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Midterm Exam #1 Math 121 February 22, 2012 1. [30 Points] Evaluate each of the following limits. Please justify your answers. Be clear if the limit equals a value, +∞ or −∞, or Does Not Exist. (a) 0 0 −2 sinh(2x) 0 L’H −4 cosh(2x) −4 1 − cosh(2x) 0 L’H = lim = lim = 4 = 1 1 x→0 x→0 x + ln(1 − x) −1 1 − 1−x − (1−x)2 ! lim x→0 (b) ! lim ln 1 4 x 1∞ x lim e − = lim e x→∞ x→∞ x (c) lim = e 3 3 1 + (3x)2 = 4 4 p 2 1 − (4x) 0 arctan(3x) 0 L’H = lim x→0 arcsin(4x) x→0 ∞·0 ! 1 ln e x − x→∞ 4 x 1 x 1 ex − lim L’H ( 00 ) = e x→∞ 4 x !x ! 1 1 ex − lim ln = ex→∞ 4 e − x 1 x 1 · e x − x12 + 4 x 4 x2 x 4 lim x ln e − x→∞ x =e 1 x − x12 1 1 1 4 1 1 lim · e x − 2 + 2 (−x2 ) · ex − 4 lim 1 1 4 4 x→∞ x→∞ x x ex − ex − x x = e−3 =e =e 2. (a) [30 Points] Z ln 7 0 Compute the following definite integral. Please simplify your answer. ln 7 Z x sinh x dx = x cosh x − 0 ln 7 0 ln 7 ln 7 cosh x dx = x cosh x − sinh x 0 = ln 7 cosh(ln 7) − 0 − sinh(ln 7) + sinh 0 = ln 7 cosh(ln 7) − sinh(ln 7) + 0 ! ! ln 7 ln 7 7 + 71 7 − 17 e + e− ln 7 e − e− ln 7 = ln 7 − = ln 7 − 2 2 2 2 ! ! 48 50 24 25 7 7 − = ln 7 − = ln 7 2 2 7 7 u=x dv = sinh xdx I.B.P. du = dx v = cosh x 1 0 2. (Continued) Compute the following definite integral. Please simplify your answer. Z √3 Z √3 x+1 x 1 √ √ √ dx = dx + dx (b) 2 2 4−x 4−x 4 − x2 1 1 1 Z u=1 Z √3 Z u=1 Z √3 1 1 1 1 1 r √ du + r dx = − = − 21 u− 2 du + 2 u=3 u 2 u=3 1 1 4 1 − x4 2 1− √ Z 3 √3 2 √ 1 1 √ dw = − u + arcsin w 1 1 − w2 w= 12 3 3 2 √ ! √ √ √ √ 3 1 π π π − arcsin = −1 + 3 + − = −1 + 3 + = − 1 + 3 + arcsin 2 2 3 6 6 1 Z 1 = − 2 · 2u + 1 2 Here w= √ 3 2 u = 4 − x2 du = −2xdx − 21 du = xdx and x= √1 =⇒ u = 3 x = 3 =⇒ u = 1 x 2 1 dw = dx 2 and x = 1 =⇒ w = √ x = 3 =⇒ w = w = Also here 1 2 √ 3 2 x 1 dx = arcsin + C. 2 4 − x2 OR you could use a trigonometric substitution: OR you can use the fact that Z √ 2 x = 2 sin θ θ √ 4 − x2 Trig. Substitute dx = 2 cos θdθ Z √ 3 1 = Z π 6 π 3 Z π 3 2 sin θ + 1 2 sin θ + 1 p 2 cos θ dθ 2 cos θ dθ = 2 π π 2 cos θ 4 − 4 sin θ 6 6 π 3 π π π π 2 sin θ + 1 dθ = −2 cos θ + θ = −2 cos + + 2 cos − 3 3 6 6 π x+1 √ dx = 4 − x2 π 3 x = −1 + Z 6 π 3 + √ 3− π 6 = −1 + √ 3+ π 6 2 x 2 2 dx 2. (Continued) Compute the following definite integral. Please simplify your answer. Z 1 1 −2x 1 1 x3 1 2x −x