File name: Calculus 2 Problems And Solutions Pdf

Rating: 4.8/5 (Based on 8437 votes)

27748 downloads

========================

Calculus 2 Problems And Solutions Pdf

========================

Calculus II Practice Final Exam, Answers 1. Differentiate: a) f x ln sin e2x. Answer. This is an exercise in the chain rule: f x 1 sin e2x cos e2x 2e2x 2e2x cot e2x b) g x xtan 1 x2. Answer. . Calculus II Practice Problems 2: Answers 1. Solve the initial value problem: 4y 3y ex y 0 7 Answer. First solve the homogeneous equation, which can be written as dy 3 y 3 4 dx, which . Calculus II Final Exam Practice Problems 1. (a) Sketch the conic section. Find and label any foci, vertices, and asymptotes. Find the equation of the ellipse with foci (0, 2) and semi-major . This collection of solved problems covers elementary and intermediate calculus, and much of advanced calculus. We have aimed at presenting the broadest range of problems that you are likely to encounter—the. Sep 12, · Find a set of practice problems for the Calculus II notes with links to the solutions. The problems cover topics such as integration techniques, improper integrals, applications of integrals, parametric equations and polar coordinates. Calc II: Practice Final Exam 1 PART I. Integration. 1. Compute the following integrals. (a) Z (lnx)2dx Integrate by parts twice, the rst time choose: f= (lnx)2; g= 1 and the second time choose: f= lnx; g= 1: Then Z (lnx)2dx= x(lnx)2 2 Z x lnx x dx= x(lnx)2 2fxlnx Z x 1 x dxg= x(lnx)2 2xlnx+ 2x+ C: (b) Z dx x2 p 4x2 + 1 Perform the trig. 4. (a) Find the equation of the sphere that’s centered at (6;4;2), and passes through the point (3;5;7). (b) Circle the correct answer: the intersection of the above sphere with the plane x= 0 is (i) a point; (ii) a circle; (iii) empty; that is, the plane and the sphere do not intersect. Exam III will be based on Sections , , , , , , , , , , 1. Review all de nitions from Sections , , , , , , , , , , Calc II: Practice Final Exam 1 PART I. Integration. 1. Compute the following integrals. (a) Z (lnx)2dx Integrate by parts twice, the rst time choose: f= (lnx)2; g= 1 and the second time choose: f= lnx; g= 1: Then Z (lnx)2dx= x(lnx)2 2 Z x lnx x dx= x(lnx)2 2fxlnx Z x 1 x dxg= x(lnx)2 2xlnx+ 2x+ C: (b) Z dx x2 p 4x2 + 1 Perform the trig.