{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–4. Sign up to view the full content.

6 Integration 550 6.5 CONCEPT QUESTIONS, page 447 1. Approach I: We first find the indefinite integral. Let 3 1 u x = + so that 2 3 du x dx = or 2 1 3 x dx du = . Then 2 3 2 2 3 3 3 1 1 1 3 9 9 ( 1) ( 1) . x x dx u du u C x C + = = + = + + Therefore, 1 1 2 3 2 3 3 7 1 1 9 9 9 0 0 ( 1) ( 1) (8 1) . x x dx x + = + = = Approach II: Transform the definite integral in x into an integral in u : Let 3 1 u x = + , 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 0 1 1 u = + = and if x = 1, then 3 1 1 2. u = + = Therefore, 1 2 2 2 3 2 2 3 7 1 1 1 3 9 9 9 1 0 1 ( 1) (8 1) . x x dx u du 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 x dx or x dx = 1 2 du . Also, if x = 0, then u = 1 and if x = 2, then u = 3. So 3 2 3 2 3 3 4 1 8 0 1 1 1 1 1 2 8 8 ( 1) (81) (1) 10. x x dx u du 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 x x dx u du u 2 3 4 0 1 1 6 4 1 1 1 30 5 1 1 1 30 1 30 1 15 2 1 ( ) ( ) . = = = − − = z z 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 1 9 2 1/2 3/ 2 0 4 4 3 19 1 1 1 5 5 5 5 10 3 5 4 (27) (8) . x x dx u du u + = = = = 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,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document