problem8_pdf - rabbani(tar547 HW08 Radin(56635 This...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
rabbani (tar547) – HW08 – Radin – (56635) 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 0 3 8 - x 2 dx . 1. I = 1 2 π 2. I = 3 4 π 3. I = π 4. I = 1 5. I = 3 4 6. I = 1 2 002 10.0 points Evaluate the integral I = integraldisplay 1 0 x 2 (2 - x 2 ) 3 / 2 dx . 1. I = 2 parenleftBig 2 + π 3 parenrightBig 2. I = 2 parenleftBig 3 - π 3 parenrightBig 3. I = 2 parenleftBig 3 + π 3 parenrightBig 4. I = 1 + π 4 5. I = 1 - π 4 6. I = 2 - π 4 003 10.0 points Evaluate the integral I = integraldisplay 1 0 2 ( x 2 + 3) 3 / 2 dx . 1. I = 2 2. I = 2 3 3. I = 1 2 4. I = 1 3 5. I = 1 004 10.0 points Evaluate the definite integral I = integraldisplay 2 2 6 x 2 x 2 - 1 dx . 1. I = 6( 3 + 2 ) 2. I = 3 2 ( 3 - 2 ) 3. I = 3( 3 + 2 ) 4. I = 6( 3 - 2 ) 5. I = 3( 3 - 2 ) 6. I = 3 2 ( 3 + 2 ) 005 10.0 points Which one of the following functions is an antiderivative of f when f ( x ) = 1 x 2 - 6 x + 10 ? 1. F ( x ) = ln( x 2 - 6 x + 10)
Image of page 1

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

View Full Document Right Arrow Icon
rabbani (tar547) – HW08 – Radin – (56635) 2 2. F ( x ) = tan - 1 ( x - 3) 3. F ( x ) = sin - 1 ( x - 3) 4. F ( x ) = ln vextendsingle vextendsingle vextendsingle vextendsingle x - 10 x + 3 vextendsingle vextendsingle vextendsingle vextendsingle 5. F ( x ) = - 10 ( x - 3) ( x 2 - 6 x + 10) 2 006 10.0 points Rewrite the expression f ( x ) = 15 x ( x - 1) ( x 2 + x + 1) using partial fractions.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern