Find the area of the region enclosed between the curves $y = x^{3}$ and $y = x$ from $x = 0$ to $x = 1$ .

easy 2 marks Class 12 · Mathematics · Application of Integrals

Answer

The area is

\frac{1}{4}

square units.

The correct answer is The area is $\frac{1}{4}$ square units. — State the final answer.

Step-by-step solution

Set up the integral for the area as $A = \int_{0}^{1} x - x^{3} d x$ because the upper curve is $y = x$ and the lower curve is $y = x^{3}$ on $[0, 1]$ .
Since $x \geq x^{3}$ for $0 \leq x \leq 1$ , drop the absolute value: $A = \int_{0}^{1} (x - x^{3}) d x$ .
Integrate term‑by‑term: $\int_{0}^{1} x d x = [\frac{x ^{2}}{2}]_{0}^{1} = \frac{1}{2}$ and $\int_{0}^{1} x^{3} d x = [\frac{x ^{4}}{4}]_{0}^{1} = \frac{1}{4}$ .
Subtract the results: $A = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ .
State the final answer.

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

Concept tested: Area between two curves