`y = e^x, y = x(e^x), x = 0` Sketch the region enclosed by the given curves and find its area.

Graph each equation to determine the bounded region. 

( Red curve is the graph of `y=e^x` . Blue curve is the graph of `y=xe^x` . And green line is the graph of x=0.)

Base on the graph,  the bounded region of the three equations starts at x=0 and ends at x =1. Then, draw a vertical strip inside the region. Apply the formula:

`A=int_a^b (y_U - y_L)dx`

(Refer to the attached figure below.)

The upper...

Graph each equation to determine the bounded region. 

( Red curve is the graph of `y=e^x` . Blue curve is the graph of `y=xe^x` . And green line is the graph of x=0.)

Base on the graph,  the bounded region of the three equations starts at x=0 and ends at x =1. Then, draw a vertical strip inside the region. Apply the formula:

`A=int_a^b (y_U - y_L)dx`

(Refer to the attached figure below.)

The upper end of the vertical strip touches the curve `y =e^x` . And its lower end touches the curve `y=xe^x` . So the integral will be:

`A=int_0^1 (e^x - xe^x) dx`

`A = int_0^1 e^x(1-x)dx`

To take the integral of this, apply integration by parts  `int udv = u*v - int vdu` .

In the integrand of area,  u , du , v and dv are:

`u=1-x`

`du=-dx`

`dv=e^xdx`

`v=e^x`

Plugging them to the formula integration by parts, then A becomes:

`A = (1-x)*e^x - int e^x *(-dx) = (1-x)e^x + int e^x dx`

`A = ((1-x)e^x + e^x )|_0^1 = (e^x-xe^x + e^x)_0^1 `

`A = (2e^x - xe^x )|_0^1`

`A= (2e^1-1*e^1) - (2e^0-0*e^0)`

`A= e -2`

`A=0.72`

Therefore, the area of the bounded region is 0.72 square units.