MATH 140 Lecture Notes - Lecture 32: Negative Number, Scilab
23 Nov 2015
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=g (f (x))=[g"( f (x))] f "( x)=g(f ( x)) g(f ( x))f "( x) dx= g(u) du=g (u)+c=g ( f ( x))+c: substitute u, then substitute f(x) for u. Examples: ex1: sin7 x cos xdx, let u=sin x du=cos x dx, u7 du= 1. 8 sin 8 x+c: ex2: (3z+1)18 dz, let u=3z +1 du=3dz 1. 57 o du=dz (3x +1)19+c: ex3: t et 2dt, let u=t 2 du=2t 1. 3 +c u: ex5: x3 1 x2 dx, let u=1+x2 x2=u 1 du=2x o. 2)+c: note: you must write the entire integram (the expression that is being integrated) in terms of u for u substitution to work, ex5": x2 1+x2dx, let u=1+x2 u 1=x du=2xdx. 0 sin 4 x cos x dx: let u=sin x du=cos x dx, finding the new bounds: Lower: x=0 sin 0=u=0: thus, the new integral is: . T +1dt: let u=t +1 du=dt, new bounds:
