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2 6 points 2. MULTIPLE CHOICE: Circle the best answer. 2.(a). (1 point) Is the integral Z 1 1 1 x2 dx an improper integral? YesNo 2.(b). (5 points) Evaluate the integral: Z 1 1 1 x2 dx = SOLUTION: The function 1/x2 is undefined at x = 0, so we we must evaluate the im-proper integral as a limit. Z 1!1 1 x 2 dx = lim c 0 Z c 1 1 x dx + lim c 0 ...
Calculus II Final Exam Practice Problems 1. (a) Sketch the conic section. Find and label any foci, vertices, and asymptotes. (x−3) −9y2 =36 (b) Find the equation of the ellipse with foci (0, 2) and semi-major axis length 3. 2. (a) Find the area of one petal of the rose r = 4sin(3 ). ... root test) (b) =1 − 2 1.5 1 3 k k
16 lis 2023 · The textbook includes examples, questions, and practice problems that will help students to review and sharpen their knowledge of the subject and enhance their performance in the classroom. The material covered in the book includes applications of integration, sequences and series and their applications, polar coordinate systems, and complex ...
FINAL EXAM CALCULUS 2 MATH 2300 FALL 2018 Name ... (2 points) Sketch the graph of the solution that satisfies following initial condition. Label the solution as (a). y(0) = 1 7.(b). ... n2 +5 (n+2)! convergent divergent Test: 10. 10 6 points 10. MULTIPLE CHOICE: Circle the best answer below.
12 wrz 2019 · Here are a set of practice problems for the Calculus II notes. Click on the "Solution" link for each problem to go to the page containing the solution. Note that some sections will have more problems than others and some will have more or less of a variety of problems. Most sections should have a range of difficulty levels in the problems ...
Calc II: Practice Final Exam 7 Part III. Power series. 1. Find the radius of convergence and interval of convergence of the series X1 n=1 (n1) xn n25n: By the ratio test, jan+1j=janj!jxj=5 <1; hence R= 5:The interval of convergence is [ 5;5], extrema in-cluded because x= 5) X1 n=1 ( 1)n 1 n2 which converges by the alternating series test (bn= 1 ...
N and solutions are Aexp M N x sin p D N x! + Bexp M N x cos p D N x!. An example of a series: X1 n=0 rn = 1 1 r if jrj<1 And another example: the alternating harmonic series 1 1 2 + 1 3 1 4 + ::: Reminder: the root test for P n a n uses lim n!1 ja nj1=n while the ratio test uses lim n!1 ja n+1j ja nj: Other tests include: comparison test ...
