{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

5.2 - mora-boellstor(erm823 5.2 Arledge(55875 This...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
mora-boellstorff (erm823) – 5.2 – Arledge – (55875) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Evaluate the definite integral I = integraldisplay 0 - 4 parenleftBig 2 + radicalbig 16 - x 2 parenrightBig dx by interpreting it in terms of known areas. 1. I = 4 π + 4 2. I = 8 π + 4 3. I = 8 π + 8 4. I = 4 π + 8 correct 5. I = 16 π + 8 6. I = 16 π + 4 Explanation: Since the graph of y = 2 + radicalbig 16 - x 2 is the upper half of the circle centered at (0 , 2) having radius r = 4, the value of I is the area of the shaded region in - 4 2 bounded by the graph of y = 2 + radicalbig 16 - x 2 , the vertical lines x = - 4, and x = 0 as well as the interval [ - 4 , 0]. Consequently, the area is the sum of the area of a quarter-circle of radius r = 4 and a 4 × 2 rectangle, i.e. , I = 4 π + 8 . 002 10.0points When f has graph R 1 R 2 a b c express the sum I = integraldisplay c a f ( x ) dx - integraldisplay b a 4 f ( x ) dx in terms of the areas A 1 = area( R 1 ) , A 2 = area( R 2 )
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
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