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

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2 Fll 7 Exm. Fill-in the blnks... rccos( - ) = π 3 (Your nswers should be n ngle in RADIANS.).. rcsin( - ) = -π 6 (Your nswers should be n ngle in RADIANS.).3/.4. Double-ngle Formuls. Your nswer should involve trig functions of θ, nd not of θ. cos(θ) = cos θ sin θ nd sin(θ) = sin θ cos θ.5/.6. Hlf-Angle Formul. Your nswer should involve cos(θ). cos + cos (θ) (θ) = nd sin (θ) = cos (θ)..7. du u u = ln u + C u n du n = u n+ n+ +C e u du = e u +C.. cos u du = sin u + C.. sec u du = tn u + C.. sec u tn u du = sec u + C.3. sin u du = - cos u + C.4. csc u du = - cot u + C.5. csc u cot u du = - csc u + C.6. tn u du = ln sec u or = - ln cos u + C.7. cot u du = - ln csc u or = ln sin u + C.8. sec u du = ln sec u + tn u or = - ln sec u tn u + C.9. csc u du = - ln csc u + cot u or = ln csc u cot u + C Prof. Girrdi Pge of 3 Mth 4

3 Fll 7 Exm.. du > = sin ( ) - u u.. +u du > =.. u u du > = ( ) tn- u ( ) u sec-.3. Trig sub.: if the integrnd involves u, then one mkes the substitution u = sin θ.4. Trig sub.: if the integrnd involves u, then one mkes the substitution u = sec θ.5. Integrtion by prts formul: u dv = uv vdu + C + C + C Improper Integrls Below,, b, c R with < c < b..6. If f : [, ) R is continuous, then we define the improper integrl f (x) dx by f (x) dx = t lim t f (x) dx..7. If f : (, ) R is continuous, then we define the improper integrl f (x) dx = [ lim t t ] f (x) dx + [ lim s s ] f (x) dx f (x) dx by..8. If f : [, c) (c, b] R is continuous, then we define the improper integrl b f (x) dx = [ lim t c t ] f (x) dx + [ lim s c + b s b ] f (x) dx f (x) dx by..9. An improper integrl s bove converges precisely when ech of the limits involves converges to finite number..3. An improper integrl s bove diverges precisely when the improper integrl does not converge. Prof. Girrdi Pge 3 of 3 Mth 4

4 Fll 7 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, 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 4 4 4b 4c 4d 4e 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 b c d e 5 Extr Credit: Prof. Girrdi Pge 4 of 3 Mth 4

5 Fll 7 Exm. Show your work below the box then put your nswer in the box. Your finl nswer should not hve trig function in it.. Complete the squre. x 4x 5 = (x ) 9 b. ( x 4 ) = (x ) = x 4x + 4 x 4x 5 = (x 4x + 4 4) 5 = (x 4x + 4) + ( 4 5) = (x ) x x dx = 9 ( ) x rcsin 3 + (x ) 5 + 4x x + C. Derive reduction formul for x n e x dx where n N = {,, 3, 4,...}. Show your work below the box then put your nswer in the box. x n e x dx = x n e x n x n e x dx Integrtion by Prts Key Ide: For x n f (x) dx where f (x) dx is esy, try u = x n nd dv = f (x) dx. (Note tht then v = dv = f (x) dx.) This often reduces x n to x n. u = x n du = nx n dx dv = e x dx v = e x So Integrtion by Prts u dv = uv v du gives us tht x n e x dx = x n e x n x n e x dx. Prof. Girrdi Pge 5 of 3 Mth 4

6 Fll 7 Exm STATEMENT OF MULTIPLE CHOICE PROBLEMS These sheets of pper re not collected. Hint. For definite integrl problems b f(x) dx. () First do the indefinite integrl, sy you get f(x) dx = 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 > nd r R, then: ln b ln = ln ( ) b nd ln( r ) = r ln.. Evlute the integrl soln. x x + 9 dx. Prof. Girrdi Pge 6 of 3 Mth 4

7 Fll 7 Exm. Evlute the integrl soln. 4 x x + 9 dx. 3. Evlute ln(π) e x cos (e x ) dx 3soln. Let u = e x. So du = e x dx. So e x cos (e x ) dx = cos u du = sin u + C = sin (e x ) + C. Next check indefinite integrl: D x sin (e x ) = [cos (e x )] D x e x = [cos (e x )] e x. So ln(π) e x cos (e x ) dx = sin e x x=ln(π) x= = sin e ln(π) sin e = sin (π) sin = sin = sin Prof. Girrdi Pge 7 of 3 Mth 4

