Find the total area enclosed by y=−x²+3x and y=2x³−x²−5x.

01 / Standard#Your turn

Official paper · jm02-2021 · 2(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Solve for all intersection abscissae.

Hint 2

The upper curve changes at x=0; split the integral.

Worked solution

Set the two ordinates equal and factor.
$2x^3-8x=2x(x-2)(x+2)=0\implies x=-2,0,2$
On (−2,0) the cubic is above; on (0,2) the quadratic is above.
$A=\int_{-2}^0(2x^3-8x)\,dx+\int_0^2(8x-2x^3)\,dx$
Evaluate both nonnegative contributions.
$A=[x^4/2-4x^2]_{-2}^0+[4x^2-x^4/2]_0^2=8+8=16$

Total area 16.

Checks and common pitfalls: A single signed integral over [−2,2] would cancel the two regions.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Set the two ordinates equal and factor.
$2x^3-8x=2x(x-2)(x+2)=0\implies x=-2,0,2$
On (−2,0) the cubic is above; on (0,2) the quadratic is above.
$A=\int_{-2}^0(2x^3-8x)\,dx+\int_0^2(8x-2x^3)\,dx$
Evaluate both nonnegative contributions.
$A=[x^4/2-4x^2]_{-2}^0+[4x^2-x^4/2]_0^2=8+8=16$

Think first. Reveal a hint when the class is ready.

2021 JM02