Question

How do you find the area between the curves `y=4x-x^2` and `y=x`?

Answer 1

First, we need to find where the two functions intersect.

`4x-x^2=x`

`x^2-4x+x=0`

`x^2-3x=0`

`x(x-3)=0`

`x=0` and `x=3`

So our interval of integration is

`0<=x<=3`

`y=4x-x^2` is a parabola that is concave down

`y=x` is the line that passes through the origin with slope 1

The integral for the area is

`int4x-x^2-xdx=int3x-x^2dx`

integrating we have

`3/2x^2-1/3x^3`

Evaluating at we have

`(3/2)3^2-(1/3)3^3-0=27/2-27/3=27/2-9`

`81/6-54/6=27/6=9/2`

Answer 2

Find the points of intersection, `(a,y_a)` and `(b,y_b)`
between
`y = 4x - x^2` and `y = x`
then subtract from the integral of the first (between `a` and `b`)
the integral of the second (again, between `a` and `b`)

Part 1:
Points of intersection occurs when
`4x - x^2 = x`
This occurs when either `x = 0` or `x = 3`
(we could, but don't actually need to calculate `y_a` and `y_b`)

Part 2:
`int_a ^b (4x-x^2) dx - int_a^b x dx`

`= ((2x^2 - x^3/3) |_a^b) - (((x^2)/2) |_a^b)`

with `a = 0` and `b = 3`
this becomes
`((2*9 - 9/3) - (0)) - ( (9/2) - (0) )`
`= 10.5`