1. Evaluate the integrals. a. (9 pts) x e x/2 dx. Solution: Using integration by parts, let u = x du = dx and dv = e x/2 dx v = 2e x/2.

Size: px

Start display at page:

Download "1. Evaluate the integrals. a. (9 pts) x e x/2 dx. Solution: Using integration by parts, let u = x du = dx and dv = e x/2 dx v = 2e x/2."

Cuthbert McCormick
5 years ago
Views:

1 MATH 8 Test -SOLUTIONS Spring 4. Evaluate the integrals. a. (9 pts) e / Solution: Using integration y parts, let u = du = and dv = e / v = e /. Then e / = e / e / e / = e / + e / = e / 4e / + c

2 MATH 8 Test -SOLUTIONS Spring 4 π/. (9 pts) sec 4 π/ sec 4 = lim π sec 4 = lim π sec sec = lim π ( tan +)sec Let u = tan du = sec. Also, when =, u =, and when =, u = tan. Therefore, π/ π/ sec 4 = lim π = lim π ( tan +)sec tan ( u +)du = lim 3 u3 + u π tan = lim 3 tan3 + tan π sec 4 = The improper integral diverges.

3 MATH 8 Test -SOLUTIONS Spring 4 3 c. (9 pts) To receive credit for this prolem you must use an integration technique other than a u- sustitution with Complete the square: = ( ). Use a trigonometric sustitution with = secθ for θ < π. Then = secθtanθdθ. Oserve that = = sec θ = tan θ = tanθ = tanθ for θ < π. Then 4 = = 4 + π/3 π/4 π/3 ( ) secθ secθ tanθdθ tanθ = sec θdθ π/4 π/3 = tanθ π/4 4 = tan π 3 tan π 4 =

4 MATH 8 Test -SOLUTIONS Spring 4 4 d. (9 pts) cos ( arctan ) + Let u = arctan. Then du = +. We have tanu = so our right triangle looks like: cos arctan = + cos u du = ( + cosu)du = u + 4 sinu + C = u + sinucosu + C = arctan + cos arctan = + arctan C C

5 MATH 8 Test -SOLUTIONS Spring 4 5. (9 pts) Find all the antiderivatives of the function f () = sin( ln ). Using integration y parts, let u = sin ln cos ln Then du = and v =. Applying the transformation: sin( ln ) = sin( ln ) = sin ln cos ln and dv =. cos ln Applying y parts again with u = cos ln sin ln Then du = and v =. Applying the transformation: cos ln sin( ln ) = sin( ln ) = sin ln = sin ln cos ln and dv =. cos( ln ) sin( ln ) = sin( ln ) cos ( ln ) sin ln sin( ln ) = sin( ln ) cos( ln ) sin( ln ) = sin ( ln ) cos ( ln ) + C sin ln

6 MATH 8 Test -SOLUTIONS Spring Prolem three is divided into two parts. These two parts are related. a. ( pts) Find the partial fraction decomposition of the rational epression. Be sure to set up the entire general system of equations to find the undetermined coefficients and then solve this system of equations. 4 4 = ( +) + Apply Partial Fraction Decomposition to the rational epression: ( ) ( +)( +) = A + B + + C + D + Equate Numerators: = A + + = A( +) + + B( ) ( +) + ( C + D) ( ) ( ) ( +)( +) + B( ) ( +) + ( C + D) ( ) To Solve for A, B and C we need to Equate Coefficients and give the general system of equations: = A + B + C 3 + A B + D + A + B C + A B D For 3 : = A + B + C For : = A B + D For : = A + B C For the constant : = A B D Solve for A, B, C, & D: One way is the following: Oserve that for =, = A( +) ( +) + B( ) ( +) + ( C + D) ( ) gives = 4A or A = / 4. For =, = A( +) B + C + D gives = 4B or B = / 4. Now sustitute A = / 4 and B = / 4 into the equation = A + B + C to get C =. Then sustitute A = / 4 and B = / 4 into the equation = A B + D to get D = /. Thus the decomposition of the rational epression is: 4 = (6 pts) Evaluate the integral. Hint: First use a sustitution (not a trigonometric sustitution). Let = e t. Then = e t dt. e t dt = e 4t e t = = 4 ln 4 ln + arctan + C + dt = e 4t 4 ln et 4 ln et + arctanet + C

7 MATH 8 Test -SOLUTIONS Spring Prolem four has three parts. Consider the integral where <. a. (9 pts) Evaluate this integral y using a trigonometric sustitution. Let = sinθ. Then = cosθdθ and = cos θ. Thus = cosθ dθ cos θ = secθ dθ = ln secθ + tanθ + C = ln + = ln + + C + C

8 MATH 8 Test -SOLUTIONS Spring Consider the integral where <.. (9 pts) Evaluate this integral y using partial fraction decomposition. = A + B + = A + Equate Numerators: = A( + ) + B( ) Equate Coefficients: A B = A + B = + B( ) ( + ) A = and B = Hence, = And = + ( + ) + + = ln + ln + + C = + ln + C

9 MATH 8 Test -SOLUTIONS Spring Consider the integral where <. c. (3 pts) Show that your answers in parts a. and. are the same. Starting with the answer to part a: ln + = ln + ln = ln + ln + ln = ln + ln ln + = + ln which is the answer to part. 5. (9 pts) Evaluate the limit. Clearly identify all indeterminate forms in your solution. Solution: lim ln a/ ( +ln ) + lna/ +ln lim + where a is a positive constant = lim eln + ln a/ +ln = e lim ln ln a + +ln = e lim + ln a/ / = e ln a = a Therefore, lim ln a/ ( +ln ) + = a Solution : lim ln a/ ( +ln ) + where = lim eln + ln a/ +ln = e lim ln ln a + +ln = e L L = lim ln a ln + + ln = lim ln a/ + / = ln a Therefore, lim( + a) / = e ln a = a

10 MATH 8 Test -SOLUTIONS Spring 4 6. (9 pts) Let R e the region ounded y y = and the -ais for shown in the + e graph elow. Determine if the solid formed y revolving R aout the -ais has a finite volume. If it does, find that volume. e V = π e + e = π lim e + e = π lim e ( + e ) Let u = + e du = e du = e. Also, when =, u =, and when =, u = + e. Therefore, π lim e ( + e ) = π lim = π lim u +e +e u du = π lim + e e V = π + e = π 4 The solid has a finite volume π 4 unit3.

Math Calculus II Homework # Due Date Solutions

Math Calculus II Homework # Due Date Solutions Math 35 - Calculus II Homework # - 007.08.3 Due Date - 007.09.07 Solutions Part : Problems from sections 7.3 and 7.4. Section 7.3: 9. + d We will use the substitution cot(θ, d csc (θ. This gives + + cot