This set has twelve original questions, ordered from easier to harder, covering all five lessons in areas and motion. Questions 1 to 4 are plain areas, 5 to 7 involve roots and sign changes, 8 to 9 are displacement and distance, and 10 to 12 mix the whole chain.
Attempt each question on paper before opening the answer. Write the limits, the antiderivative and each substitution as separate lines.
Mark the ones you got wrong and use the routing list at the end. The mistake log and retest queue can hold the error types for a later retest.
Questions
1. Find the area between y = x², the x-axis, x = 0 and x = 3.
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F(x) = x³/3. F(3) = 9 and F(0) = 0.
9 square units
2. Find the area between y = 4x³ + 1, the x-axis, x = 1 and x = 2.
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F(x) = x⁴ + x. F(2) = 16 + 2 = 18 and F(1) = 1 + 1 = 2.
18 − 2 = 16 square units
3. Find the area between y = 3√x, the x-axis, x = 1 and x = 4.
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Write 3√x as 3x1/2. F(x) = 2x3/2.
F(4) = 2 × 8 = 16 and F(1) = 2 × 1 = 2.
16 − 2 = 14 square units
4. Find the area between y = 1/x², the x-axis, x = 1 and x = 4.
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Write 1/x² as x−2. F(x) = −x−1 = −1/x.
F(4) = −1/4 and F(1) = −1.
−1/4 − (−1) = 3/4 square unit
5. The curve y = x² − 6x + 8 meets the x-axis at two points. Find the area enclosed between the curve and the x-axis.
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Solve (x − 2)(x − 4) = 0, so the roots are x = 2 and x = 4. The curve is below the axis between them.
F(x) = x³/3 − 3x² + 8x. F(4) = 64/3 − 48 + 32 = 16/3. F(2) = 8/3 − 12 + 16 = 20/3.
F(4) − F(2) = −4/3. The area is the positive value.
4/3 square units
6. For the same curve y = x² − 6x + 8, find the total area between the curve and the x-axis from x = 0 to x = 4.
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The root x = 2 is inside the interval, so split there. Use F(x) = x³/3 − 3x² + 8x with F(0) = 0, F(2) = 20/3, F(4) = 16/3.
From 0 to 2: 20/3 − 0 = 20/3, above the axis.
From 2 to 4: 16/3 − 20/3 = −4/3, so the area is 4/3.
Total = 20/3 + 4/3 = 24/3 = 8 square units.
Note that one integral over 0 to 4 would give 16/3, which is wrong for an area.
7. Show that the integral of y = x³ − x from x = −1 to x = 1 is 0, then find the total area between the curve and the x-axis.
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F(x) = x⁴/4 − x²/2. F(−1) = 1/4 − 1/2 = −1/4, F(0) = 0, F(1) = −1/4.
Whole interval: F(1) − F(−1) = −1/4 − (−1/4) = 0.
The roots are x = −1, 0 and 1, so split at 0. From −1 to 0: 0 − (−1/4) = 1/4. From 0 to 1: −1/4 − 0 = −1/4, so the area is 1/4.
Total area = 1/4 + 1/4 = 1/2 square unit
8. A particle moves in a straight line with velocity v = 6t − 2 m/s. Find its displacement from t = 0 to t = 3.
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F(t) = 3t² − 2t. F(3) = 27 − 6 = 21 and F(0) = 0.
Displacement = 21 m
9. A particle has velocity v = 2t − 6 m/s for 0 ≤ t ≤ 5. Find its displacement and the distance it travels.
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v = 0 at t = 3, inside the interval. s = t² − 6t, so s(0) = 0, s(3) = 9 − 18 = −9, s(5) = 25 − 30 = −5.
Displacement = s(5) − s(0) = −5 m.
From 0 to 3 the particle moves 9 m in the negative direction. From 3 to 5 it moves −5 − (−9) = 4 m in the positive direction.
Distance = 9 + 4 = 13 m
10. A particle has acceleration a = 2t − 4 m/s². When t = 0 its velocity is 3 m/s. Find its velocity when t = 5, and the times when it is instantaneously at rest.
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v = t² − 4t + c. At t = 0, v = 3, so c = 3. So v = t² − 4t + 3.
At t = 5: v = 25 − 20 + 3 = 8 m/s.
v = (t − 1)(t − 3) = 0, so the particle is at rest at t = 1 and t = 3.
11. Continue with the particle in question 10, and suppose it starts at the origin O. Find its position at t = 3 and the distance it travels from t = 0 to t = 3.
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s = t³/3 − 2t² + 3t, with s(0) = 0.
s(1) = 1/3 − 2 + 3 = 4/3. s(3) = 9 − 18 + 9 = 0.
The position at t = 3 is 0 m, so it is back at O.
From 0 to 1 it moves 4/3 m forward. From 1 to 3 it moves 4/3 m backward, because v is negative between 1 and 3 (for example v(2) = −1).
Distance = 4/3 + 4/3 = 8/3 m (2⅔ m), while the displacement is 0.
12. A particle has position s = t³ − 9t² + 24t metres from O for t ≥ 0. Find its velocity and acceleration at each time it is at rest, then find the distance it travels from t = 0 to t = 5.
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v = 3t² − 18t + 24 = 3(t − 2)(t − 4), so v = 0 at t = 2 and t = 4. a = 6t − 18.
At t = 2: a = −6 m/s². At t = 4: a = 6 m/s².
Positions: s(0) = 0, s(2) = 8 − 36 + 48 = 20, s(4) = 64 − 144 + 96 = 16, s(5) = 125 − 225 + 120 = 20.
From 0 to 2: 20 m forward. From 2 to 4: 4 m backward. From 4 to 5: 4 m forward.
Distance = 20 + 4 + 4 = 28 m, and the displacement is 20 m.
If you got these wrong
Match the kind of error to the lesson that fixes it.
- Questions 1 to 4, wrong area or forgot the lower limit: revisit finding area between a curve and an axis. Write F(upper) − F(lower) every time.
- Questions 5 to 7, area came out zero or too small: revisit splitting an interval when signed areas change. Solve y = 0 and sketch before you integrate.
- Questions 8 and 10, velocity multiplied by time or wrong operation: revisit calculating displacement from velocity, then connecting position, velocity and acceleration.
- Questions 9, 11 and 12, distance equals displacement: revisit distinguishing distance travelled from displacement. List the moments v = 0 first.
- Questions 10 to 12, constants missing or chain in the wrong direction: revisit connecting position, velocity and acceleration.
- Fraction slips in the limits: rehearse with the non-calculator working trainer, and use the quadratic structure explorer to check roots.
When a pattern repeats across several questions, our teachers can help you find the habit behind it in online one-to-one Additional Mathematics tuition.