Calculus of a Single Variable, Chapter 7, 7.5, Section 7.5, Problem 9

Hooke's law is written as F = kx
where:
F = force
k = proportionality constant or spring constant
x = length displacement from its natural length
Apply Hooke's Law to the integral application for work: W = int_a^b F dx , we get:
W = int_a^b kx dx
W = k * int_a^b x dx
Apply Power rule for integration: int x^n(dx) = x^(n+1)/(n+1).
W = k * x^(1+1)/(1+1)|_a^b
W = k * x^2/2|_a^b
From the required work 18 ft-lbs, note that the units has "ft" instead of inches. To be consistent, apply the conversion factor: 12 inches = 1 foot then:
4 inches = 1/3 ft
7 inches = 7/12 ft

To solve for k, we consider the initial condition: W =18 ft-lbs to stretch a spring 4 inches or 1/3 ft from its natural length. Stretching 1/3 ft of it natural length implies the boundary values: a=0 to b=1/3 ft.
Applying W = k * x^2/2|_a^b , we get:
18= k * x^2/2|_0^(1/3)
Apply definite integral formula: F(x)|_a^b = F(b)-F(a) .
18 =k [(1/3)^2/2-(0)^2/2]
18 = k * [(1/9)/2 -0]
18 = k *[1/18]
18 = k/18
k =18*18
k= 324
To solve for the work need to stretch the spring with additional 3 inches, we plug-in: k =324 , a=1/3, and b = 7/12 on W = k * x^2/2|_a^b .
Note that stretching "additional 3 inches" from its initial stretch of 4 inches is the same as stretching 7 inches from its natural length.
W= 324 * x^2/2|_((1/3))^((7/12))
W =324 [ (7/12)^2/2 -(1/3)^2/2]
W = 324 [ 49/288 -1/18]
W = 324[11/96]
W=297/8 or 37.125 ft-lbs