HW07 Solution

# HW07 Solution - badruddin(ssb776 HW07 Radin(57410 This...

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

badruddin (ssb776) – HW07 – Radin – (57410) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine the integral I = integraldisplay 1 1 + 16( x - 7) 2 dx . 1. I = 4 sin 1 parenleftBig x - 7 4 parenrightBig + C 2. I = 1 4 sin 1 4( x - 7) + C 3. I = tan 1 4( x - 7) + C 4. I = 4 tan 1 parenleftBig x - 7 4 parenrightBig + C 5. I = sin 1 4( x - 7) + C 6. I = 1 4 tan 1 4( x - 7) + C correct Explanation: Since d dx tan 1 x = 1 1 + x 2 , the substitution u = 4( x - 7) is suggested. For then du = 4 dx , in which case I = 1 4 integraldisplay 1 1 + u 2 du = 1 4 tan 1 u + C , with C an arbitrary constant. Consequently, I = 1 4 tan 1 4( x - 7) + C . keywords: 002 10.0 points Determine the integral I = integraldisplay 1 0 4 4 - x 2 dx . 1. I = 1 2. I = 4 3 π 3. I = 4 3 4. I = 2 3 5. I = π 6. I = 2 3 π correct Explanation: Since integraldisplay 1 1 - x 2 dx = sin 1 x + C , we need to reduce I to an integal of this form by changing the x variable. Indeed, set x = 2 u . Then dx = 2 du while x = 0 = u = 0 and x = 1 = u = 1 2 . In this case I = 8 integraldisplay 1 / 2 0 1 2 1 - u 2 du = 4 integraldisplay 1 / 2 0 1 1 - u 2 du . Consequently, I = bracketleftBig 4 sin 1 u bracketrightBig 1 / 2 0 = 2 3 π . keywords: 003 10.0 points

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

View Full Document
badruddin (ssb776) – HW07 – Radin – (57410) 2 Evaluate the definite integral I = integraldisplay π/ 2 0 4 sin θ 1 + cos 2 θ dθ . 1. I = 1 2 π 2. I = 3 4 π 3. I = 5 4 π 4. I = 3 2 π 5. I = π correct Explanation: Since d cos θ = - sin θ , the substitution u = cos θ is suggested. For then du = - sin θ dθ , while θ = 0 = u = 1 , θ = π 2 = u = 0 , so that I = - 4 integraldisplay 0 1 1 1 + u 2 du = 4 integraldisplay 1 0 1 1 + u 2 du , which can now be integrated using the fact that d du tan 1 u = 1 1 + u 2 . Consequently, I = 4 bracketleftBig tan 1 u bracketrightBig 1 0 = π since tan 1 1 = π 4 . 004 10.0 points Determine the integral I = integraldisplay 1 0 8 tan 1 x 1 + x 2 dx . 1. I = π 2 8 2. I = π 16 3. I = π 2 16 4. I = π 2 4 correct 5. I = π 4 6. I = π 8 Explanation: Set u = tan 1 x . Then du = 1 1 + x 2 dx , while x = 0 = u = 0 , x = 1 = u = π 4 . In this case, I = integraldisplay π/ 4 0 8 u du = bracketleftBig 4 u 2 bracketrightBig π/ 4 0 . Consequently, I = 1 4 π 2 . keywords: 005 10.0 points Determine the integral I = integraldisplay 4 + x 1 - x 2 dx .
badruddin (ssb776) – HW07 – Radin – (57410) 3 1. I = 4 sin 1 x - 1 2 radicalbig 1 - x 2 + C 2. I = 2 tan 1 x + 1 2 radicalbig 1 - x 2 + C 3. I = 4 sin 1 x - radicalbig 1 - x 2 + C correct 4. I = 4 tan 1 x + radicalbig 1 - x 2 + C 5. I = 2 tan 1 x - radicalbig 1 - x 2 + C 6. I = 2 sin 1 x + radicalbig 1 - x 2 + C Explanation: We deal with the two integrals I 1 = integraldisplay 4 1 - x 2 dx, I 2 = x 1 - x 2 dx separately. Now d dx sin 1 x = 1 1 - x 2 , so we see that I 1 = 4 sin 1 x + C . On the the other hand, to evaluate I 2 set u = 1 - x 2 . Then du = - 2 x dx , and so I 2 = - 1 2 integraldisplay u 1 / 2 du = - u + C .

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.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern