F eg 2 find the point at which the function f x e x

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f.E.g. 2. Find the point at which the functionf(x) =exhas its value equal to its averaged value in theinterval[0,2].
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6. Arc length - length of curvesConsider an arbitrary curvey=f(x), recall that when we introduced differentiation we discussed thata sufficiently small portion of this curve can be approximated by a straight line. Then the derivativedydxof the functiony=f(x)at the midpoint of this curve portion can be approximated by the fractiontriangleytrianglex,wheretrianglexandtriangleyare respectively the differences inx-coordinates andy-coordinates of the endpointsof the curve portion. In this case,trianglex,triangley0andtriangleytrianglexcan be interpreted as theslopeof the curve atthe midpoint of this small curve portion.As shown in the above diagram, let the curve extends fromx=atox=b.We divide the interval[a, b]intonsub-intervals so that each has a difference inx-coordinatestrianglex=b-an. Iftrianglex0fornissufficiently large (n→ ∞), in thek-th sub-interval fork= 1, . . . , n, we may approximate the length ofthe curve portion astriangleLk=radicalbigtrianglex2+triangley2k, wheretriangleykis the difference iny-coordinates of the endpointsin thek-th sub-interval.If the functionf(x)isdifferentiablein the interval[a, b],triangleykcan be approximated bytrianglexdydx|xxk=fprimexk)trianglexfortrianglex0as discussed before, where¯xkis thex-coordinate of a point within thek-thsub-interval fork= 1, . . . , n.
2.Taking the infinite Riemann sum of the length of all curve portions, we may obtain the total lengthLorusually calledarc lengthof the curve in the interval[a, b]asArc lengthL= limn→∞nk=1triangleLk= limn→∞nk=1radicalBig1 + [fprimexk)]2trianglex=´baradicalBig1 + [fprime(x)]2dx.E.g. 1. Find the arc length of the curvey=12(ex+e-x)in thex-interval[-ln 2,ln 2].[ans: 3/2]13
# Instead of integrating along thexdirection, we may also find the arc length of the curve by integrationalong theydirection. In this case, we consider the equation of the curve in the formx=g(y)whichextends fromy=ctoy=d. Similar to the previous approach, we divide the interval[c, d]intonsub-intervals and each has a difference iny-coordinatestriangley=d-cn. In thek-th sub-interval, the derivativegprimeyk)can be approximated by the fractiontrianglexktriangley, i.e.,trianglexksimilarequalgprimeyk)triangley, where¯ykis they-coordinate ofa point within thek-th sub-interval fork= 1, . . . , n. Then, the length of the curve portion at thek-thsub-interval istriangleLk=radicalbigtrianglex2k+triangley2=radicalBigtriangley2[gprimeyk)]2+triangley2=triangleyradicalBig1 + [gprimeyk)]2, andArc lengthL= limn→∞nk=1triangleLk= limn→∞nk=1radicalBig1 + [gprimeyk)]2triangley=´d

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