Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 391: 91

$$32,760$$

$$\eqalign{ & \int_2^4 {{x^{2x}}\left( {1 + \ln x} \right)} dx \cr & or \cr & \int_2^4 {{x^x}\left( {1 + \ln x} \right){x^x}} dx \cr & {\text{Use substitution:}}\cr & {\text{Let }}u = {x^{x}}\cr & {\text{Find the derivative of }}u{\text{; take the natural logaritm}} \cr & {\text{on both sides of the equation}} \cr & \,\,\,\,\,\ln u = \ln {x^x} \cr & {\text{use power property for logarithms}} \cr & \,\,\,\,\,\ln u = x\ln x \cr & {\text{Differentiate both sides with respect to }}x \cr & \,\,\,\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{d}{{dx}}\left[ {x\ln x} \right] \cr & {\text{use product rule for }}\frac{d}{{dx}}\left[ {x\ln x} \right] \cr & \,\,\,\frac{d}{{dx}}\left[ {\ln u} \right] = x\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ x \right] \cr & {\text{solving derivatives}} \cr & \,\,\,\frac{1}{u}\frac{{du}}{{dx}} = x\left( {\frac{1}{x}} \right) + \ln x\left( 1 \right) \cr & \,\,\,\frac{1}{u}\frac{{du}}{{dx}} = 1 + \ln x \cr & \,\,\,\frac{{du}}{{dx}} = u\left( {1 + \ln x} \right) \cr & \,\,\,\frac{{du}}{{dx}} = {x^x}\left( {1 + \ln x} \right) \cr & \,\,\,du = {x^x}\left( {1 + \ln x} \right)dx \cr & \cr & {\text{changing the limits of integration}} \cr & \,\,\,\,\,\,{\text{For }}x = 4,{\text{ }}u = {4^4} = 256 \cr & \,\,\,\,\,\,{\text{For }}x = 2,{\text{ }}u = {2^2} = 4 \cr & {\text{write the integral in terms of }}u \cr & \int_2^4 {{x^x}\left( {1 + \ln x} \right){x^x}} dx = \int_4^{256} {udu} \cr & {\text{integrate using the power rule}} \cr & = \left( {\frac{{{u^2}}}{2}} \right)_4^{256} \cr & {\text{use fundamental theorem of calculus: }}\cr & \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr & = \left( {\frac{{{{\left( {256} \right)}^2}}}{2} - \frac{{{{\left( 4 \right)}^2}}}{2}} \right) \cr & {\text{simplifying}} \cr & = 32,760 \cr} $$