AC6.5 - 6.5 CONCEPT QUESTIONS, page 447 1. Approach I: We...

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6 Integration 550 6.5 CONCEPT QUESTIONS, page 447 1. Approach I: We first find the indefinite integral. Let 3 1 ux = + so that 2 3 du x dx = or 2 1 3 x dx du = . Then 23 2 2 3 3 3 11 1 39 9 (1 ) ) . x xd xu d u u C x C += = + = + + ∫∫ Therefore, 1 1 2 3 3 7 99 9 0 0 ) ) ( 8 1 ). xx d x x + = = Approach II: Transform the definite integral in x into an integral in u : Let 3 1 =+ , so that 2 3 du x dx = or 2 1 3 x dx du = . Next, find the limits of integration with respect to u : If x = 0, then 3 011 u = += and if x = 1, then 3 112 . u =+= Therefore, 12 2 2 2 3 7 1 9 9 1 01 ) ( 8 1 d x ud u u == = 2. See the definition on page 445 of the text. EXERCISES 6.5, page 447 1. Let u = x 2 1 so that du = 2 xdx or = 1 2 du . Also, if x = 0, then u = 1 and if x = 2, then u = 3. So 3 3 4 1 8 1 1 28 8 ( 1) (81) (1) 10. dx udu u −= = =− = 2. Let u = 2 x 3 1 so that du = 6 x 2 dx or x 2 dx = 1 6 du . Also, if x = 0, u = 1, and if x = 1, then u = 1. So d x u d u u 4 0 1 1 6 4 1 1 1 30 5 1 1 1 30 1 30 1 15 21 () ( ) . = = = zz 3. Let u = 5 x 2 + 4 so that du = 10 x dx or x dx = 1 10 du . Also, if x = 0, then u = 4, and if x = 1, then u = 9. So 9 19 / 2 3 / 2 04 4 31 9 1 55 5 5 10 35 4 ( 2 7 ) ( 8 xx d x = = 4. Let u = 3 x 2 2 so that du = 6 x dx or x dx = 1 6 du . Also, if x = 1, then u = 1, and if x = 3, then u = 25. So,
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6 Integration 551 xx d x ud u u 32 1 6 1 9 1 9 125 1 9 1 124 9 21 2 3 2 1 25 1 3 1 25 −= = = = z z // () ( ) . 5. Let u = x 3 + 1 so that du = 3 x 2 dx or x 2 dx = 1 3 du . Also, if x = 0, then u = 1, and if x = 2, then u = 9. So, 9 29 23 3 / 2 3 / 2 5 / 2 01 1 484 222 555 5 3 ( 1) (243) (1) . d x u d u u += ==− = ∫∫ 6. Let u = 2 x 1 so that du = 2 dx or dx = 1 2 du . Also, if x = 1, then u = 1 and if x = 5 then u = 9. So ( ) ( ) . / 2 1 2187 1 52 72 1 9 1 5 1 9 1 2 1 7 1 7 1 7 2186 7 xd xu d u u == = z z 7. Let u = 2 x + 1 so that du = 2 dx or dx = 1 2 du . Also, if x = 0, then u = 1 and if x = 2 then u = 5. So 5 25 5 1/2 1 1 11 1 1 51 . 22 dx du u du u = = + 8. Let u = x 2 + 5 so that du = 2 x dx or x dx = 1 2 du . Also, if x = 0 then u = 5 and if x = 2 then u = 9. So x x dx du u u 2 12 5 9 5 9 0 2 5 1 2 35 + = X Z Y X Z Y / . 9. . 4 1 2 x u x z Put so that du = 2 dx or dx = 1 2 du . Then if x = 1, u = 1 and if x = 2, then u = 3. Then ( ) . 2 1 243 1 24 4 1 2 4 1 10 5 1 3 1 2 1 3 1 10 121 5 1 5 x u d u u = = z z 10. Let u = x 2 + 4 x 8 so that du = (2 x + 4) dx . Also, if x = 1 then u = 3 and if x = 2, then u = 4. So ( ) ( ) ( 8 1 ) . 2 4 4 8 256 233 1 4 4 3 4 3 4 1 2 1 4 1 4 175 4 d u u ++ = = = z z 11. Let u = x 3 +1 ±so ±that± du = 3 x 2 dx or x 2 dx = 1 3 du . Also, if x = 1,
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6 Integration 552 then u = 0 and if x = 1, then u = 2. So xx d x ud u u 23 4 1 3 4 1 15 5 0 2 1 1 0 2 32 15 1 () . += == z z 12. Let u = x 4 + 3 x so that du = (4 x 3 + 3) dx = 4( x 3 + 3 4 ) dx or dx = x 3 3 4 + = 1 4 du . Also, if x =1 ,±then ± u = 4 and if x = 2, then u = 22. So 22 22 2 34 2 2 3 9 11 2 1 1 48 8 1 6 1 7 6 1 7 6 14 4 (3 ) .
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This note was uploaded on 03/26/2011 for the course ACCOUNTING 102 taught by Professor Albert during the Spring '11 term at Karachi Institute of Economics & Technology.

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AC6.5 - 6.5 CONCEPT QUESTIONS, page 447 1. Approach I: We...

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