The area (in square units) bounded by the curve y2 = 4x and the line…

Given the curve y² = 4x

line x = 1

y² = 4x

$y=±\sqrt{4x}$

$y=±2\sqrt{x}$

As AOCA is in 1st quadrant

$∴y=2\sqrt{x}$

Area of AOBC = 2 × [Area of  AOCA]

$=2×∫_0^1y×dx⇒2∫_0^12\sqrt{x}dx$

$=4∫_0^1(x)^{\frac{1}{2}}dx$

$⇒4\begin{bmatrix}\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\end{bmatrix}_0^1⇒\frac{4×2}{3}\begin{bmatrix}x^{\frac{3}{2}}\end{bmatrix}_0^1⇒\frac{8}{3}[(1)^{\frac{3}{2}0}]$

$⇒\frac{8}{3}[(1)^{\frac{3}{2}0}]⇒\frac{8}{3}$