CUET UG 2024 — Mathematics Calculus

First, let's find the points of intersection:

$$4x^2 = 4$$4x2=4

$$x^2 = 1$$x2=1

$$x = \pm 1$$x=±1

So the curves intersect at $(-1,4)$(−1,4) and $(1,4)$(1,4).

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Rearranging the parabola equation: $4x^2=y$4x2=y becomes $y=4x^2$y=4x2

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To find the area, we integrate the difference between the upper curve ($y=4$y=4) and lower curve ($y=4x^2$y=4x2) from $x=-1$x=−1 to $x=1$x=1:

$$\text{Area} = \int_{-1}^{1} [4 - 4x^2] \, dx$$Area=∫−11​[4−4x2]dx

$$= \int_{-1}^{1} 4 \, dx - \int_{-1}^{1} 4x^2 \, dx$$=∫−11​4dx−∫−11​4x2dx

$$= 4 \int_{-1}^{1} \, dx - 4 \int_{-1}^{1} x^2 \, dx$$=4∫−11​dx−4∫−11​x2dx

$$= 4[x]_{-1}^{1} - 4[\dfrac{x^3}{3}]_{-1}^{1}$$=4[x]−11​−4[3x3​]−11​

$$= 4(1-(-1)) - 4(\dfrac{1^3}{3}-\dfrac{(-1)^3}{3})$$=4(1−(−1))−4(313​−3(−1)3​)

$$= 4 \cdot 2 - 4(\dfrac{1}{3}-\dfrac{-1}{3})$$=4⋅2−4(31​−3−1​)

$$= 8 - 4 \cdot \dfrac{2}{3}$$=8−4⋅32​

$$= 8 - \dfrac{8}{3}$$=8−38​

$$= \dfrac{24 - 8}{3}$$=324−8​

$$= \dfrac{16}{3}$$=316​

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Therefore, the area of the region enclosed between the curves is $\dfrac{16}{3}$316​ square units.