The area enclosed between the curves $y^{2} = x$ and $y = ∣ x ∣$ is

Answer (Detailed Solution Below)View answer ▾

Answer: Option C — $\frac{1}{6}$ sq. units

Detailed Solution

$y = ∣ x ∣ \geq 0$ , so only the upper branch $y = x$ ( $x \geq 0$ ) can meet it, and there $∣ x ∣ = x$ .

$x = x$ at $x = 0$ and $x = 1$ , with $x \geq x$ in between.

Area $= \int_{0}^{1} (x - x) d x = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$ .

✓ Check

x = \frac{1}{4}

x = \frac{1}{2} > \frac{1}{4}

, so the parabola lies above the line on

(0, 1)

⚠ Common trap

(D)

\frac{1}{3}

doubles the area by also counting the lower branch

y = - x

, which meets

y = ∣ x ∣

only at the origin.

Quick tip

$\int_{0}^{1} (x - x) d x = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$

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Application of Definite Integration — Preparation

This question tests your understanding of Application of Definite Integration, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

This topic uses definite integrals to find the area of regions in the plane. It covers the area under a curve and above the x-axis, the area between a curve and the y-axis, the area between two curves, and the areas of standard regions bounded by circles, parabolas, ellipses and straight lines. Always start with a rough sketch of the region and find where the curves intersect, since those points become the limits of integration. For the area between two curves, integrate (upper curve - lower curve) between their intersection points. If part of a curve lies below the x-axis, take the absolute value of that part separately rather than integrating straight across, or the positive and negative areas will cancel. Useful standard results include the area of the ellipse x^2/a^2 + y^2/b^2 = 1, which is pi ab. Practise the area under the curve questions below, each with a sketch and a step-by-step solution.

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