# B let u x 2 1 and then du 2 xdx x 2 u 1 and z 2 4 x 3

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(b) Let u = x 2 + 1 and then du = 2 xdx, x 2 = u - 1 and Z 2 0 4 x 3 ( x 2 + 1) 2 dx = Z 5 1 2 . u - 1 u 2 du = Z 5 1 (2 u - 1 - 2 u - 2 ) du = (2 ln | u | + 2 u - 1 ) 5 1 = 2 ln 5 - 8 5 44. (a) Z π/ 4 0 sec xdx . 88 using midpoint evalu- ation with n 10 . (b) Z π/ 4 0 sec 2 xdx = (tan x ) π/ 4 0 = 1 . 45. 1 2 Z 4 0 f ( u ) du. 46. 1 3 Z 8 1 f ( u ) du. 47. Z 1 0 f ( u ) du. 48. Z 4 0 f ( x ) x dx = 2 Z 2 0 f ( u ) du. 49. Z a - a f ( x ) dx = Z 0 - a f ( x ) dx + Z a 0 f ( x ) dx Let u = - x and du = - dx in the first integral. Then Z a - a f ( x ) = - Z 0 - a f ( - u ) du + Z a 0 f ( x ) dx = Z a 0 f ( - u ) du + Z a 0 f ( x ) dx If f is even, then f ( - u ) = f ( u ) , and so Z a - a f ( x ) dx = Z a 0 f ( u ) du + Z a 0 f ( x ) dx = Z a 0 f ( x ) dx + Z a 0 f ( x ) dx = 2 Z a 0 f ( x ) dx If f is odd, then f ( - u ) = - f ( u ) , and so Z a - a f ( x ) dx = - Z a 0 f ( u ) du + Z a 0 f ( x ) dx = - Z a 0 f ( x ) dx + Z a 0 f ( x ) dx = 0 50. First, let u = x - T, then for any a, Z a + T T f ( x ) dx = Z a 0 f ( u + T ) du = Z a 0 f ( u ) du = Z a 0 f ( x ) dx If we let a = T, then we get Z T a f ( x ) dx = Z 2 T T f ( x ) dx. If we let a = 2 T, then we get Z 2 T 0 f ( x ) dx = Z 3 T T f ( x ) dx and then Z T 0 f ( x ) dx = Z 2 T T f ( x ) dx = Z 2 T 0 f ( x ) dx - Z T 0 f ( x ) dx = Z 3 T T f ( x ) dx - Z 2 T T f ( x ) dx = Z 3 T 2 T f ( x ) dx It is straight forward to see that for any integer i , Z T 0 f ( x ) dx = Z ( i +1) T iT f ( x ) dx Now suppose 0 a T , then Z T 0 f ( x ) dx - Z a + T a dx
4.6. INTEGRATION BY SUBSTITUTION 281 = Z a 0 f ( x ) dx - Z a + T T f ( x ) dx So Z T 0 f ( x ) dx = Z a + T a dx Now suppose a is any number. Then a must lie in some interval [ iT, ( i +1) T ] for some interger i . Use the similar method as in above, we shall get Z ( i +1) T iT f ( x ) dx = Z a + T a f ( x ) dx And since Z ( i +1) T iT f ( x ) dx = Z T 0 f ( x ) dx, we get Z T 0 f ( x ) dx = Z a + T a f ( x ) dx 51. (a) Let u = 10 - x , so that du = - dx . Then, I = Z 10 0 x x + 10 - x dx = - Z x =10 x =0 10 - u 10 - u + u du = - Z u =0 u =10 10 - u 10 - u + u du = Z u =10 u =0 10 - u 10 - u + u du I = Z x =10 x =0 10 - x 10 - x + x dx The last equation follows from the previ- ous one because u and x are dummy vari- ables of integration. Now note that x x + 10 - x = x + 10 - x - 10 - x x + 10 - x = 1 - 10 - x x + 10 - x Thus, Z 10 0 x x + 10 - x dx = Z 10 0 1 - 10 - x x + 10 - x dx = Z 10 0 1 dx - Z 10 0 10 - x x + 10 - x dx I = Z 10 0 1 dx - I 2 I = 10 I = 5 (b) Let u = a - x , so that du = - dx Then, I = Z a 0 f ( x ) f ( x ) + f ( a - x ) dx = - Z 0 a f ( a - u ) f ( a - u ) + f ( u ) du = Z a 0 f ( a - u ) f ( a - u ) + f ( u ) du I = Z a 0 f ( a - x ) f ( a - x ) + f ( x ) dx The last equation follows from the previ- ous one because u and x are dummy vari- ables of integration. Now note that f ( x ) f ( x ) + f ( a - x ) = f ( x ) + f ( a - x ) - f ( a - x ) f ( x ) + f ( a - x ) = 1 - f ( a - x ) f ( a - x ) + f ( x ) Thus, Z a 0 f ( x ) f ( x ) + f ( a - x ) dx = Z a 0 1 - f ( a - x ) f ( a - x ) + f ( x ) dx = Z a 0 1 dx - Z a 0 f ( a - x ) f ( a - x ) + f ( x ) dx 2 I = a I = a/ 2 52. (a) Let u = 6 - x , so that du = - dx . Then, I = Z 4 2 sin 2 (9 - x ) sin 2 (9 - x ) + sin 2 ( x + 3) dx = - Z 2 4 sin 2 ( u + 3) sin 2 ( u + 3) + sin 2 (9 - u ) du = Z 4 2 sin 2 ( u + 3) sin 2 ( u + 3) + sin 2 (9 - u ) du = Z 4 2 sin 2 ( x + 3) sin 2 ( x + 3) + sin 2 (9 - x ) dx = Z 4 2 1 - sin 2 (9 - x ) sin 2 ( x + 3) + sin 2 (9 - x ) dx I = Z 4 2 1 dx - I 2 I = 2 I = 1 (b) Let u = 6 - x , so that du = - dx . Then, I = Z 4 2 f (9 - x ) f (9 - x ) + f ( x + 3) dx = - Z 2 4 f ( u + 3) f ( u + 3) + f (9 - u ) du = Z 4 2 f ( u + 3) f ( u + 3) + f (9 - u ) du = Z 4 2 f ( x + 3) f ( x + 3) + f (9 - x ) dx
282 CHAPTER 4. INTEGRATION = Z 4 2 1 - f (9 - x ) f ( x + 3) + f (9 - x ) dx I = Z 4 2 1 dx - I 2 I = 2 I = 1 53. Let 6 - u = x +4; that is, let u = 2 - x , so that du = - dx .