The area of the region bounded by the curves x^2=4y and x=4y is Option 1: 3/8 Sq.unit Option 2: 5/8 Sq.unit Option 3: 7/8 Sq.unit Option 4: 1/24 Sq.unit

Answer: 1/24 Sq.unit. Solution: The curves meet where x^2/4=x/4, i.e. x=0 and x=1. On [0,1] the line lies above the parabola, so area =_0^1(x/4-x^2/4)dx=14(12-13)=1/24.

Calculus

Application of Integrals

Medium

core

The area of the region bounded by the curves $x^2=4y$ and $x=4y$ is

$\dfrac{3}{8}$ Sq.unit

$\dfrac{5}{8}$ Sq.unit

$\dfrac{7}{8}$ Sq.unit

$\dfrac{1}{24}$ Sq.unit

✅ Correct Option: 4

The curves meet where $\dfrac{x^2}{4}=\dfrac{x}{4}$ , i.e. $x=0$ and $x=1$ . On $[0,1]$ the line lies above the parabola, so area $=\displaystyle\int_0^1\left(\dfrac{x}{4}-\dfrac{x^2}{4}\right)dx=\dfrac14\left(\dfrac12-\dfrac13\right)=\dfrac{1}{24}$ .

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The area of the region bounded by the curves $x^2=4y$ and $x=4y$ is