The linear density of a rod of length 4 m is given by ρ(x) = 8 + 5sqrtx measured in kilograms per meter...?

8+5sqrt(4)=8+5*2=8+10=18

We have been given the linear density ρ (x) = 8 + 5√(x)  ( in kg/m)

Where x is the length of the rod measured in meters from one end of the rod.

Since total length of the rod is 4m (given)

So we have to find the total mass of the rod from x=0 to x=4

If m(x) is the mass of the rod then linear density is ρ (x) =m'(x)

So mass of the rod between x=0 and x=4 is

m(4) -m(0) =  ∫ (ρ (x), 0, 4)

= ∫ ( 8x + 5√(x) ,0,4)

m(4) -m(0) =[8x + 5x^(3/2)/(3/2), 0, 4]   [x^(3/2)/(3/2) is an anti derivative of √(x)

= [8x + 5x^(3/2)/(3/2), 0, 4]  

=[8 *4  + 5*4^(3/2)/(3/2), 0, 4]  

=[32 + 80/3]

=[96/3 +80/3]

=176/3= 58.666667kg

so total mass is 58(2/3) kg