Solutions to homework exercises - University of Exeter

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