CUET Mathematics 2024 - The area of the region enclosed between the curves 4 x^2=y and y=4 is : | PYQs + Solutions

CUET Mathematics 2024

Calculus

Application of Integrals

Medium

The area of the region enclosed between the curves $4 x^{2}=y$ and $y=4$ is :

16

sq. units

\frac{32}{3}

sq. units

\frac{8}{3}

sq. units

\frac{16}{3}

sq. units

✅ Correct Option: 4

First, let's find the points of intersection:

4x^2 = 4

x^2 = 1

x = \pm 1

So the curves intersect at

(-1,4)

and

(1,4)

Rearranging the parabola equation:

4x^2=y

becomes

y=4x^2

To find the area, we integrate the difference between the upper curve (

y=4

) and lower curve (

y=4x^2

) from

x=-1

x=1

\text{Area} = \int_{-1}^{1} [4 - 4x^2] \, dx

= \int_{-1}^{1} 4 \, dx - \int_{-1}^{1} 4x^2 \, dx

= 4 \int_{-1}^{1} \, dx - 4 \int_{-1}^{1} x^2 \, dx

= 4[x]_{-1}^{1} - 4[\dfrac{x^3}{3}]_{-1}^{1}

= 4(1-(-1)) - 4(\dfrac{1^3}{3}-\dfrac{(-1)^3}{3})

= 4 \cdot 2 - 4(\dfrac{1}{3}-\dfrac{-1}{3})

= 8 - 4 \cdot \dfrac{2}{3}

= 8 - \dfrac{8}{3}

= \dfrac{24 - 8}{3}

= \dfrac{16}{3}

Therefore, the area of the region enclosed between the curves is

\dfrac{16}{3}

square units.