xe dx − e +2 x + x + 2x dx = dx = (c) e e 3 0 2 0 0 0 0 ! Z 1 1 1 −2x 1 x3 1 1 −2x x3 −x −x −x −x = e dx − + + 2 −xe e = + 2 −xe − e − e 3 0 2 3 2 0 0 0 1 1 1 1 1 1 + 2 −e−1 − e−1 − e−2 − 0 − 0 − 2 − e0 = − 4e−1 − e−2 + 2 + 3 2 2 3 2 2 1 1 x+ x e Z = 2 Z 1 2 17 4 1 − − 2 6 e 2e u=x dv = e−x dx I.B.P. above (on middle term) du = dx v = −e−x 3. [40 Points] (a) Compute the following indefinite integral. Z Z 1 x2 1 sin2 θ x2 x2 √ p arcsin x − arcsin x − x arcsin x dx = dx = · cos θ dθ 2 2 2 2 1 − x2 1 − sin2 θ Z Z Z 1 1 x2 1 sin2 θ sin2 θ x2 √ arcsin x− arcsin x− ·cos θ dθ = arcsin x− sin2 θdθ ·cos θ dθ = 2 2 2 2 cos θ 2 2 cos θ Z Z 1 − cos(2θ) 1 x2 1 1 − cos(2θ)dθ arcsin x − dθ = arcsin x − 2 2 2 4 1 x2 1 1 1 arcsin x − arcsin x − θ + sin(2θ) + C θ − sin(2θ) + C = 4 2 2 4 8 Z = x2 2 = x2 2 = x2 2 = x2 1 1 1 1 p x2 arcsin x − θ + 2 sin θ cos θ + C = arcsin x − arcsin x + x 1 − x2 + C 2 4 8 2 4 4 u = arcsin x du = √ dv = xdx x2 1 dx v = 2 1 − x2 1 x = sin θ θ √ 1 − x2 Trig. Substitute dx = cos θdθ (b) Z 1 (x2 + 4) 5 2 dx = Z x 1 (4 tan2 θ + 4) 5 2 · 2 sec2 θ dθ = 3 Z 1 5 (4 sec2 θ) 2 · 2 sec2 θ dθ = = = = = 1 1 · 2 sec2 θ dθ 5 2 5 (2 sec θ) ( 4 sec θ) Z Z Z 1 1 1 sec2 θ 1 dθ = 4 dθ = cos3 θ dθ 24 sec5 θ 2 sec3 θ 16 Z Z Z 1 1 1 2 2 cos θ cos θ dθ = (1 − sin θ) cos θ dθ = (1 − w2 ) dw 16 16 16 3 ! 1 w3 1 sin3 θ 1 x 1 x √ √ w− +C = sin θ − +C = +C − 16 3 16 3 16 x2 + 4 3 x2 + 4 ! 1 x3 x √ +C − 3 16 x2 + 4 3(x2 + 4) 2 Z √ · 2 sec2 θ dθ = Z √ x = 2 tan θ x2 + 4 x θ Trig. Substitute dx = 2 sec2 θdθ 2 Standard w substitution for odd trig. integral Z cos3 θ dθ technique: w = sin θ dw = cos θ dθ Z 2 x2 x +1−1 2 (c) ln(x + 1) dx = x ln(x + 1) − 2 dx = x ln(x + 1) − 2 dx 2 x +1 x2 + 1 Z Z 2 Z Z 1 1 x +1 2 2 1 dx − dx − dx = x ln(x + 1) − 2 dx = x ln(x + 1) − 2 x2 + 1 x2 + 1 x2 + 1 Z 2 2 Z = x ln(x2 + 1) − 2 (x − arctan x) + C = x ln(x2 + 1) − 2x + 2 arctan x + C u = ln(x2 + 1) I.B.P. du = dv = dx 2x dx v = x +1 x2 OR you could do a trigonometric substitution on = Z tan2 θ sec2 x dθ = sec2 x Z 2 tan θ dθ = Z Z x2 dx = x2 + 1 Z tan2 θ sec2 x dθ tan2 θ + 1 sec2 θ − 1 dθ = tan θ − θ + C = x − arctan x + C 4 √ x2 + 1 x = tan θ x θ Trig. Substitute dx = sec2 θdθ 1 *************************************************************************** OPTIONAL BONUS Do not attempt these unless you are completely