Let #A# be the given net area. Then,
#A = int_(x_1)^(x_2) xe^x - 3x dx#.
Now, since the area wanted is the area above the x-axis, that means we have find the minimum and maximum values of #x in [1,5]#, #x_1# and #x_2# respectively, such that #f(x_1)# and #f(x_2)# are both bigger than #0#.
Let's plug in #x_1 = 1# and #x_2 = 5# firstly;
#f(1) = e-3<0#
#f(5) = 5e^5 - 15 > 0#
Clearly, #x_1# has to be bigger than #1#. Because #xe^x# increases exponentially at a much faster rate then #3x#, it means that #f(x)# is monotonic on the interval #[1, +oo]#. We don't need the exact interval of monotony to be able to find #A#.
In order to find the exact value of #x_1#, we must solve the equation #alphae^alpha - 3alpha = 0# :
#alpha(e^alpha - 3) = 0#
The two possible solutions are #0# and #ln 3#. However, #x_1 in [1,5] => x_1 = ln 3#. #A# has now become
#A = int_ln 3^5 xe^x - 3x dx#.
Given the basic properties of integrals, we have
#A = color(red)(int_ln3^5 xe^x dx) - 3(color(blue)(int_0^5 x dx - int_0^ln 3 xdx))#
- #color(red)("Solving the integral highlighted in red")# :
Let #u = e^x => x = ln u#.
#(du)/dx = e^x => dx = (du)/e^x = (du)/u#
#int_ln3^5 xe^x dx = int_ln3^5 ln(u)*u (du)/u#
#= int_ln3^5 ln u du#
We will solve this with integration by parts :
#int_a^b f(u)g'(u) = f(u)g(u)|_a^b - int_a^b g(u)f'(u)#
In our case, #f(u) = ln(u) => f'(u) = 1/u# and #g'(u) = 1 => g(u) = u#, #b = 5# and #a=ln 3#.
#int_ln3^5 ln u du = 5ln(5) - ln(3)ln(ln 3) - int_ln3^5 u*1/u du#
#= 5ln(5) - ln(3)ln(ln3) - int_0^5 1 du + int_0^ln 3 1 du#
#= 5ln(5) - ln(3)ln(ln3) - 5 + ln3#
#color(red)(= 5(ln5 - 1) - ln(3)(ln(ln3) - 1))#
- #color(blue)("Solving the difference of integrals :")#
We know #int_0^k x dx = k^2/2#, so
#color(blue)(int_0^5 x dx - int_0^ln 3 xdx = (25 - ln^2 3)/2)#
And finally,
#color(purple)
(A) = color(red)(5(ln5 - 1) - ln(3)(ln(ln3) - 1)) + color(blue)((3ln^2 3 - 75)/2)#
