Find the area enclosed between the curves $y = x^{3}$ and $y = 4 x^{2} - 3 x$ between their intersection points.

hard 4 marks Class 12 · Mathematics · Integrals

Answer

\frac{37}{12}

The correct answer is $\frac{37}{12}$ — Thus the area enclosed by the two curves is $\frac{37}{12}$ square units.

Step-by-step solution

Set the two expressions equal to find intersection points: $x^{3} = 4 x^{2} - 3 x \Rightarrow x (x^{2} - 4 x + 3) = 0 \Rightarrow x = 0, 1, 3$ .
On $[0, 1]$ the cubic $x^{3}$ lies above $4 x^{2} - 3 x$ ; on $[1, 3]$ the quadratic lies above the cubic.
Compute the area as the sum of two definite integrals: $A = \int_{0}^{1} (x^{3} - (4 x^{2} - 3 x)) d x + \int_{1}^{3} ((4 x^{2} - 3 x) - x^{3}) d x .$
First integral: integrand simplifies to $x^{3} - 4 x^{2} + 3 x$ . Antiderivative: $\frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{3 x ^{2}}{2}$ . Evaluate from $0$ to $1$ to get $\frac{5}{12}$ .
Second integral: integrand simplifies to $4 x^{2} - 3 x - x^{3}$ . Antiderivative: $\frac{4 x ^{3}}{3} - \frac{3 x ^{2}}{2} - \frac{x ^{4}}{4}$ . Evaluate from $1$ to $3$ to get $\frac{8}{3}$ .
Add the two areas: $\frac{5}{12} + \frac{8}{3} = \frac{5}{12} + \frac{32}{12} = \frac{37}{12}$ .
Thus the area enclosed by the two curves is $\frac{37}{12}$ square units.

Common mistake: Failing to identify which curve is upper on each subinterval

Concept tested: Area between curves with multiple intervals