MAT1322, Final exam Practice sheet:IntegrationArea, Volumes, Applications of IntegralsCompute the area of the following regions:1. Bounded region delimited byy=ex, y= 1, x= 2.2. Bounded region delimited byy= sin(x), y= cos(x), x= 0, y= 0, x=π2.3. Bounded region delimited byy= 3x-6, y= 9/x, x= 4.Solution :1. This region is fromx= 0 tox= 2. Throughout this interval, the curvey=exlies abovey= 1. Thus,the areaAisA=20(ex-1)dx=ex-x20=e2-2-(e0-0) =e2-32. This region consists of two parts: one fromx= 0 tox=π4in whichy= cos(x) lies abovey= sin(x),and the other fromx=π4tox=π2in whichy= sin(x) lies abovey= cos(x). Thus, the total area ofthese two regions isA=π/40(cos(x)-sin(x))dx+π/2π/4(sin(x)-cos(x))dx=sin(x) + cos(x)π/40+-cos(x)-sin(x)π/2π/4= (sin(π/4) + cos(π/4))-(sin(0) + cos(0)) + (-cos(π/2)-sin(π/2))-(-cos(π/4)-sin(π/4))=√22+√22-(0 + 1) + (0-1)--√22-√22= 2√2-23.(corrected April 9)This region spans the interval fromx= 3 (the positive point of intersection ofy= 3x-6andy= 9/x) tox= 4, and throughout this interval, the liney= 3x-6 lies above the curvey= 9/x.Thus, the areaAisA=43(3x-6-9x)dx=3x22-6x-9 ln|x|43=(3·162-6(4)-9 ln(4))-(3·92-6(3)-9 ln(3))≈1.91Compute the volume of the following solids:1.y=ex, y= 1, x= 2 rotated aroundy=-2.2.y= sin(x), y= cos(x), x= 0, y= 0, x=√22abouty=-1.3.y= 3x-6, y= 9/x, y= 0, x= 4 aboutx=-2.Solution :1. 133.06309752. 4.1269407661
3. We use the method of cylindrical shells. The functionsy1(x) = 3x-6 andy2(x) = 9/xintersect at-1and 3, however we only need the positive intersect. The functiony1crosses thex-axis atx= 2 (drawa sketch). The volume has to be compute in two parts: usingy1fromx= 2 tox= 3 andy2fromx= 3 tox= 4. We rotate around the axisx=-2, hence the radius for the method of cylindricalshells becomesr(x) = (x-(-2)) =x+ 2. Volume 1:V1= 2π32(x+ 2)(3x-6)dx= 2π323x2-12dx= 14π= 43.98229716.Volume 2:V2= 2π43(x+ 2)·9xdx= 89.08474367.So the total volume isV=V1+V2= 133.0670408.Applications of integrals:1. A cable that weighs 2 kg/m is used to lift 800 kg of coal up to a mine shaft which is 500 m deep. Findthe work done.Solution :(corrected April 10)The total work is the sum of the work from lifting the coal and thecable. We denote byxthe depth from the top of the mine.The mass of a tiny segment of the cable of length Δxis 2Δxkg. Hence its weight is 2gΔxN, wheregis the gravitational constant.Also the work required to lift the segment of cable of length Δxto the heightxisΔW= (2gΔx)×xJ.Finally, the work required to lift the cable isWcable=50002gxdx= 2g×12x25000=g×50022 450 000 Jby usingg= 9.8 m/s2.The work to lift the coal is justWcoal= 800×500×9.8 = 3 920 000 J(since the force of the bag of coal does not change with the depth).The total work is henceW=Wcable+Wcoal= 6 370 000 J.