Chapter 7 - Principles Of Integral Evaluation - 7.2 Integration By Parts - Exercises Set 7.2 - Page 499: 55

$$\eqalign{ & \left( {\text{a}} \right)A = 1 \cr & \left( {\text{b}} \right)A = \pi \left( {e - 2} \right) \cr} $$

$$\eqalign{ & \left( {\text{a}} \right){\text{Enclosed by }}y = \ln x,{\text{ }}x = e{\text{ and }}y = 0 \cr & {\text{The area is given by}} \cr & A = \int_1^e {\ln x} dx \cr & {\text{Integrating by parts}} \cr & u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr & dv = dx,{\text{ }}v = x \cr & \int {\ln xdx} = x\ln x - \int {x\left( {\frac{1}{x}} \right)} dx \cr & \int {\ln xdx} = x\ln x - \int {dx} \cr & \int {\ln xdx} = x\ln x - x + C \cr & {\text{Therefore,}} \cr & A = \int_1^e {\ln x} dx \cr & A = \left[ {x\ln x - x} \right]_1^e \cr & A = \left( {e\ln e - e} \right) - \left( {1\ln 1 - 1} \right) \cr & A = e - e + 1 \cr & A = 1 \cr & \left( {\text{b}} \right){\text{Generated by }}y = \ln x,{\text{ }}x = e{\text{ and }}y = 0,{\text{ revolved about }}x \cr & {\text{The volume is given by}} \cr & A = \int_1^e {\pi {{\left( {\ln x} \right)}^2}dx} \cr & {\text{Integrating by parts}} \cr & u = {\left( {\ln x} \right)^2},{\text{ }}du = 2\left( {\ln x} \right)\left( {\frac{1}{x}} \right)dx \cr & dv = dx,{\text{ }}v = x \cr & \int {\ln xdx} = x{\left( {\ln x} \right)^2} - \int {2x\left( {\ln x} \right)\left( {\frac{1}{x}} \right)} dx \cr & \int {\ln xdx} = x{\left( {\ln x} \right)^2} - 2\int {\ln x} dx \cr & \int {\ln xdx} = x{\left( {\ln x} \right)^2} - 2\left( {x\ln x - x} \right) \cr & \int {\ln xdx} = x{\left( {\ln x} \right)^2} - 2x\ln x + 2x + C \cr & {\text{Therefore,}} \cr & A = \pi \int_1^e {{{\left( {\ln x} \right)}^2}dx} \cr & A = \pi \left[ {x{{\left( {\ln x} \right)}^2} - 2x\ln x + 2x} \right]_1^e \cr & A = \pi \left[ {e{{\left( {\ln e} \right)}^2} - 2e\ln e + 2e} \right] - \pi \left[ {1{{\left( {\ln 1} \right)}^2} - 2\ln 1 + 2\left( 1 \right)} \right] \cr & A = \pi \left( {e - 2} \right) \cr} $$