Question

How do you find the area between `f(x)=x^2-4x, g(x)=0`?

Answer 1

Area: `32/3 =10 2/3` square units.

Explanation:

Note that `f(x)` and `g(x)` intersect when `f(x)=g(x)`
`rarr f(x)=x^2-4x=0color(white)("xxx")x in {0,4}`

The area enclosed by `f(x)` and `g(x)` is therefore
`abs(int_0^4(f(x)-g(x)) dx)`

`color(white)("XXX")=abs(int_0^4 (x^2-4x) dx)`

`color(white)("XXX")=abs(color(white)("x")((x^3)/3-2x^2)"]"_0^4)`

`color(white)("XXX")=abs(64/3-32)`

`color(white)("XXX")=32/3`