NDA 2019 Maths · Mathematics · Application of Integrals · hard

The area of the region enclosed by the parabolas y^2=4x and x^2=4y is

Single correct option

The area of the region enclosed by the parabolas

y^2=4x

and

x^2=4y

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Answer: (C) $\frac{16}{3}$ sq. units

The parabolas meet at

(0,0)

and

(4,4)

. Between

x=0

and

4

, the upper curve is

y=2\sqrt x

and the lower is

y=\frac{x^2}{4}

. Area

=\int_0^4\left(2\sqrt x-\frac{x^2}{4}\right)dx=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}

. Using only the integral of

2\sqrt x

gives the trap value

\frac{32}{3}

Original question written in the NDA 2019 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake