## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 8: Techniques of Integration - Practice Exercises - Page 519: 54

#### Answer

$$- 1$$

#### Work Step by Step

\eqalign{ & \int_0^1 {\ln x} dx \cr & {\text{The domain of the function }}f\left( x \right) = \ln x{\text{ is }}\left( {0,\infty } \right),\cr & {\text{so the integral is not defined for x=0}} \cr & {\text{By the definition of the improper integrals, we have}} \cr & \int_0^1 {\ln x} dx = \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\ln x} dx \cr & \cr & {\text{Integrate }}\int {\ln x} dx,{\text{ Using the method of integration by parts}} \cr & \,\,\,\,\,\,\,\,\,{\text{Let }}u = \ln x,\,\,\,\,\,du = \frac{1}{x}dx \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = dx,\,\,\,\,\,v = x \cr & \int {udv} = uv - \int {vdu} \,\,\,\, \to \,\,\,\,\int {\ln x} dx = x\ln x - \int {x\left( {\frac{1}{x}} \right)dx} \cr & \int {\ln x} dx = x\ln x - \int {dx} \cr & \int {\ln x} dx = x\ln x - x + C \cr & \cr & {\text{Then}}{\text{,}} \cr & \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\ln x} dx = \mathop {\lim }\limits_{a \to {0^ + }} \left( {x\ln x - x} \right)_a^1 \cr & = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\left( {1\ln 1 - 1} \right) - \left( {a\ln a - a} \right)} \right] \cr & = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\left( { - 1} \right) - \left( {a\ln a - a} \right)} \right] \cr & = \mathop {\lim }\limits_{a \to {0^ + }} \left( { - 1} \right) - \mathop {\lim }\limits_{a \to {0^ + }} \left( {a\ln a} \right) + \mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) \cr & \cr & {\text{Write }}\mathop {\lim }\limits_{a \to {0^ + }} \left( {a\ln a} \right){\text{ as }}\mathop {\lim }\limits_{a \to {0^ + }} \frac{{\ln a}}{{\left( {1/a} \right)}} \cr & = \mathop {\lim }\limits_{a \to {0^ + }} \left( { - 1} \right) - \mathop {\lim }\limits_{a \to {0^ + }} \frac{{\ln a}}{{\left( {1/a} \right)}} + \mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) \cr & {\text{Using the L'Hopital's rule for }}\mathop {\lim }\limits_{a \to {0^ + }} \frac{{\ln a}}{{\left( {1/a} \right)}} \cr & = \mathop {\lim }\limits_{a \to {0^ + }} \left( { - 1} \right) - \mathop {\lim }\limits_{a \to {0^ + }} \frac{{\left( {1/a} \right)}}{{ - \left( {1/{a^2}} \right)}} + \mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) \cr & = \mathop {\lim }\limits_{a \to {0^ + }} \left( { - 1} \right) + \mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) + \mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) \cr & = \mathop {\lim }\limits_{a \to {0^ + }} \left( { - 1} \right) + 2\mathop {\lim }\limits_{a \to {0^ + }} \left( a \right) \cr & {\text{Evaluate the limit when }}a \to {0^ + } \cr & = \left( { - 1} \right) + 2\left( 0 \right) \cr & = - 1 \cr & \cr & {\text{Then}}{\text{,}} \cr & \int_0^1 {\ln x} dx = - 1 \cr}

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