Question

Make a rough sketch of the graph of the function y = 4 − x², 0 ≤ x ≤ 2 and determine the area enclosed by the curve, the x-axis and the lines x = 0 and x = 2.

Answer 1

\[y = 4 - x^2 , 0 \leq x \leq 2\text{ represents a half parabola with vetex at }(4, 0) \]
\[x = 2\text{ represents a line parallel to y - axis and cutting x - axis at } (2, 0)\]
\[\text{ In quadrant OABO, consider a vertical strip of length }= \left| y \right|,\text{ width }= dx\]
\[ \therefore\text{ Area of approximating rectangle }= \left| y \right| dx \]
\[ \text{ The approximating rectangle moves from }x = 0\text{ to }x = 2\]
\[ \Rightarrow \text{ A = Area OABO }= \int_0^2 \left| y \right| dx \]
\[ \Rightarrow A = \int_0^2 y dx .....................\left[ As, y > o, \left| y \right| = y \right]\]
\[ \Rightarrow A = \int_0^2 \left( 4 - x^2 \right) dx \]
\[ \Rightarrow A = \left[ 4x - \frac{x^3}{3} \right]_0^2 \]
\[ \Rightarrow A = 8 - \frac{8}{3}\]
\[ \Rightarrow A = \frac{16}{3}\text{ sq . units }\]
\[ \therefore\text{ The area enclosed by the curve and }x - \text{ axis and given lines }= \frac{16}{3} \text{ sq . units }\]