Question

What is the net area between `f(x) = xlnx-xe^x` and the x-axis over `x in [2, 4 ]`?

Answer 1

`14 ln 2-3 e^4+e^2-3=-149.7` areal units, nearly. The negative integral indicates that area below the x-axis is quite large. Really, f < 0 and x > 8, and the graph is below the x-axis

Explanation:

`y = f(x) = x(ln x-e^x)`

The area `= int y dx`, with x from 2 to 4.

`= intx(ln x - e^x) dx`, with x from 2 to 4

`= int (ln xd(x^2/2)-xd(e^x))`, with x from 2 to 4

`=[x^2/2 ln x-int x^2/2 d(ln x) -(xe^x-int e^x dx)]`, x from 2 to 4.

`=[x^2/2 ln x -1/2 int x dx-xe^x+e^x]`, with x from 2 to 4

#=[1/4x^2(2 ln x -1 )-e^x(x-1)]@, between the limits x = 2 and 4.

Upon substitution of limits and simplification, this

`=14 ln 2-3 -3e^4+e^2`.

The graph inserted is suitably scaled to revveal the relevant portion of the curve

graph{x(ln x-e^x) [0, 4, -300, -0]}