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SOLUTION 7:    If y=(1/2)(ex+ex) for 0xln2, then dydx=(1/2)(ex+ex(1))=(1/2)(exex) so that ARC=ln201+(dydx)2 dx =ln201+((1/2)(exex))2 dx =ln201+(1/2)2(exex)2 dx =ln201+(1/4)((ex)22exex+(ex)2) dx =ln201+(1/4)(e2x2e0+e2x) dx =ln201+(1/4)(e2x2(1)+e2x) dx =ln201+(1/4)e2x(1/2)+(1/4)e2x dx =ln20(1/4)e2x+(1/2)+(1/4)e2x dx =ln20e2x4+12+14e2x dx =ln20e2x4e2xe2x+122e2x2e2x+14e2x dx =ln20e4x+2e2x+14e2x dx =ln20e4x+2e2x+14e2x dx =ln20(e2x+1)2(2ex)2 dx =ln20e2x+12ex dx =ln20(e2x2ex+12ex) dx =ln20((1/2)ex+(1/2)ex) dx =((1/2)ex+(1/2)ex1) |ln20 =(ex212ex) |ln20 =(eln2212eln2)(e0212e0) (Recall that elnz=z.) =(2212(2))(1212(1)) =114 =34

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