Hint

Locate any axis crossing inside the interval before calculating geometric area.

Worked solution

(a)

x²−4=0 gives x=±2, but only x=2 is in [0,3]. The expression is negative on [0,2) and positive on (2,3]. It equals zero at the crossing.

Answer: Crossing x=2; below for 0≤x<2, above for 2<x≤3.

(b)

Using F(x)=x³/3−4x gives F(3)−F(0)=9−12=−3. This is a signed integral.

Answer: −3.

(c)

F(2)=−16/3 and F(3)=−3. The below-axis piece has area −[F(2)−F(0)]=16/3, while the above-axis piece has area F(3)−F(2)=7/3. Their sum is 23/3 square units, not |−3|.

Answer: \(23/3\) square units.

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 2 marks

• 1 mark: Give the only in-interval crossing x=2.
• 1 mark: Give both correct sign intervals.

Alternative methods: A labelled sketch with correct endpoint status is accepted.

Common error: Using the out-of-interval root −2 as a boundary.

(b) · 2 marks

• 1 mark: Use the correct antiderivative and endpoint order.
• 1 mark: Give −3.

Alternative methods: Any equivalent exact endpoint calculation is accepted.

Common error: Changing the integral to a positive number because an area is anticipated.

(c) · 3 marks

• 1 mark: Split at x=2 and reverse the sign of the below-axis contribution.
• 1 mark: Obtain the two positive areas 16/3 and 7/3.
• 1 mark: Give total area 23/3 square units.

Alternative methods: Integrating 4−x² on [0,2] and x²−4 on [2,3] is accepted.

Common error: Taking the absolute value only after the two signed contributions cancel.

Total: 7 marks.