What is the volume of the solid produced by revolving f(x)=e^x-xlnx, x in [1,5] f(x)=ex−xlnx,x∈[1,5]around the x-axis?
1 Answer
The volume is
V = pi/2(e^(10) - e^2) - 1696.084pi + (125pi)/3[ln^2(5) - 2/3ln5 + 2/9] - (2pi)/27V=π2(e10−e2)−1696.084π+125π3[ln2(5)−23ln5+29]−2π27 "u"^3u3
To do these kinds of solid of revolution problems around the
The volume is then given by:
V = pi int_(a)^(b) f^2(x)dxV=π∫baf2(x)dx
The
In this case, we have:
V = pi int_(1)^(5) (e^x - xlnx)^2dxV=π∫51(ex−xlnx)2dx
= pi int_(1)^(5) e^(2x) - 2xe^xlnx + x^2ln^2xdx=π∫51e2x−2xexlnx+x2ln2xdx
= pi overbrace(int_(1)^(5) e^(2x)dx)^("Integral 1") - 2pi overbrace(int_(1)^(5) xe^xlnxdx)^"Integral 2" + pi overbrace(int_(1)^(5) x^2ln^2xdx)^"Integral 3"
- The first integral is
1/2e^(2x) evaluated fromx = 1 to5 , which is1/2(e^(10) - e^2) . - The second integral is not possible with elementary functions, and the numerical solution is
848.042 . - The third integral is hard but doable here.
int x^2ln^2xdx = ?
Let:
u = ln^2x dv = x^2dx du = (2lnx)/xdx v = x^3/3
Thus,
int x^2ln^2xdx
= (x^3ln^2x)/3 - int x^3/3 (2lnx)/xdx
= 1/3 x^3ln^2x - 2/3int x^2 lnxdx
Repeat with:
u = lnx dv = x^2dx du = 1/xdx v = x^3/3
And we get:
=> 1/3 x^3ln^2x - 2/3[1/3x^3lnx - int x^3/3 1/xdx]
= 1/3 x^3ln^2x - 2/3[1/3x^3lnx - 1/3 int x^2dx]
= 1/3 x^3ln^2x - [2/9x^3lnx - 2/9 int x^2dx]
= 1/3 x^3ln^2x - 2/9x^3lnx + 2/9 int x^2dx
= {:[1/3 x^3ln^2x - 2/9x^3lnx + 2/9 x^3/3]|:}_(1)^(5)
= [1/3 (5)^3ln^2(5) - 2/9(5)^3ln(5) + 2/9(5)^3/3] - [cancel(1/3 (1)^3ln^2(1))^(0) - cancel(2/9(1)^3ln(1))^(0) + 2/9 (1)^3/3]
= [1/3 125ln^2(5) - 2/9(125ln5 - 125/3)] - 2/9(1/3)
= 125/3[ln^2(5) - 2/3ln5 + 2/9] - 2/27
So, adding up all the results,
V = pi { 1/2(e^(10) - e^2)} - 2pi {848.042} + pi {125/3[ln^2(5) - 2/3ln5 + 2/9] - 2/27}
= color(blue)(pi/2(e^(10) - e^2) - 1696.084pi + (125pi)/3[ln^2(5) - 2/3ln5 + 2/9] - (2pi)/27)
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