8 Fll 7 Exm 4. Evlute x= 3π x= e x cos x dx. 4soln. Below we show tht So x= 3π x= e x cos x dx = ex (sin x + cos x) e x cos x dx = ex (sin x + cos x) + C. 3π/ = e3π/ ( ) e () = e3π/. To find the indefinite integrl, use two integrtion by prts nd the bring to the other side ide. For the two integrtion by prts, put the expontentil function with either the u s both times or the dv s both times. Wy # For this wy, for ech integrtion by prts, we let the u involve the expontenil function. u = e x dv = cos x dx du = e x dx v = sin x. So by integrtion by prts e x cos x dx = ex sin x e x sin x dx. Now let to get u = e x du = e x dx dv = sin x dx v = cos x. e x cos x dx = ex sin x [ ex cos x ] e x cos x dx = ex sin x + ex cos x e x cos x dx. Now solving for e x cos x dx (use the bring to the other side ide) we get ] [ + e x cos x dx = ex sin x + ex cos x + K nd so [ ] ( e x cos x dx = ex sin x + ) ex cos x + K = ex sin x + [ ] K ex cos x + [ ] = ex K ( sin x + cos x) +. Prof. Girrdi Pge 8 of 3 Mth 4

9 Fll 7 Exm Thus e x cos x dx = ex ( cos x + sin x) + C. Wy # For this wy, for ech integrtion by prts, we let the dv involve the expontenil function. u = cos x dv = e x dx du = sin x dx v = ex. So, by integrtion by prts e x cos x dx = ex cos x Now let to get u = sin x dv = e x dx du = cos x dx v = ex. e x sin x dx. e x cos x dx = ex cos x + [ ex sin x ] e x cos x dx = ex cos x + ex sin x e x cos x dx. Now solving for e x cos x dx (use the bring to the other side ide) we get ] [ + e x cos x dx = ex cos x + ex sin x + K nd so Thus [ ] ( e x cos x dx = + ex cos x + ) ex sin x + K = ex cos x + [ ] K ex sin x + + [ ] = ex K ( cos x + sin x) + + e x cos x dx = ex ( cos x + sin x) + 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 =, which is true but not helpful. Prof. Girrdi Pge 9 of 3 Mth 4

10 Fll 7 Exm 5. Investigte the convergence of x= x= e x dx. x 5soln. This is Exmple 9 from 8.8 of our book by Thoms. 6. Evlute x= sin 4 x dx. 6soln. x= From Clss Hndout on Trig. Substitution: Exmple 4. sin 4 x dx. u-du sub does not work (why? e.g.: sin 4 x dx u=cos = x sin 3 x [ sin x dx]). For sin n x cos m x dx, with both m, n {,, 4, 6,...}, use the hlf-ngle formuls. [ ] sin 4 x dx lg. = [sin x] dx ( = ) cos(x) dx lg. = [ cos(x) + cos (x)] dx 4 ( = ) [ cos(x) + + cos(4x) ] dx 4 = dx cos(x) dx dx + 4 cos(4x) dx = ( + ) dx ( 4 4 cos(x) [dx] + 4 ) ( ) cos(4x) [4dx] 4 So, since sin =, = 3 8 x 4 sin(x) + sin(4x) + C. 3 x= x= [ 3 sin 4 x dx = 8 x 4 sin(x) + ] 3 sin(4x) = [ sin + ] 3 sin 4 = sin + sin 4. 3 x= x= [ + ] Prof. Girrdi Pge of 3 Mth 4

11 Fll 7 Exm 7. Evlute x= x 5 x=5 x dx AND specify the initil substitution. 7soln. The integrnd hs u, so we let u = sec θ. x = 5secθ so dx = 5 sec θ tn θ dθ Prof. Girrdi Pge of 3 Mth 4

12 Fll 7 Exm 8. Evlute 8soln. x=3 x= 5x + 3x x 3 + x dx. 9. For which vlue of p does 9soln. dx =.5? xp First compute dx = lim xp t + t x p dx p x p = lim t + p t = p lim t + x p t = If p >, or equivlently > p, then lim t + t p = nd so x dx = p p lim [ ] t p = [ ] = t + p p. If p <, or equivlently < p, then lim t + t p = nd so x dx = p p lim [ ] t p =. t + p lim [ ] t p. t + If p =, then So we need Note x p dx = p =.5 = 5 4 = 4 5 lim t + t x dx = p < nd lim ln t + x t = lim [ ln t] =. t + p = 4 5 p =.5. p = 4 5 = 5 =.. Prof. Girrdi Pge of 3 Mth 4

13 Fll 7 Exm. Evlute the integrl x= x= x 3 dx. soln. Prof. Girrdi Pge 3 of 3 Mth 4

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