x + 1)[sup 5]/10] + C
  2. ∫ from 0 to 1 of e2x dx = [e2x] from 0 to 1 = e2 − e0 = [frac e[sup 2] − 1/2] ≈ 3.19

Find the area between y = x2 − 4 and the x-axis from x = 0 to x = 2

  1. On 0 ≤ x ≤ 2 the curve is below the axis (for example y = −4 at x = 0), so expect a negative integral.
  2. ∫ from 0 to 2 of (x2 − 4) dx = [ − 4x] from 0 to 2 = ( − 8) − 0 = −
  3. The integral is −, so the area is [frac 16/3] square units (about 5.33). State the area as a positive number.

Find the area enclosed between y = x + 2 and y = x2

  1. Intersections: x2 = x + 2, so x2 − x − 2 = 0, (x − 2)(x + 1) = 0, giving x = −1 and x = 2.
  2. Between these, the line is on top (test x = 0: line gives 2, parabola gives 0).
  3. Area = ∫ from −1 to 2 of (x + 2 − x2) dx = [ + 2x − ] from −1 to 2
  4. At x = 2: 2 + 4 − = . At x = −1: − 2 + = −.
  5. Area = − (−) = + = = [frac 9/2] square units

Trapezoidal rule: approximate ∫ from 0 to 2 of √(1 + x3) dx using 2 subintervals

  1. h = = 1. Function values: f(0) = √1 = 1, f(1) = √2 ≈ 1.4142, f(2) = √9 = 3.
  2. Trapezoidal rule: ∫ ≈ [f(0) + 2f(1) + f(2)] = [1 + 2(1.4142) + 3]
  3. = × 6.8284 = 3.41 (to 2 decimal places)
  4. The function is concave up here, so the trapezoids slightly overestimate the true area.
Common mistake: forgetting to divide by the coefficient of x when integrating (ax + b)n or eax. Check by differentiating your answer: of = = (2x + 1)4. Correct.
Common mistake: integrating straight across an x-intercept. If the curve crosses the axis inside the interval, the positive and negative parts partly cancel and the integral is smaller than the area. Find the intercepts first and split the integral there.

Check yourself

  1. Find ∫ (4x3 − 2x) dx.
  2. Evaluate ∫ from 0 to 3 of 2x dx.
  3. Find ∫ (3x − 2)2 dx.
  4. Evaluate ∫ from 0 to 1 of ex dx, exactly and to 2 decimal places.
  5. Find the area between y = x3 and the x-axis from x = −1 to x = 0.
  6. Find the area enclosed between y = 4 − x2 and the x-axis.

Answers: 1. x4 − x2 + C 2. [x2] from 0 to 3 = 9 3. + C 4. e − 1 ≈ 1.72 5. ∫ = [] from −1 to 0 = 0 − = −, so the area is square units 6. x-intercepts ±2; ∫ from −2 to 2 of (4 − x2) dx = [4x − ] from −2 to 2 = (8 − ) − (−8 + ) = square units

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Last reviewed
1 October 2026
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