MATH100 Study Guide - Final Guide: Even And Odd Functions, Scilab, The Sketch
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MATH100 Full Course Notes
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Solutions for math 100 2003 final: evaluating the integrals (a) the substitution x = u2 =) dx = 2u du, gives. 1 + u2 = 2 tan(cid:0)1 (u) + c = 2 tan(cid:0)1(cid:0)px(cid:1) + c: (b) observing that the integrand is an odd function and the integration domain is symmetric about x = 0, implies. Z 1 (cid:0)1 sin (x + tan (x)) dx = 0: (c) the substitution u = exp (x) =) du = exp (x) dx, gives p1 (cid:0) u2 pe(cid:0)2x (cid:0) 1 p1 (cid:0) e2x. = sin(cid:0)1 (u) + c = sin(cid:0)1 (ex) + c: (d) the substitution u = tan(cid:0)1 (x) =) u =(cid:0)1 + x2(cid:1)(cid:0)1 dx, gives. 0 eu du = exp(cid:0)tan(cid:0)1 ((cid:25)=4)(cid:1) (cid:0) 1: sawdust is falling onto a pile at a rate of 0:5 m3= min. The volume of a right circular cone is given by v = (cid:25)r2h=3 where r is the radius of the base and h is the height.