What is the volume of the solid produced by revolving #f(x)=xe^x-(x/2)e^x, x in [2,7] #around the x-axis?

1 Answer
Nov 28, 2016

The volume of the solid produced by revolving around the x-axis the trapezoid of the function #f(x)# is calculated using the formula:

#V = pi int_a^b f^2(x)dx#

Explanation:

#V= pi int_2^7 (xe^x-(x/2)e^x)^2dx#

#V= pi int_2^7 (x^2e^(2x)-2xe^x(x/2)e^x+x^2/4e^(2x))dx#

#V= pi int_2^7 (x^2e^(2x)-x^2e^(2x)+x^2/4e^(2x))dx = pi int_2^7 x^2/4e^(2x)dx #

Substitute #2x=t#:

#V = pi/32 int_1^(7/2) t^2e^tdt #

The integral is solved iteratively by parts:

#int t^2e^tdt = int t^2d(e^t) =t^2e^t - int 2te^tdt = t^2e^t - 2te^t +2inte^tdt = e^t(t^2-2t+2)#

Finally:

#V = pi/32[e^(7/2)(49/4-7+2)-e(1-2+2)] = pi/32[e^(7/2)(49-28+8)/4-e] = (pi*e)/32 (29/4e^(5/2)-1) #