Q239 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
Let the function $f(x) = x^2 - 1$.
What is the area bounded by the function $f(x)$ and the $x$-axis?
Let the function $f(x) = x^2 - 1$.
What is the area bounded by the function $f(x)$ and the $x$-axis?
- A$\dfrac{1}{3}$ square unit
- B$\dfrac{2}{3}$ square unit
- Correct answerC$\dfrac{4}{3}$ square units
- D$2$ square units
Solution
The curve intersects the $x$-axis when $f(x) = 0$.
$x^2 - 1 = 0 \Rightarrow x = -1, 1$
The required area is given by $A = \int_{-1}^{1} |x^2 - 1| dx$.
Since $x^2 - 1 \le 0$ for $x \in [-1, 1]$, we have:
$A = \int_{-1}^{1} (1 - x^2) dx$
$A = \left[ x - \dfrac{x^3}{3} \right]_{-1}^{1}$
$A = \left( 1 - \dfrac{1}{3} \right) - \left( -1 + \dfrac{1}{3} \right)$
$A = \dfrac{2}{3} - \left( -\dfrac{2}{3} \right) = \dfrac{4}{3}$ square units.
Answer: $\dfrac{4}{3}$ square units
$x^2 - 1 = 0 \Rightarrow x = -1, 1$
The required area is given by $A = \int_{-1}^{1} |x^2 - 1| dx$.
Since $x^2 - 1 \le 0$ for $x \in [-1, 1]$, we have:
$A = \int_{-1}^{1} (1 - x^2) dx$
$A = \left[ x - \dfrac{x^3}{3} \right]_{-1}^{1}$
$A = \left( 1 - \dfrac{1}{3} \right) - \left( -1 + \dfrac{1}{3} \right)$
$A = \dfrac{2}{3} - \left( -\dfrac{2}{3} \right) = \dfrac{4}{3}$ square units.
Answer: $\dfrac{4}{3}$ square units
Hence, the correct answer is option: 'C'.