Find the area enclosed between the curves $y = x^{3}$ and $y = 2 x^{2} - x$ . Give the exact value and its decimal approximation up to three decimal places.

medium 4 marks Class 12 · Mathematics · Integrals

Answer

\frac{1}{12} (approximately 0.083)

The correct answer is $\frac{1}{12} (approximately 0.083)$ — Decimal value: $\frac{1}{12} = 0.08333 \dots$ , so up to three decimals $0.083$ .

Step-by-step solution

Set the two expressions equal to locate intersection points: $x^{3} = 2 x^{2} - x$ .
Rewrite as $x^{3} - 2 x^{2} + x = 0 \Rightarrow x (x^{2} - 2 x + 1) = 0$ .
Factor further: $x (x - 1)^{2} = 0$ , giving $x = 0$ and $x = 1$ (the double root $x = 1$ does not create a new interval).
Determine which curve is above on $[0, 1]$ . Test $x = \frac{1}{2}$ : $x^{3} = 0.125$ , $2 x^{2} - x = 0$ , so $x^{3}$ is larger.
Hence the upper function is $y_{up} = x^{3}$ and the lower is $y_{low} = 2 x^{2} - x$ .
Write the area integral: $A = \int_{0}^{1} [x^{3} - (2 x^{2} - x)] d x = \int_{0}^{1} (x^{3} - 2 x^{2} + x) d x$ .
Integrate term by term: $\int x^{3} d x = \frac{x ^{4}}{4}$ , $\int - 2 x^{2} d x = - \frac{2 x ^{3}}{3}$ , $\int x d x = \frac{x ^{2}}{2}$ .
Combine: $A = [\frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} + \frac{x ^{2}}{2}]_{0}^{1}$ .
Evaluate at $x = 1$ : $\frac{1}{4} - \frac{2}{3} + \frac{1}{2} = \frac{3 - 8 + 6}{12} = \frac{1}{12}$ .
Evaluate at $x = 0$ : $0$ .
Thus the exact area is $\frac{1}{12}$ square units.
Decimal value: $\frac{1}{12} = 0.08333 \dots$ , so up to three decimals $0.083$ .

Common mistake: Reversing the order of subtraction, which leads to a negative area.

Concept tested: Area between two curves