Board 2025 Class 12 Maths Q17 Solution: Area bounded by y² = x, x = 4 and the x-axis

The question

The area of the shaded region bounded by the curves $y^2 = x$ ⁠, $x = 4$ and the $x$ -axis is given by

(A) $\displaystyle\int_{0}^{4} x\,dx$

(B) $\displaystyle\int_{0}^{2} y^2\,dy$

(C) $\displaystyle 2\int_{0}^{4} \sqrt{x}\,dx$

(D) $\displaystyle\int_{0}^{4} \sqrt{x}\,dx$

Exam answer

From $y^2 = x$ ⁠, $y = \pm\sqrt{x}$ ⁠.

The shaded region lies above the x-axis, under the upper branch $y = \sqrt{x}$ of $y^2 = x$ ⁠, from $x = 0$ to $x = 4$ ⁠.

Why?

The parabola $y^2 = x$ has two halves: $y = \sqrt{x}$ above the x-axis and $y = -\sqrt{x}$ below it. Only the part above the axis is shaded, so we keep $y = \sqrt{x}$ ⁠.

The region starts at the vertex, where $x = 0$ ⁠, and ends at the line $x = 4$ ⁠, where the curve is at height $\sqrt{4} = 2$ ⁠.

\text{Area} = \int_{0}^{4} y\,dx = \int_{0}^{4} \sqrt{x}\,dx

Why?

Area between a curve $y = f(x) \ge 0$ and the x-axis, from $x = a$ to $x = b$ ⁠:

\text{Area} = \int_a^b y\,dx

Think of thin vertical strips of height y and width dx, added up from $x = 0$ to $x = 4$ ⁠.

\begin{aligned} \int_{0}^{4} \sqrt{x}\,dx &= \left[\frac{2}{3}x^{\frac{3}{2}}\right]_0^4 \\ &= \frac{2}{3}(8) = \frac{16}{3} \end{aligned}

(C) is twice this area, (A) is the area under $y = x$ ⁠, and (B) is the area between the curve and the y-axis.

Why?

By the power rule:

\begin{aligned} \int x^{\frac12}\,dx &= \frac{x^{\frac32}}{\frac32} + C \\ &= \frac23 x^{\frac32} + C \end{aligned}

Also $4^{\frac32} = (\sqrt{4})^3 = 8$ ⁠.

Put in the upper limit, then the lower limit, and subtract:

\begin{aligned} &\left[\frac{2}{3}x^{\frac{3}{2}}\right]_0^4 \\ &= \frac{2}{3}(4)^{\frac32} - \frac{2}{3}(0)^{\frac32} \\ &= \frac{2}{3}(8) - 0 \\ &= \frac{2 \times 8}{3} = \frac{16}{3} \end{aligned}

Watch out: $2\int_0^4\sqrt{x}\,dx = \frac{32}{3}$ ⁠, option (C), is the area of both halves, above and below the x-axis. Option (B) gives $\frac{8}{3}$ ⁠: the unshaded part of the 4 by 2 rectangle, since $8 - \frac{8}{3} = \frac{16}{3}$ ⁠.

\begin{aligned} 2 \times \frac{16}{3} &= \frac{32}{3} \\[4pt] \int_0^2 y^2\,dy &= \left[\frac{y^3}{3}\right]_0^2 \\ &= \frac{(2)^3}{3} - \frac{(0)^3}{3} = \frac{8}{3} \\[4pt] 8 - \frac{8}{3} &= \frac{24}{3} - \frac{8}{3} = \frac{16}{3} \end{aligned}

Answer: (D) $\displaystyle\int_{0}^{4} \sqrt{x}\,dx$

Where this comes from

Chapter 8, Application of Integrals: Area under y² = x by integration.

Question type: Area under a curve

How to spot it: click to see

Type 1. Area under a curve: vertical strips

Spot it: a curve $y = f(x)$ above the $x$ -axis, and two lines such as $x = 1$ and $x = 2$ ⁠.

Here: The shaded part lies under the top branch $y = \sqrt x$ of $y^2 = x$ ⁠, from $x = 0$ to the line $x = 4$ ⁠. So the area is $\int_0^4\sqrt x\,dx$ ⁠: option (D).

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

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Q17: Area bounded by y2=xy^2 = xy2=x⁠, x=4x = 4x=4 and the x-axis

Q17: Area bounded by $y^2 = x$ ⁠, $x = 4$ and the x-axis