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SOLUTION 7: If y=(1/2)(ex+e−x) for 0≤x≤ln2, then
dydx=(1/2)(ex+e−x(−1))=(1/2)(ex−e−x)
so that
ARC=∫ln20√1+(dydx)2 dx
=∫ln20√1+((1/2)(ex−e−x))2 dx
=∫ln20√1+(1/2)2(ex−e−x)2 dx
=∫ln20√1+(1/4)((ex)2−2exe−x+(e−x)2) dx
=∫ln20√1+(1/4)(e2x−2e0+e−2x) dx
=∫ln20√1+(1/4)(e2x−2(1)+e−2x) dx
=∫ln20√1+(1/4)e2x−(1/2)+(1/4)e−2x dx
=∫ln20√(1/4)e2x+(1/2)+(1/4)e−2x dx
=∫ln20√e2x4+12+14e2x dx
=∫ln20√e2x4e2xe2x+122e2x2e2x+14e2x dx
=∫ln20√e4x+2e2x+14e2x dx
=∫ln20√e4x+2e2x+1√4e2x dx
=∫ln20√(e2x+1)2√(2ex)2 dx
=∫ln20e2x+12ex dx
=∫ln20(e2x2ex+12ex) dx
=∫ln20((1/2)ex+(1/2)e−x) dx
=((1/2)ex+(1/2)e−x−1) |ln20
=(ex2−12ex) |ln20
=(eln22−12eln2)−(e02−12e0)
(Recall that elnz=z.)
=(22−12(2))−(12−12(1))
=1−14
=34
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