Make a rough sketch of the graph of the function `y=4-x^2,\ 0lt=xlt=2` and determine the area enclosed by the curve, the x-axis and the lines `x=0\ a n d\ x=2.`

Explanation

`y=4-x^{2}, 0 leq x leq 2` represents half parabola with vetex at `(4,0)`

`x=2` represents line parallel to `y` - axis and cutting `x` - axis at `(2,0)`

In quadrant `O A B O`, consider a vertical strip of length `=|y|`, width `=d x`

`therefore` Area of approximating rectangle `=|y| d x`

The approximating rectangle moves from `x=0` to `x=2`

`Rightarrow A=` Area `O A B O=int_{0}^{2}|y| d x`

`Rightarrow A=int_{0}^{2} y d x . . . . . . .[A s, y>o,|y|=y]`

`Rightarrow A=int_{0}^{2}left(4-x^{2}right) d x`

...