Solutions to homework exercises - University of Exeter

BEE1024 –Mathematics for EconomistsHomework -SolutionWeek 8Juliette StephensonAmr AlgarhiDepartment of EconomicsUniversity of ExeterProblem 1 Solve by partial integration. Check your answers by di¤erentiation.Za) xe x 2 dxZb) (3 2x) e x dxSolution 1 a) We set(Check:u = x v = 2e x 2u 0 = 1 v 0 = e x 2ddx 2e x de y d x2 = 22dy dx = ey = e x 2where y = x 2 ) Then Zxe x 2 dx = 2xex2Z2e x 2 dx = 2xex2 4e x 2 + C:Check:b) We setddx 2xe x2 4e x x x2 + C = xe 2 + 2e 2 2e x x2 = xe 2u = 3 2x v = e xu 0 = 2 v 0 = e xThen Z Z(3 2x) e x dx = (3 2x) e x ( 2) e x dxZ= (3 2x) e x 2 e x dx = (3 2x) e x + 2e x + CCheck:= 2xe x e x + Cddx 2xe x e x + C = 2xe x + 2e x + e x = (3 2x) e xProblem 2 Solve by substitution using u = x 2 1 and u = 3x + 5 respectively. Checkyour answers by di¤erentiation.Za) 2xe x2 1 dxZ1b)3x + 5 dx

Solution 2 a) u = x 2Z2xe x2 1 dx =1 implies dudxZ1= 2x or dx = du: Hence2xZ2xe u 12x du =e u du = e u + C = e x2 1 + C:Test:d e x2 1 + C = ddxdu (eu ) ddx x2 1 = e u 2x = 2xe x2 1b) u = 3x + 5 implies du = 3dx or dx = 1 du. Hence3ZZ 1 1 13x + 5 dx = u 3 du = 1 Z3u 1 dx = 1 3 ln juj + C = 1 ln j3x + 5j + C3Test: d 1dx 3 ln j3x + 5j + CProblem 3 “Multiply”Solution 32 1 2 3 4 5 64= 1 d3 dx (ln j3x + 5j) = 1 3= 1 13 u 3 = 13x + 52 1 2 3 4 5 6454321Problem 4 Consider the matricesA = 1 2 34 5 6354321375 =?d ln juj d (3x + 5)du dx75 = 1 5 + 2 4 + 3 3 + 4 2 + 5 1 = 35 5 4 3; B =2 1 02; C = 41 0 10 1 01 0 1a) Which of the following products is de…ned: AC 0 , CA, CB, CC, CA 0 , CB 0 ?b) Calculate BC 0 and CC 0 .Solution 4 a)AC 0 CA CB CC CA 0 CB 02 3, 3 3 3 3, 2 3 2 3, 2 3 3 3, 3 3 3 3, 3 2 3 3, 3 2yes(2 3) no no yes (3 3) yes (3 2) yes (3 2)235