# FinalPracticeSheetSolutions.pdf - MAT1322 Final exam...

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MAT1322, Final exam Practice sheet: Integration Area, Volumes, Applications of Integrals Compute the area of the following regions: 1. Bounded region delimited by y = e x , y = 1 , x = 2. 2. Bounded region delimited by y = sin( x ) , y = cos( x ) , x = 0 , y = 0 , x = π 2 . 3. Bounded region delimited by y = 3 x - 6 , y = 9 /x, x = 4. Solution : 1. This region is from x = 0 to x = 2. Throughout this interval, the curve y = e x lies above y = 1. Thus, the area A is A = 2 0 ( e x - 1) dx = e x - x 2 0 = e 2 - 2 - ( e 0 - 0) = e 2 - 3 2. This region consists of two parts: one from x = 0 to x = π 4 in which y = cos( x ) lies above y = sin( x ), and the other from x = π 4 to x = π 2 in which y = sin( x ) lies above y = cos( x ). Thus, the total area of these two regions is A = π/ 4 0 (cos( x ) - sin( x )) dx + π/ 2 π/ 4 (sin( x ) - cos( x )) dx = sin( x ) + cos( x ) π/ 4 0 + - cos( x ) - sin( x ) π/ 2 π/ 4 = (sin( π/ 4) + cos( π/ 4)) - (sin(0) + cos(0)) + ( - cos( π/ 2) - sin( π/ 2)) - ( - cos( π/ 4) - sin( π/ 4)) = 2 2 + 2 2 - (0 + 1) + (0 - 1) - - 2 2 - 2 2 = 2 2 - 2 3. This region spans the interval from x = 3 (the positive point of intersection of y = 3 x - 6 and y = 9 /x ) to x = 4, and througout this interval, the line y = 3 x - 6 lies above the curve y = 9 /x . Thus, the area A is A = 4 3 ( 3 x - 6 - 9 x ) dx = 3 x 2 2 - 6 x + 9 x 2 4 3 = ( 3 · 16 2 - 6(4) + 9 16 ) - ( 3 · 9 2 - 6(3) + 9 9 ) = 65 16 Compute the volume of the following solids: 1. y = e x , y = 1 , x = 2 rotated around y = - 2. 2. y = sin( x ) , y = cos( x ) , x = 0 , y = 0 , x = 2 2 about y = - 1. 1
3. y = 3 x - 6 , y = 9 /x, y = 0 , x = 4 about x = - 2. Solution : 1. 133 . 0630975 2. 4 . 126940766 3. We use the method of cylindrical shells. The functions y 1 ( x ) = 3 x - 6 and y 2 ( x ) = 9 /x intersect at - 1 and 3, however we only need the positive intersect. The function y 1 crosses the x -axis at x = 2 (draw a sketch). The volume has to be compute in two parts: using y 1 from x = 2 to x = 3 and y 2 from x = 3 to x = 4. We rotate around the axis x = - 2, hence the radius for the method of cylindrical shells becomes r ( x ) = ( x - ( - 2)) = x + 2. Volume 1: V 1 = 2 π 3 2 ( x + 2)(3 x - 6) dx = 2 π 3 2 3 x 2 - 12 dx = 14 π = 43 . 98229716 . Volume 2: V 2 = 2 π 4 3 ( x + 2) · 9 x dx = 89 . 08474367 . So the total volume is V = V 1 + V 2 = 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. Find the work done. Solution : The total work is the sum of the work from lifting the coal and the cable. We denote by x the depth from the top of the mine. The mass of a tiny segment of the cable of length Δ x is 2Δ x kg. Hence its weight is 2 g Δ x N, where g is the gravitational constant. Also the work required to lift the segment of cable of length Δ x to the height x is Δ W = (2 g Δ x ) × x J . Finally, the work required to lift the cable is W cable = 500 0 2 gxdx = 2 g × 1 2 x 2 500 0 = g × 500 2 2 , 45 × 10 6 J by using g = 9 . 81 m/s 2 . The work to lift the coal is just W coal = 800 × 500 = 400000 J (since the force of the bag of coal does not change with the depth). The total work is hence W = W cable + W coal 2 . 85 × 10 6 J . 2
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