Lee Cantrell, Ben Arrants, Jack Blahouse, Marcus Lynch Chapter 6 practice test 6.1 1. Find the general solution to the exact differential equation: ππ¦ 1 = 5π₯ ln 5 + 2 ππ₯ π₯ +1 Solution: π¦ = 5π₯ + tanβ1 π₯ + π Find the anti-derivative 2. Solve the initial value problem explicitly: ππ£ = 4 sec π‘ tan π‘ + π π‘ + 6π‘ πππ π£ = 5 π€βππ π‘ = 0 ππ₯ Solution: π£ = 4 sec π‘ + π π‘ + 3π‘ 2 + π Find the anti-derivative 5 = 4 sec(0) + π (0) + 3(0)2 + π Plug in the points given 5= 4+1+0+π Simplify 0=π Solve for c π π Answer: [π¦ = 4 sec π‘ + π π‘ + 3π‘ 2 (β 2 < 0 < 2 )] 3. Solve the initial value problem using the Fundamental Theorem. (Your answer will contain definite integral.) πΉ β² (π₯) = π cos π₯ πππ πΉ(2) = 9 Solution: π₯ πΉ(π₯) = β«2 π cos π‘ ππ‘ + 9 4. Use Eulerβs method with increments of βπ₯ = 0.1 to approximate the value of y when π₯ = 1.3. ππ¦ = π¦ β π₯ πππ π¦ = 2 π€βππ π₯ = 1 ππ₯ Solution: (x, y) (1, 2) (1.1, 2.1) (1.2, 2.2) ππ¦ =π¦βπ₯ ππ₯ 1 1 1 βπ₯ ππ¦ βπ₯ ππ₯ 0.1 0.1 0.1 βπ¦ = 0.1 0.1 0.1 (π₯ + βπ₯, π¦ + βπ¦) (1.1, 2.1) (1.2, 2.2) (1.3, 2.3) Answer: [πΉ(1.3) = 2.3] 5. Use Eulerβs method with increments of βπ₯ = β0.1 to approximate the value of y when π₯ = 1.7 ππ¦ = π₯ β π¦ πππ π¦ = 2 π€βππ π₯ = 2 ππ₯ Solution: (x, y) (2, 2) (1.9, 2) (1.8, 2.01) ππ¦ =π₯βπ¦ ππ₯ 0 -0.1 -0.21 βπ₯ -0.1 -0.1 -0.1 ππ¦ βπ₯ ππ₯ 0 0.01 0.021 βπ¦ = (π₯ + βπ₯, π¦ + βπ¦) (1.9, 2) (1.8, 2.01) (1.7, 2031) Answer: [πΉ(1.7) = 2.031] Use the indicated substitution to evaluate the integral. Confirm your answer by differentiation 1. sec(2x)tan(2x)dx, u=2x Solution: 2. sin(3x)dx, u=3x Solution: Use substitution to evaluate the integral 3. Solution: 4. Solution:
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