The area (in square units) bounded by the curve y2 = 4x and the line x = 1 is: |
$\frac{4}{3}$ $\frac{8}{3}$ $\frac{11}{3}$ $\frac{16}{3}$ |
$\frac{8}{3}$ |
Given the curve y2 = 4x line x = 1 y2 = 4x $y=±\sqrt{4x}$ $y=±2\sqrt{x}$ As AOCA is in 1st quadrant $∴y=2\sqrt{x}$ $=2×∫_0^1y×dx⇒2∫_0^12\sqrt{x}dx$ $⇒4\begin{bmatrix}\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\end{bmatrix}_0^1⇒\frac{4×2}{3}\begin{bmatrix}x^{\frac{3}{2}}\end{bmatrix}_0^1⇒\frac{8}{3}[(1)^{\frac{3}{2}0}]$ $⇒\frac{8}{3}[(1)^{\frac{3}{2}0}]⇒\frac{8}{3}$ |