# Spring 2017 Exam 1 MARK BOX HAND IN PART PIN: 17

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2 Spring 07 Exm 0. Fill-in-the blnks/boxes. 0.. If u 0, then du u = ln u + C 0.. tn u du = ln sec u or = - ln cos u + C 0.3. sec u du = ln sec u + tn u or = - ln sec u tn u + C 0.. sec u du = tn u + C 0.5. sec u tn u du = sec u + C 0.6. Integrtion by prts formul: u dv = uv v du 0.7. Trig sub.: if the integrnd involves u, then one mkes the substitution u = sin θ 0.8. Trig sub.: if the integrnd involves u, then one mkes the substitution u = sec θ 0.9. trig formul... your nswer should involve trig functions of θ, nd not of θ: cos(θ) = cos θ sin θ trig formul... your nswer should involve trig functions of θ, nd not of θ: sin(θ) = sin θ cos θ. 0.. trig formul... cos(θ) should pper in the numertor: cos (θ) = ( + cos (θ)). 0.. trig formul... cos(θ) should pper in the numertor: sin (θ) = ( cos (θ)) trig formul... since cos θ + sin θ =, we know tht the corresponding reltionship beween tngent (i.e., tn) nd secnt (i.e., sec) is + tn θ = sec θ. 0.. rctn( - ) = RADIANS. (your nswer should be n ngle) 0.5. If f : [, b) R is continuous, then we define the improper integrl b f (x) by b f (x) = lim t b t f (x). Prof. Girrdi Pge of Mth

3 Spring 07 Exm MULTIPLE CHOICE PROBLEMS Indicte (by circling) directly in the tble below your solution to the multiple choice problems. You my choice up to nswers for ech multiple choice problem. The scoring is s follows. For problem with precisely one nswer mrked nd the nswer is correct, 5 points. For problem with precisely two nswers mrked, one of which is correct, points. For problem with nothing mrked (i.e., left blnk) point. All other cses, 0 points. Fill in the number of solutions circled column. (Worth totl of point of extr credit.) Tble for Your Muliple Choice Solutions Do Not Write Below problem b c d e number of solutions circled B x b c d e 3 3 3b 3c 3d 3e b c d e 5 5 5b 5c 5d 5e 6 6 6b 6c 6d 6e 7 7 7b 7c 7d 7e 8 8 8b 8c 8d 8e 9 9 9b 9c 9d 9e 0 0 0b 0c 0d 0e b c d e 5 0 Extr Credit: Prof. Girrdi Pge 3 of Mth

4 Spring 07 Exm For problems -, put your nswer in the box nd show your work below the box.. x + = ln x + + x + C lso cceptble is ln x + + x + C On this problem, your finl nswer should not hve trig function in it. This is number 0 from the 00 Integrls. Why re both solutions cceptble? Well, they re both correct since x + ln + x ( ) + K = ln ( ) x + + x + K = ln + ln x + + x + K = ln ( x + + x + K + ln ). Prof. Girrdi Pge of Mth

5 Spring 07 Exm 3. e 3x cos 5x = e3x 3 (3 cos 5x + 5 sin 5x) + C Wy # For this wy, for ech integrtion by prts, we let the u involve the expontenil function. u = e 3x dv = cos 5x du = 3e 3x v = sin 5x. 5 So by integrtion by prts e 3x cos 5x = 5 e3x sin 5x 3 e 3x sin 5x. 5 Now let to get u = e 3x du = 3e 3x dv = sin 5x v = cos 5x. 5 e 3x cos 5x = 5 e3x sin 5x 3 [ 5 5 e3x cos 5x 3 e 3x cos 5x 5 = 5 e3x sin 5x e3x cos 5x 3 e 3x cos 5x. 5 Now solving for e 3x cos 5x (use the bring to the other side ide) we get [ + 3 e 3x cos 5x = 5 e3x sin 5x e3x cos 5x + K nd so Thus 5 [ ( 5 e 3x cos 5x = 3 5 e3x sin 5x + 3 ) 5 e3x cos 5x + K = 5 3 e3x sin 5x + 3 [ K5 3 e3x cos 5x + 3 [ = e3x K5 (5 sin 5x + 3 cos 5x) e 3x cos 5x = e3x 3 (3 cos 5x + 5 sin 5x) + C. Prof. Girrdi Pge 5 of Mth

6 Spring 07 Exm Wy # For this wy, for ech integrtion by prts, we let the dv involve the expontenil function. u = cos 5x dv = e 3x du = 5 sin 5x v = 3 e3x. So, by integrtion by prts e 3x cos 5x = 3 e3x cos 5x 5 3 Now let to get u = sin 5x dv = e 3x du = 5 cos 5x v = 3 e3x. e 3x sin 5x. e 3x cos 5x = 3 e3x cos 5x + 5 [ 3 3 e3x sin 5x 5 e 3x cos 5x 3 = 3 e3x cos 5x e3x sin 5x 5 e 3x cos 5x. 3 Now solving for e 3x cos 5x (use the bring to the other side ide) we get [ + 5 e 3x cos 5x = 3 e3x cos 5x e3x sin 5x + K nd so Thus 3 [ ( e 3x 3 cos 5x = e3x cos 5x + 5 ) 3 e3x sin 5x + K = 3 3 e3x cos 5x + 5 [ K3 3 e3x sin 5x [ = e3x K3 (3 cos 5x + 5 sin 5x) e 3x cos 5x = e3x 3 (3 cos 5x + 5 sin 5x) + C. Doesn t Work Wy If you try two integrtion by prt with letting the exponentil function be with the u one time nd the dv the other time, then when you use the bring to the other side ide, you will get 0 = 0, which is true but not helpful. Prof. Girrdi Pge 6 of Mth

