Question

Find the area of the region enclosed by `y=x^2-2x` and `y=3`?

`y=x^2-2x` and `y=3`

Answer 1

`32/3` unit`"s"^2`

Explanation:

Find the points of intersection `M` and `N`:

`x^2-2x=y=3`

`x^2-2x-3=0`

`(x-3)(x+1)=0`

So `M` is `x=3rarry=3`

And `N` is `x=-1rarry=3`

The area bounded by two curves is given by:

`int_a^b[f(x)-g(x)]dx`, where `f(x) and g(x)` are functions, and `a and b` are the `x` coordinates of the intersection points.

Here, `b=3` and `a=-1`

`f(x)=x^2-2x`

`g(x)=3`

Inputting:

`int_-1^3[(x^2-2x)-(3)]dx`

`int_-1^3(x^2-2x-3)dx`

`int_-1^3(x^2)dx-int_-1^3(x)dx-3int_-1^3(x^0)dx`

`(x^3/3-x^2/2-3x)_-1^3`

`(-9)-(5/3)`

`-32/3`

But since area cannot be negative, we have `32/3` unit`"s"^2` as the area.