Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 15

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.