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# Calculus Techniques MATH 1100

Utah State University

GPA 3.66

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This 4 page Class Notes was uploaded by Benton Yundt on Wednesday October 28, 2015. The Class Notes belongs to MATH 1100 at Utah State University taught by Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/230422/math-1100-utah-state-university in Mathematics (M) at Utah State University.

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Date Created: 10/28/15

Problem De nition Problem 25 Describe the intervals on which the following function is con tinuous 3 C c 5 2 f c21 3Cgt2 Solution Step 1 First we should look at the function de nition for C S 2 and C gt 2 For values of C S 2 the function is de ned to be the polynomial c 3 which is continuous for all real numbers So the function de ned above is continuous on the interval 00 2 Also the function is de ned to be the polynomial 2 1 for all C gt 2 So our function is continuous on the interval 2 00 Solution Step 2 The next step is to investigate what happens where the function de nition changes that is at C 2 Since the function is de ned at C 2 by f2235 The next step is to test to see if the limit exists For this problem we will need to evaluate the left and right hand limits The left hand limit is given by lim lim 63 23 5 ma ma The right hand limit is computed follows I I Q Q y it 135566 1 2 1 0 Since the right and left hand limits exist and are equal to each other the limit exists and we can write me a The second property that needs to be satis ed is m m Since this is also true the function is continuous at C 2 Solution Step 3 Finally we can combine our results to see that the function is continuous on the following 00 so 00 so So this function is continuous for all real numbers Problem De nition Problem 23 Find the inde nite integral Hint Integration by parts is not necessary in all cases in this section of the text Solution Step 1 In this case it is not necessary to compute the inde nite integral using integration by parts Substituting for the argument in the square root will give an easier integral with which to work We can use 3 1 and require 13 11 and y 1 The new form of the integral is y 1W3 dy 932 912 dy Solution Step 2 The inde nite integral is then computed follows 52 32 32 12 y y 9 J M 5232C 2 y52Ey32C o 3 Solution Step 3 The last step is to return to the original variables 2 2 2 2 y527 y32C 2 1527 x l32C o 3 o 3 Note that integration by parts will work but is a bit more dif cult Solution Step 4 To check the work we can differentiate the solution in an effort to repro duce the integrand 1 113 213 1 5 52 LII 132C 2 52 d 2 x 152 x132 lt2gtltltw 1gt32gtltgtlt2gt 1 2 d c ampwxw a 1W2 AGING AA ax This is the same function we started with

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