Transcription of AP Calculus—Integration Practice
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AP Calculus—Integration Practice
AP Calculus Integration by Idea:Ifu=f(x), thendu=f (x) have xdxx4+ 1u=x2=dx= 2xdx12 duu2+ 1=12tan 1u+C=12tan 1x2+CPractice Problems:1. x3 4 +x4dx2. dxxlnx3. (x+ 5)dx x+ 44. In each integral below, find the integernthat allows for an integration bysub-stitution. Then perform the integration.(a) xn 1 x4dx(b) xn 1 x4dx(there are two very natural choices forn).(c) xn1 +x10dx(there are two very natural choices forn).(d) x61 +xndx(e) xne x2dx(f) xne2x5dx(g) x5 1 xndx(h) x6 1 xndx(i) dxxnlnx(j) dxxn(lnx)7(k) xnsin(x6)dx(l) sinnxcosx 3 + sin4xdx(m) sin3xcosx 3 + by Parts:Basic Idea: udv=uv v du(Try to substituteuso thatdudxis simpler thanuand so thatvis no more complicatedthandv.) have xsinxdxu=x, dv= sinxdx=du=dx, v= cosxdx xcosx+ cosxdx= xcosx+ sinxNotice that in the above, settingu=xyieldsdudx= 1( ,du=dx), which issimpleranddv= sinxdxwhich givesv= cosx, which is no more Problems:1.
AP Calculus—Integration Practice I. Integration by substitition. Basic Idea: If u= f(x), then du= f0(x)dx: Example. We have Z xdx x4 +1 u= x2 dx= 2xdx 1 2 Z du u2 +1 1 2 tan 1 u+C 1 2 tan 1 x2 +C Practice Problems: 1. Z x3 p
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