# A and x b as your limits of integration while you are

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aandx=bas your limits of integration while you aredoing a change of variable.This is a good method for those having trouble withmethod 2.setu= 9 +x2,thendu= 2x dx40x9 +x2dx=40129 +x2(2x)dx=x=4x=012u du=12·23u3/2x=4x=0=13(9 +x2)3/240=13(9 + 16)3/2-13(9 + 0)3/2=3223151
Example 1:Findx2(2x3+ 1)4dxu=2x3+ 1du=6x2dx10x2(2x3+ 1)4dx=1016(2x3+ 1)46x2dxx= 0=u(0) = 0 + 1 = 1x= 1=u(1) = 2 + 1 = 310x2(2x3+ 1)4dx=3116u4du=16·15u531=130u531=130(35-15)=130(243-1)=24230=12115152
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Example 2:Findπ/20cosx esinxdxu=sinxdu=cosx dxx= 0=u(0) = sin 0 = 0x=π/2=u(π/2) = sin(π/2) = 1π/20cosx esinxdx=π/20esinxcosx dx=10eudu=[eu]10=e1-e0=e-1153
Example 3:Find212x+ 1x2+xdxu=x2+xdu=2x+ 1dxu(1)=2 + 1 = 3u(2)=4 + 2 = 6212x+ 1x2+xdx=211x2+x(2x+ 1)dx=631udu=11/2u1/263=2u1/263=2(6-3)154
Example 4:Find4 + 2xxdxu=4 + 2xdu=1xdxu(4)=4 + 2(2) = 8u(9)=4 + 2(3) = 10944 + 2xxdx=944 + 2x1xdx=108u du=13/2u3/2108=23u3/2108=23103/2-83/2155
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Integrals with absolute valuesRecall thatf(x) =|x|=xifx0-xifx <0Example 5:Find1-1|x|dxWe split the integral into 2 pieces and change the boundsof integration:1-1|x|dx=0-1|x|dx+10|x|dx=0-1-x dx+10x dx=-12x20-1+12x210=-1202-(-1)2+1212-02=-12(-1) +12(1)=1156
Example 6:Find4-2|x-3|dx(This example across 3 slides)We split the integral atx= 3:4-2|x-3|dx=3-2|x-3|dx+43|x-3|dx=3-2-(x-3)dx+43(x-3)dxTo avoid confusion, do these 2 integrals separately!157
First integral:3-2-(x-3)dx=3-2-x+ 3dx=-12x2+ 3x3-2=-322+ 9--(-2)22-6=-92+ 9 +42+ 6=252158
Second integral:43(x-3)dx=12x2-3x43=1216-12-129-9=-4-92+ 9 =12Putting it all together:4-2|x-3|dx=3-2-(x-3)dx+43(x-3)dx=252+12=262= 13159
45. Using symmetry to simplify integralsSupposefis continuous on [-a, a]. Then(i) Iffis even (that is,f(-x) =f(x)), thena-af(x)dx= 2a0f(x)dx(ii) Iffis odd (that is,f(-x) =-f(x), thena-af(x)dx= 0160
Example 1:Find2-2(x6+ 1)dxSolution:The functionf(x) =x6+ 1 is even sincef(-x) = (-x)6+ 1 =x6+ 1 =f(x).Hence2-2(x6+ 1)dx=220(x6+ 1)dx=217x7+x20=2277+ 2-0=21287+ 2=2847161
Example 2:Find1-1tanx1 +x2+x4dxSolution:The functionf(x) =tanx1+x2+x4is odd sincef(-x) =tan(-x)1 + (-x)2+ (-x)4=-tanx1 +x2+x4=-f(x).Hence1-1tanx1 +x2+x4dx= 0162
46. Areas between curvesDefinition:The areaAof the region bounded byy=f(x),y=g(x), and the linesx=aandx=b, wherefandgare continuous andf(x)g(x) for allxin [a, b] isA=ba[f(x)-g(x)]dxThis is often thought of asA=ba[top-bottom]dxNote: When solving for such areas, we start by drawinga picture!!163
Example 1:Find the area bounded above byy=exand bounded below byy=xbetweenx= 0 andx= 1.First we draw a picture!!ThenArea=ba[top-bottom]dx164
Example 2:Find the area bounded by thexaxis, thegraph ofy=-x2and the linesx= 0 andx= 3.First we draw a picture!!Top Function:Bottom Function:165
ThenArea=ba[top-bottom]dx166
Example 3:Find the area of the region bounded byy= 2x-1 andy=x2-4 and the linesx= 1 andx= 2.