done with the rest of the exam. *************************************************************************** OPTIONAL BONUS #1 1. Z e √ √ 1+ x dx = 4 Z 2 Compute the following indefinite integral. w(w − 1)e dw = 4 w Z 3 (w − w)e dw = 4 w Z 3 w w e dw − Z w we dw w3 ew − 3w2 ew + 6wew − 6ew − (wew − ew ) + C = 4 w3 ew − 3w2 ew + 6wew − 6ew − wew + ew + C = 4 w3 ew − 3w2 ew + 5wew − 5ew + C p p √ √ p √ 3 √1+√x √ 2 √1+√x √ √1+√x +5 − 5e 1+ x + C =4 1+ x e 1+ x e 1+ x e −3 = 4 ((∗) − (∗∗)) = 4 = 4e √ √ √ 1+ x p p p √ 3 √ 2 √ 1+ x −3 1+ x +5 1+ x −5 +C p i p √ √ √ 1 + x − 3 (1 + x) + 5 1+ x −5 +C p i √ √ hp √ √ p √ √ √ = 4e 1+ x 1+ x+ x 1+ x −3−3 x+5 1+ x −5 +C = 4e = 4e Here √ h 1+ x (1 √ + √ √ h p 1+ x 6 1 x) + √ x+ √ p √ √ i x 1+ x −8−3 x +C p √ √ √ w = 1 + x =⇒ w2 = 1 + x =⇒ w2 − 1 = x 1 √ √1 √ dx dw = 2 x 2 p √ √ 4 1+ x xdw = dx 1+ x 4w(w2 − 1)dw = dx ***************************************************************************** Z Z Z 3 w 3 w 2 w 3 w 2 w w (*)Aside: w e dw = w e − 3 w e dw = w e − 3 w e − 2 we dw = w3 ew − 3w2 ew + 6 Z Z wew dw = w3 ew − 3w2 ew + 6 wew − ew dw 5 = w3 ew − 3w2 ew + 6 Z wew dw = w3 ew − 3w2 ew + 6wew − 6ew + C u = w3 u = w2 dv = ew dw First I.B.P. dv = ew dw Second I.B.P. du = 3w2 dw v = ew u=w du = 2wdw v = ew dv = ew dw Third I.B.P. du = dw v = ew ***************************************************************************** Z Z (**)Aside: wew dw = wew − ew dw = wew − ew + C u=w dv = ew dw I.B.P. du = dw v = ew ***************************************************************************** OPTIONAL BONUS #2 Z Z Z IBP sec x dx = sec x sec x dx = sec x tan x − sec x tan2 x dx Z Z 2 = sec x tan x − sec x(sec x − 1) dx = sec x tan x − sec3 x − sec x dx Z Z Z (∗∗∗) = sec x tan x − sec3 x dx + sec x dx = sec x tan x + ln | sec x + tan x| − sec3 x dx 2. 3 Compute the following indefinite integral. 2 Regroup: Z Z sec3 x dx = sec x tan x + ln | sec x + tan x| − sec3 x dx which yields Z 2 sec3 x dx = sec x tan x + ln | sec x + tan x| + C and then Z 1 sec3 x dx = (sec x tan x + ln | sec x + tan x|) + C 2 u = sec x dv = sec2 xdx I.B.P. du = sec x tan x v = tan x (***) Aside 6 Z sec x dx = Z sec x sec x + tan x sec x + tan x = ln | sec x + tan x| + C Here dx = Z sec2 x + sec x tan x dx = sec x + tan x w = sec x + tan x dw = sec x tan x + sec2 xdx 7 Z 1 dw w