Calculus of a Single Variable, Chapter 7, 7.5, Section 7.5, Problem 38

Work is done when a constant force F is applied to move an object a distance D. It is defined with a formula W = FD.
For expanding gas, we denote the work done as W =P* DeltaV.
With the stated assumption pressure is inversely proportional to volume, we let P =k/V where k is the proportionality constant.
Then plug-in P = k/V on W =P* DeltaV , we get: W =k/V* DeltaV or (kDeltaV)/V
The integral of work done will be W=int_(V_1)^(V_2)(kdV)/V
To solve for the proportionality constant (k) , we plug-in the initial condition:
P= 2500 pounds and V_1= 1 on P= k/V.
2500 = k/1
k = 2500*1 =2500
To solve for the work done by the gas to expand the volume, we plug-in k=2500 , V_1=1 , and V_2=3 on W=int_(V_1)^(V_2)(kdV)/V .
W=int_(1)^(3)(2500(dV))/V
Apply basic integration property: int cf(x)dx = c int f(x)dx .
W=2500int_(1)^(3)(dV)/V .
Apply basis integration formula for logarithm.
W=2500 ln(V)|_(1)^(3)
Apply definite integral formula: F(x)|_a^b = F(b)-F(a) .
W=2500 ln(3)-2500 ln(1)
W=2500 [ ln(3)- ln(1)]
Apply natural logarithm property: ln(x/y) = ln(x)- ln(y) .
W=2500 [ ln(3/1)]
W=2500 ln(3) or 2746.53 ft-lbs.