Respuesta :
Answer:
ρ(ave) = 29.1 kg/m
Explanation:
The average value of a function over an interval = definite integral computed with the boundaries of the interval divided by the extent of the integral
ρ(ave) = [∫⁵₀ (11x + 4) dx]/(5-0)
ρ(ave) = [(11x²/2) + 4x]⁵₀ ÷ (5)
ρ(ave) = [(11(5²)/2) + (4×2)]/5
ρ(ave) = (137.5 + 8)/5
ρ(ave) = 29.1 kg/m
Corrected Question:
The linear density ρ in a rod 5 m long is (11x+4) kg/m, where x is measured in meters from one end of the rod. Find the average density ρ(ave)
Answer:
157.5kg/m
Explanation:
The linear density, ρ, of a rod is given by;
ρ = [tex]\frac{m}{l}[/tex]
Where;
m = mass of the rod
l = length of the rod
The average density ρ(ave) is then, the average of the linear density taken over the entire length of the rod. i.e
ρ(ave) = [tex]\int\limits^a_b {p} \, dx[/tex] -----------------(ii)
Where;
a and b are the limits of the length of the rod.
b = 0
a = 5m
p = (11x + 4) kg/m [as given in the question]
Substitute these values into equation (ii) as follows;
ρ(ave) = [tex]\int\limits^5_0 {(11x + 4)} \, dx[/tex]
ρ(ave) = [[tex]\frac{11x^2}{2}[/tex] + 4x ]₀⁵
Substitute the integral limits into the equation;
ρ(ave) = [ [tex]\frac{11(5)^2}{2}[/tex] + 4(5) ] - [[tex]\frac{11(0)^2}{2}[/tex] + 4(0)]
ρ(ave) = [ [tex]\frac{11(5)^2}{2}[/tex] + 4(5)]
ρ(ave) = [ 137.5 + 20)]
ρ(ave) = 157.5
Therefore, the average density of the rod is 157.5kg/m
