Question

Draw a rough sketch to indicate the region bounded between the curve y² = 4x and the line x = 3. Also, find the area of this region.

Answer 1

\[y^2 = 4x \text{ represents a parabola with vertex at (0, 0) and axis of symmetry along the + ve direction of } x\text{ axis }\]
\[x = 3\text{ is a line parallel to } y \text{ axis and cutting} x \text{ axis at (3, 0) }\]
\[\text{ Since }y^2 = 4x\text{ is symmetrical about }x\text{ axis , }\]
\[ \therefore\text{ Required area A }= \text{ OA }\hspace{0.167em} \text{CO }= 2 \times\text{ area OABO}\]
\[\text{ Slicing the area above }x \text{ axis into vertical strips of length }= \left| y \right|\text{ and width }= dx \]
\[\text{ Area of corresponding rectangle }= \left| y \right| dx\]
\[\text{ The corresponding rectangle moves from }x = 0\text{ to }x = 3\]
\[A = 2 \times\text{ area OABO }\]
\[ \Rightarrow A = 2 \int_0^3 \left| y \right| dx = 2 \int_0^3 \left| y \right| dx ..................\left[ As, \left| y \right| = y, y > 0 \right]\]
\[ \Rightarrow A = 2 \int_0^3 \sqrt{\left( 4x \right)} dx \]
\[ \Rightarrow A = 4 \int_0^3 \sqrt{x} dx \]
\[ \Rightarrow A = 4 \left[ \frac{x^\frac{3}{2}}{\frac{3}{2}} \right]_0^3 = \frac{8}{3} \left[ x^\frac{3}{2} \right]_0^3 = \frac{8}{3} \times 3\sqrt{3} = 8\sqrt{3}\text{ sq . units }\]
\[ \therefore \text{ Area of region bound by curve }y^2 = 4x \text{ and }x = 3\text{ is }8\sqrt{3}\text{ sq . units }\]