# T bardbl dt integraldisplay 2 t radicalbig 4 t 2 1 dt

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(t)bardbldt=integraldisplay20tradicalbig4t2+ 1dt .This last integral can be evaluated by substi-tution: setu2= 4t2+ 1. Then2u du= 8t dt ,14u du=t dt ,soI=14integraldisplay31u2du=bracketleftBig112u3bracketrightBig31.Consequently,I=136.00510.0 pointsUse the fact thatF= (2xy+ 4 cosy)i+ (x24xsiny)j,is a gradient vector field to evaluate the lineintegralI=integraldisplayCF·ds
Version 039 – EXAM03 – gilbert – (56690)3along a smooth curveCfrom(1, π)toparenleftBig2,π2.1.I=π+ 4correct2.I= 2π+ 43.I=π44.I= 2π45.I=π+ 26.I= 2π2Explanation:Iff(x, y) is a potential function for thegradient vector fieldF= (2xy+ 4 cosy)i+ (x24xsiny)j,then∂f∂x= 2xy+ 4 cosy ,∂f∂y=x24xsiny .Now by the first equation,f(x, y) =x2y+ 4xcosy+D(y)for an arbitrary functionD(y), which by thesecond equation satisfiesx24xsiny+D(y) =x24xsiny ,i.e.,D(y) =Kfor an arbitrary constantK.Thusf(x, y) =x2y+ 4xcosy+K .But thenf(1, π) =π4 +K ,whilefparenleftBig2,π2parenrightBig= 2π+K .Consequently,fparenleftBig2,π2parenrightBigf(1, π) =π+ 4.Evaluate the integralI=integraldisplayC(3y dxx dy)whenCis the graph ofy= cosxon [0, π/2].1.I= 72.I= 4correct3.I= 64.I= 55.I= 8Explanation:WhenCis the graph ofy= cosx, then3y dxx dy= (3 cosx+xsinx)dx .ThusI=integraldisplayπ/20(3 cosx+xsinx)dx=I1+I2.NowI1=integraldisplayπ/203 cosx dx= 3bracketleftBigsinxbracketrightBigπ/20= 3.On the other hand, to evaluateI2we integrateby parts:I2=integraldisplayπ/20xsinx dx=bracketleftBigx(cosx)bracketrightBigπ/20integraldisplayπ/20(cosx)dx= 0 +bracketleftBigsinxbracketrightBigπ/20= 1,Consequently,I=I1+I2= 4.
parenrightBig
00610.0 points
Version 039 – EXAM03 – gilbert – (56690)400710.0 pointsWhich vector fieldFhas graph1. F(x, y) = (2x+ 2)i+yj2. F(x, y) =yi+ cosxj3. F(x, y) = (x+y)i+ (xy)j4. F(x, y) = (xy)i+ (x+y)j correct5. F(x, y) = 2i+xjExplanation:We determineF(x, y) by looking at a fewpoints on the graph. Now on thex-axis,F(x,0) =xi+xj,while on they-axis,F(0, y) =yi+yj.The only one of the given vector fields satis-fying these conditions isF(x, y) = (xy)i+ (x+y)j.008
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