Calculus

Application of Integrals

Medium

core

The area (in square units) of the region bounded by the curve $x^2 = y$ and the straight line $y = 4$ in the first quadrant is equal to

$\frac{3}{2}$

$\frac{16}{3}$

$\frac{32}{3}$

$\frac{7}{2}$

✅ Correct Option: 2

The region is bounded by the parabola $x^2 = y$ (opening upward, vertex at origin) and the horizontal line $y = 4$ , in the first quadrant where $x \geq 0$ and $y \geq 0$ .

Setting $y = 4$ in $x^2 = y$ :

$x^2 = 4$

$x = 2$ (first quadrant)

The curves meet at $(2, 4)$ .

In the region from $x = 0$ to $x = 2$ :

Top curve: $y = 4$

Bottom curve: $y = x^2$

$A = \int_0^2 (4 - x^2)\, dx$

$A = \left[4x - \dfrac{x^3}{3}\right]_0^2$

$= \left(4(2) - \dfrac{(2)^3}{3}\right) - \left(0 - 0\right)$

$= 8 - \dfrac{8}{3}$

$= \dfrac{24 - 8}{3}$

$= \dfrac{16}{3}$

The area of the region $= \dfrac{16}{3}$ square units.

2025: 14 May Shift 2

Consider the region bounded by the lines $y - 1 = x, x = -2, x = 3$ and $x$ - axis. Then

(A) The area of the bounded region is given by $\int_{-2}^{3}(x + 1)dx$

(B) The numerical value of the area is $\frac{15}{2}$ sq. units

(D) The numerical value of the area is $\frac{17}{2}$ sq. units

Choose the correct answer from the options given below:

2025: 14 May Shift 1

The area (in sq. units) of the region bounded by $y = 2\sqrt{1 - x^2}$ , $x \in [0, 1]$ and $x$ -axis is equal to

2026: 15 May Shift 1

The area (in sq. units) bounded between the curves $y = x^2$ and $y = 1$ in the first quadrant is

2026: 22 May Shift 1

Area of the region bounded by the curve $y=\sin x, y=\cos x$ and $x-$ axis, $\left(0\le x\le\frac{\pi}{2}\right)$ is;

2022: 6 Aug Shift 2

The value of $\frac{dy}{dx}$ when $x = \frac{1}{6}$ , is

2025: 27 May Shift 1

The area of the region bounded by the parabola $y^2 = x$ and the straight line $2y = x$ is

The area (in square units) of the region bounded by the curve x2=yx^2 = yx2=y and the straight line y=4y = 4y=4 in the first quadrant is equal to

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The area (in square units) of the region bounded by the curve $x^2 = y$ and the straight line $y = 4$ in the first quadrant is equal to