Study Guide for Prelim 1

Study Guide for Prelim 1 - 1 Study Guide for Prelim 1 Math...

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

View Full Document Right Arrow Icon
1 Study Guide for Prelim 1 Math 192 - Spring 1997 Written by Don Allers; Revised by Sean Carver 1 Improper Integrals Idea: A proper integral is an integral R b a f ( x ) dx, where a and b are both finite and f ( x ) is continuous and finite on [ a, b ]. For proper integrals, the fundamental theorem of calculus tells us that if F is an antiderivative of f (i.e. if F’(x) = f(x)) then R b a f ( x ) dx = F ( b ) - F ( a ) . We will define an improper integral as a limit of proper integrals. Comment The integral R b a f ( x ) dx is improper if: (i) a or b is infinite, or (ii) f ( x ) is not finite on all of [ a, b ]. The fundamental theorem of calculus does not apply directly to this class of integrals. Indeed, the definition of a definite integral given in Chapter 4 explicitly assumes that the integral is proper. Without the definitions below, an improper integral would be an expression with no meaning. Solution Method: 1. If a is finite, and f ( x ) is continuous on [ a, ] we define R a f ( x ) dx = lim c →∞ R c a f ( x ) dx. 2. If b is finite, and f ( x ) is continuous on [ -∞ , b ] we define R b -∞ f ( x ) dx = lim c →-∞ R b c f ( x ) dx. 3. If a, b are finite, and f ( x ) is continuous on ( a, b ] we define R b a f ( x ) dx = lim c a + R b c f ( x ) dx. 4. If a, b are finite, and f ( x ) is continuous on [ a, b ) we b define R b a f ( x ) dx = lim c b - R c a f ( x ) dx. 5. If the integral is improper at multiple points on the domain [ a, b ], we divide this domain into subintervals [ a, c 1 ] , [ c 1 , c 2 ] , . . . , [ c n - 1 , c n ] , [ c n , b ] so that each subinterval contains only one problem point at one of its endpoints, and no problem points in its interior. Then define R b a f ( x ) dx = R c 1 a f ( x ) dx + R c 2 c 1 f ( x ) dx + . . . R c n c n - 1 f ( x ) dx + R b c n f ( x ) dx, applying the above rules to each piece. Eg: (a) Z 1 0 x ln x dx, (b) Z 0 dx x 2 + 1 , (c) Z -∞ 2 x dx ( x 2 + 1) 2 , (d) Z 1 - 1 dx x 2 / 3 . 2 Sequences Idea: Given a sequence of numbers { a n } = a 1 , a 2 , a 3 , . . . , the question is whether or not these numbers are converging to some limit a . We say that { a n } converges to a , written a n a , as n → ∞ , if the distance | a n - a | between the sequence points and a goes to zero as n increases. Extend the Sequence to a Real Function Idea: Compute the limit of a sequence as the limit of a function f ( x ) as x → ∞ . When? Use this method if the expression defining the sequence a n can be considered as a real function in the variable n whose limit as n → ∞ can be computed easily. Method: To compute the limit of the sequence, take take the limit of its defining expression as you would take the limit of a real function. Remember L’Hopital’s Rule. (The theorem behind this method is the following: If f ( n ) = a n and lim x →∞ f ( x ) = L then lim n →∞ a n = L.
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
2 2 SEQUENCES Continuous Function Theorem Idea: It would be nice if we could move a limit inside a function, that is, if we could say lim n →∞ f ( a n ) = f ( lim n →∞ a n ) .
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