# hw8 - padilla(tp5647 – HW08 – Gilbert –(56650 1 This...

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

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: padilla (tp5647) – HW08 – Gilbert – (56650) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the integral I = integraldisplay 2 4 √ 16- x 2 dx . 1. I = 4 3 π 2. I = 2 3 π correct 3. I = π 4. I = 1 5. I = 2 3 6. I = 4 3 Explanation: Set x = 4 sin u ; then dx = 4 cos u du and 16- x 2 = 16(1- sin 2 u ) = 8 cos 2 u , while x = 0 = ⇒ u = 0 , x = 2 = ⇒ u = π 6 . In this case I = 4 integraldisplay π/ 6 cos u cos u du = 4 integraldisplay π/ 6 du . Consequently I = 2 3 π . 002 10.0 points Evaluate the integral I = integraldisplay 1 x 2 (2- x 2 ) 3 / 2 dx . 1. I = 2 parenleftBig √ 3 + π 3 parenrightBig 2. I = 2 parenleftBig √ 3- π 3 parenrightBig 3. I = √ 2- π 4 4. I = 2 parenleftBig √ 2 + π 3 parenrightBig 5. I = 1 + π 4 6. I = 1- π 4 correct Explanation: Let x = √ 2 sin θ . Then dx = √ 2 cos θ dθ , 2- x 2 = 2 cos 2 θ , while x = 0 = ⇒ θ = 0 , x = 1 = ⇒ θ = π 4 . In this case, I = integraldisplay π/ 4 2 √ 2 sin 2 θ cos θ 2 √ 2 cos 3 θ dθ = integraldisplay π/ 4 sin 2 θ cos 2 θ dθ = integraldisplay π/ 4 tan 2 θ dθ . Now tan 2 θ = sec 2 θ- 1 , d dθ tan θ = sec 2 θ , and so I = integraldisplay π/ 4 (sec 2 θ- 1) dθ = bracketleftBig tan θ- θ bracketrightBig π/ 4 . Consequently, I = 1- π 4 . padilla (tp5647) – HW08 – Gilbert – (56650) 2 003 10.0 points Evaluate the integral I = integraldisplay 1 1 (3 x 2 + 1) 3 / 2 dx . 1. I = 1 6 2. I = 1 3 3. I = 1 4. I = 1 2 correct 5. I = 1 4 Explanation: Set √ 3 x = tan u. Then √ 3 dx = sec 2 u du , while x = 0 = ⇒ u = 0 , x = 1 = ⇒ u = π 3 . On the other hand, (3 x 2 + 1) 3 / 2 = ( tan 2 u + 1 ) 3 / 2 = sec 3 u . Thus I = integraldisplay π/ 3 sec 2 u √ 3 sec 3 u du = 1 √ 3 integraldisplay π/ 3 cos u du = 1 √ 3 bracketleftBig sin u bracketrightBig π/ 3 . Consequently I = 1 2 . keywords: 004 10.0 points Evaluate the definite integral I = integraldisplay 2 √ 2 1 x 2 √ x 2- 1 dx . 1. I = √ 3- √ 2 2. I = √ 3 + √ 2 3. I = 1 2 ( √ 3 + √ 2 ) 4. I = 1 4 ( √ 3 + √ 2 ) 5. I = 1 4 ( √ 3- √ 2 ) 6. I = 1 2 ( √ 3- √ 2 ) correct Explanation: Set x = sec u . Then dx = sec u tan u du , x 2- 1 = tan 2 u , while x = √ 2 = ⇒ u = π 4 , x = 2 = ⇒ u = π 3 . In this case, I = integraldisplay π/ 3 π/ 4 sec u tan u sec 2 u tan u du = integraldisplay π/ 3 π/ 4 cos u du = bracketleftBig sin u bracketrightBig π/ 3 π/ 4 . Consequently, I = 1 2 ( √ 3- √ 2) . 005 10.0 points padilla (tp5647) – HW08 – Gilbert – (56650) 3 Which one of the following functions is an antiderivative of f when f ( x ) = 1 x 2- 8 x + 17 ?...
View Full Document

{[ snackBarMessage ]}

### Page1 / 11

hw8 - padilla(tp5647 – HW08 – Gilbert –(56650 1 This...

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

View Full Document
Ask a homework question - tutors are online