Question

Find the area lying above the x-axis and under the parabola y = 4x − x².

Answer 1

\[\text{ The equation }y = 4x - x^2\text{ represents a parabola opening downwards and cutting the }x \text{axis at O(0, 0) and }B(4, 0)\]
\[\text{ Slicing the region above x axis in vertical strips of length }= \left| y \right|\text{ and width }= dx , \text{ area of corresponding rectangle is }= \left| y \right| dx\]
\[\text{ Since the corresponding rectangle can move from } x = 0\text{ to } x = 4, \]
\[ \therefore \text{ Required area OABO is }\]
\[A = \int_0^4 \left| y \right| dx = \int_0^4 y dx .............\left[\text{ As, } y > 0\text{ for }0 \leq x \leq 4 \Rightarrow \left| y \right| = y \right]\]
\[ \Rightarrow A = \int_0^4 \left( 4x - x^2 \right)dx \]
\[ \Rightarrow A = \left[ \frac{4 x^2}{2} - \frac{x^3}{3} \right]_0^4 \]
\[ \Rightarrow A = 32 - \frac{64}{3}\]
\[ \Rightarrow A = \frac{32}{3}\text{ square units }\]