The area lying between the curves x^2=4y and x=2y is Option 1: 2/3 Sq. units Option 2: 1/3 Sq. units Option 3: 1/4 Sq. units Option 4: 3/4 Sq. units

Answer: 1/3 Sq. units. Solution: The curves meet where x^2=4*x/2=2x, so x=0 and x=2. Area =_0^2(x/2-x^2/4)dx=[x^2/4-x^3/12]_0^2=1-23=13 square units.

Calculus

Application of Integrals

Medium

core

The area lying between the curves $x^2=4y$ and $x=2y$ is

$\frac{2}{3}$ Sq. units

$\frac{1}{3}$ Sq. units

$\frac{1}{4}$ Sq. units

$\frac{3}{4}$ Sq. units

✅ Correct Option: 2

The curves meet where $x^2=4\cdot\frac{x}{2}=2x$ , so $x=0$ and $x=2$ .

Area $=\int_0^2\left(\frac{x}{2}-\frac{x^2}{4}\right)dx=\left[\frac{x^2}{4}-\frac{x^3}{12}\right]_0^2=1-\frac23=\frac13$ square units.

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