1007_Test4_Sol

1007_Test4_Sol - MATH 1007 A Test #4 November 17, 2006...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 1007 A Test #4 November 17, 2006 SOLUTIONS 1. [2 marks] Z 3 x 2 dx Solution: Z 3 x 2 dx = Z x 2 / 3 dx = 1 5 / 3 x 5 / 3 + C = 3 5 x 5 / 3 + C 2. [2 marks] Z e- x + sin x dx Solution: Z e- x + sin x dx =- e- x- cos x + C 3. [2 marks] Z 2 1 8 x 3- 9 x 2 dx Solution: Z 2 1 8 x 3- 9 x 2 dx = 8 1 4 x 4- 9 1 3 x 3 2 1 = 2 x 4- 3 x 3 2 1 = (2 2 4- 3 2 3 )- (2 1 4- 3 1 3 ) = (32- 24)- (- 1) = 9 1 4. [3 marks] Z 4 (1 + 2 x ) 3 dx Solution: Z 4 (1 + 2 x ) 3 dx = Z 2 (1 + 2 x ) 3 2 dx let u = 1 + 2 x, du = 2 dx = Z 2 u 3 du =- u- 2 + C =- (1 + 2 x )- 2 + C 5. [3 marks] Z x 3 2 x 4- 1 dx Solution: Z x 3 2 x 4- 1 dx = Z 2 x 4- 1 1 8 8 x 3 dx let u = 2 x 4- 1 , du = 8 x 3 dx = Z u 1 8 du = 1 8 1 3 / 2 u 3 / 2 + C = 1 12 (2 x 4- 1) 3 / 2 + C 2 6. [3 marks] Z 1 5- x dx Solution: Z 1 5- x dx = Z- 1 5- x (- 1) dx let u = 5- x, du =- dx = Z- 1 u du =- ln | u | + C =- ln | 5- x | + C 7. [4 marks] Z 2 1 x 2- 4 x 3 dx Solution: Z 2 1 x 2- 4 x...
View Full Document

This note was uploaded on 03/22/2010 for the course MATH 1007 taught by Professor Unknown during the Fall '07 term at Carleton.

Page1 / 6

1007_Test4_Sol - MATH 1007 A Test #4 November 17, 2006...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online