Find the total area of both regions between the cubic and y=x.

01 / Standard#Your turn

f(x)=x^3-x^2-x

Official paper · jm02-2026 · 2(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Solve f(x)=x to locate all boundaries.

Hint 2

The upper curve changes at x=0.

Worked solution

Find the three intersection abscissas.
$f(x)-x=x(x-2)(x+1)=0\implies x=-1,0,2$
Integrate upper minus lower separately on the two intervals.
$A=\int_{-1}^0(x^3-x^2-2x)\,dx+\int_0^2(-x^3+x^2+2x)\,dx$
Evaluate the two positive areas and add.
$A=\left[\frac{x^4}4-\frac{x^3}3-x^2\right]_{-1}^0+\left[-\frac{x^4}4+\frac{x^3}3+x^2\right]_0^2=\frac5{12}+\frac83=\frac{37}{12}$

37/12 square units.

Checks and common pitfalls: One signed integral over [−1,2] would cancel part of the area.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Find the three intersection abscissas.
$f(x)-x=x(x-2)(x+1)=0\implies x=-1,0,2$
Integrate upper minus lower separately on the two intervals.
$A=\int_{-1}^0(x^3-x^2-2x)\,dx+\int_0^2(-x^3+x^2+2x)\,dx$
Evaluate the two positive areas and add.
$A=\left[\frac{x^4}4-\frac{x^3}3-x^2\right]_{-1}^0+\left[-\frac{x^4}4+\frac{x^3}3+x^2\right]_0^2=\frac5{12}+\frac83=\frac{37}{12}$

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

2026 JM02