Question 1 3.75 / 5 points Match the integral with the u needed for u-substitution. Some choices of u will be used more than once, and some integrals will not need u-substitution. f dz 2-J V16427 L= 1 [ " e 422_'_9 22U = 2T zdx -1 z2+1 3 substitution fails in this problem, use algebra to rewrite as a rule. 2 st e S dn Question 2 1.25 / 5 points Match the integral with the technigue to use first to solve it. 1 [ -2 _dz z24z+5 __2__ algebra: division or long division 2 f d:'l: ~J z?24r+5 __3__ double substitution _4 _regular u-sub 922 _4z4+5 d 5. [ 2idzts __1__ complete the square on the denominator 2z+6 = i Question 3 Match the idea with the formula to find it. 2 fab f:'(a':) dz _i_f:f(a':) dz (b) f(a) 4~ "ba __1__fab f!(a':) dz = [lff@) de 0.8 / 2 points 1. Average value of f() on [a, b] 2. Displacement of f(Z) on [a, b] 3. Total distance traveled for f(Z) on [@, ] a. Average rate of change o () on [a, b] 5. Given f:'(.'E) average velocity on [., b] Question 5 1.333 / 2 points Match the integral with the u needed for u-substitution. Some choices of u will be used more than once, and some integrals will not need u-substitution. 1 S r'+1 dx 2 f sin(2x) dx -2 Jesc(2x) cot (2x) dx 2. u = 2x 1fx . Vx2 + 1dx 3. substitution fails in this problem, use algebra to 2cdr rewrite as a rule. 1_ J (2+1) 1 f2x2 (x2 + 1) dafor substitution. Integrals Given: 1. S 3. S TED 4. fretide Substitution Options: 1. u=x + 1 2. u = 23 3. Substitution fails, algebraic rewrite needed. Correct Matches: 1. JT: u-substitution fails. This integral is standard: S= arctan(x) + C. So. no substitution works here. Answer: 3. Substitution fails. 16 47: Let u = 2x, so du = 2dr. The term 16 - 4x becomes 16 - u', resembling a trigonometric substitution. Answer: 2. u = 2x. 3. J a: Rewrite 4x2 + 9 as (2x)2 + 32. Use u = 2x, so du = 2da. Answer: 2. u = 2:x. Here, let u = a + 1, then du = 2rdx. Adjust for constants. Answer: 1. u = 2 +1. Question 2: Match the integral with the solving technique. Integrals Given: 2. f dd dx 4. J adr Techniques: 1, Complete the square. 2. Algebra: division/long division. 3. Double substitution. 4. Regular u-substitution.Correct Matches: 1. Ja ingde: Use complete the square; Rewrite - 4x + as (ar - 2) + Answer: 1. Complete the square. This integral simplifies directly using a regular substitution 1= x -4r+5. Answer: 4. Regular u substitution. 3. 2-Ar odr: Use long division since the numerator is a higher degree polynomial than the denominator. Answer: 2. Algebra: division or long division. Use u = + 3, so du = dr, and =U-3. Answer: 4. Regular u substitution. Question 3: Match the idea with the formula. Formulas Given: 2 [f(x)de 3. . Jo /(z)dx 5. [ f(x)dx Ideas: 1. Average value of f () on [a, b). 2. Displacement of f (x) on [a, b] 3. Total distance traveled for / (2) on a, b 4. Average rate of change of f(a) on [a, b. 5, Given f'(2), average velocity on [a, b). Correct Matches: 1. . f. f(2)da: This gives the average value of f'(a), which relates to velocity. Answer: 5. Average velocity given f'(:). 2. f f(x)dr: This calculates the displacement of / (3). Answer: 2. Displacement. 3. . J. f(x)de: This is the formula for the average value of f(a). Answer: 1. Average value. This gives the average rate of change of f(x) Answer: 4. Average rate of change. 5. [ /(x)dr: The integral of the absolute value gives total distance traveled. Answer: 3. Total distance traveled,Question 5: Match the integral with u -substitution. Integrals: 2. f sin(2x)da 3. [ csc(2x) cot(2x) da 4. fava' + 1dx 2xde 5. J ( 2+1). Substitution Options: 1. u = x'+ 1 2. u = 2x 3. Substitution fails. Correct Matches: Simplify algebraically: substitution fails here. Answer: 3. Substitution fails. 2. f sin(2x)dr: Use u = 2x, then du = 2dr. Answer: 2. u = 2x. 3. f csc(2x) cot(2x)dr: Use u = 2x, as the derivative of 2x simplifies the integral. Answer: 2. u = 2x. 4. favx2 + 1dx: Use u = x2 + 1, then du = 2xdx. Answer: 1. u = 23+ 1. 5. . 2ndz Use u = 2 + 1, then du = 2xdx. Answer: 1. u = x2+ 1. Summary of Key Techniques: . u-substitution simplifies integrals where derivatives match terms. . Algebraic rewrites (division, completing squares) solve non-substitution cases. . Recognizing standard integral forms (like arctan or trigonometric identities) is essential