7 Spring 07 Exm. Let n be n integer greter thn (so n {3,, 5, 6,...}). Find reduction formul for sec n x. sec n x = ( ) sec n x tn x + n n n sec n x Prof. Girrdi Pge 7 of Mth

8 Spring 07 Exm STATEMENT OF MULTIPLE CHOICE PROBLEMS These sheets of pper re not collected. Hint. For definite integrl problems b f(x). () First do the indefinite integrl, sy you get f(x) = F (x) + C. () Next check if you did the indefininte integrl correctly by using the Fundementl Theorem of Clculus (i.e. F (x) should be f(x)). (3) Once you re confident tht your indefinite integrl is correct, use the indefinite integrl to find the definite integrl. Hint. If, b > 0 nd r R, then: ln b ln = ln ( ) b nd ln( r ) = r ln.. Evlute soln. x= sin x. x= sin x = cos x x=/ x= = ( cos ) ( cos 0) = (0) () =. Evlute soln. x= x sin ( x ) First do indefinite integrl. Simple u-du substitution with u = x nd so du = x. x sin ( x ) = sin ( x ) x = sin u du = cos u cos x + C = + C. Next check indefinite integrl: ( ) cos x D x = D ( ( )) ( X cos x = sin x ) (x) = x sin x. So x= x sin ( x ) = cos (x ) x= ( cos ) = ( ) cos 0 = (0) ( ) =. 3. Evlute 3soln. x= sin x First do indefinite integrl. Prof. Girrdi Pge 8 of Mth

9 Spring 07 Exm Next check indefinite integrl: ( ) x sin x D x = D x (sin x) = ( cos x) = cos x = sin x. So x= sin x = ( ) ( x sin x x= = 8 sin ) ( 0 sin 0 ) ( = 8 ) ( 0 0 ) = 8.. Evlute x= soln. sin x cos x First do indefinite integrl. Let u = sin x so du = cos x. So sin x cos x = u du = u3 3 + C = sin3 x + C. 3 Next check indefinite integrl: ( ) sin 3 x D x = 3 3 D ( x (sin x) 3 ) = ( 3 sin 3 x ) D x sin x = sin x cos x 3 So x= sin x cos x = sin3 x 3 x= = ( sin 3 ) 3 sin3 0 ( ) = 3 ( ) 0 3 = Evlute 5soln. x= x= x + x The integrl x x + = rctn x x= = rctn () rctn (0) = 0 =. cn be evluted the following wy. Prof. Girrdi Pge 9 of Mth

10 6soln. Spring 07 Exm x cn be evluted by simple u-du substitution with u = x + x +. If u = x +, then x x + = x x + = du u = ln u + C = ln x + + C. So one cn integrte the integrl using simple u-du substitution of u = x +. Note tht if one uses Trig. Substitution, the initil substitution would be x = tn θ. Note tht the integrnd is lredy in its Prtil Frction Decomposition. 7. Let y = p (x) be polynomil of degree 5. Wht is the form of the prtil frction decomposition of p (x) (x ) (x + )? Here A, B, C, D, E nd F re constnts. 7soln. (x ) (x + ) = (x ) (x + ) (x + ) where x nd x + re liner terms while x + is n irreducible qudrtic term. Now see the prtil frction hndout from clss to see tht the PDF tkes the form A x + B x + + Cx + D x + + Ex + F (x + ). 8. Evlute 8soln. 3 5x + 3x x 3 + x. Prtil Frction Decomposition Problem. As usul, () we first find the indefinite integrl, () then check tht our indefinite integrl is correct by integrting the indefinite integrl nd mking sure we get the integrnd, (3) nd then evlute our definite integrl. Prof. Girrdi Pge 0 of Mth

11 Spring 07 Exm 9. Evlute 9soln. x= x= x= x= + x. =. From our textbook, pge 506, Exmple. +x 0. Evlute x. 0soln. diverges to. x This is n improper integrl (the integrnd y = x defined t x = 0) nd x = [ 0 [ + x 0 x = [ is continuous on (, 0) (0, ) but is not lim b 0 b [ + lim x 0 + x. Prof. Girrdi Pge of Mth

12 Spring 07 Exm Note lim 0 + x = lim 0 + The grph of y = x x = lim x 0 + = [ lim () 0 + is symmetric bout the y xis nd so lim b 0 ( ) clculte this limit similrly to the limit we just clculted. So [ b [ x = lim + lim = [ + [ =. b 0 x 0 + x b [ = lim 0 + =. = ; or, you cn x. Evlute soln. x= x= x 3. x 3 does not exist but lso does not diverge to infinity. This is n improper integrl (the integrnd y = x 3 defined t x = 0) nd x 3 = [ 0 [ + x 3 0 x 3 = [ is continuous on (, 0) (0, ) but is not lim b 0 b [ + lim x x 3 Prof. Girrdi Pge of Mth

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