Math 121 Midterm Exam #1 2014 Answer Key

Math 121 Midterm Exam #1 February 21, 2014 Answer Key 1. [10 Points] (a) Let y = arcsin x. Use implicit differentiation to PROVE that 1 dy =√ . dx 1 − x2 If y = arcsin x, then sin y = x. Implicitly differentiate both sides, d d (sin y) = (x). dx dx dy Then cos y = 1. dx 1 dy 1 1 Here we used the trig. identity to finish =√ Solve for = =p 2 dx cos y 1 − x2 1 − sin y OR finish 1 1 1 dy = = =√ dx cos y cos (arcsin x) 1 − x2 1 arcsin x √ 1 − x2 using the trig. from the triangle (b) From part (a) we now know that PROVE that Z √ Z x √ 1 dx = arcsin x + C. You may use this fact to 1 − x2 x 1 dx = arcsin +C 3 9 − x2 ←− Prove this. Z 1 1 s  r dx =   x 2 dx √ x2 9 1− 9 1− 3 9 Z Z x 1 1 1 r √ = +C du = arcsin u = arcsin dx =  x 2 3 3 1 − u2 1− 3 x u= 3 Standard u substitution to simplify: 1 du = dx 3 Z √ 1 dx = 9 − x2 Z Note: OR you can also do a trig. substitution here. 1 2. [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. 5 0 5xex + 5ex − − arctan(5x) 0 L’H 1 + (5x)2 = lim (a) lim −1 x→0 x→0 sinh x + ln(1 − x) cosh x + 1−x 5(50x) 5xex + 5ex + 5ex + 10 L’H (1 + 25x2 )2 = lim = = −10 −1 x→0 −1 sinh x + (1 − x)2 ! 5xex (b) lim x→∞ lim ∞·0 = e 1 ex −   ln 1 4 x 1∞ ex − = lim e x→∞ x  1 ln e x − x→∞ 4 x 1 x   lim L’H ( 00 ) = e x→∞   4 x 1 1 ex − !x ! 0 0 !      1 1 4 x 4 lim x ln e x − lim ln ex − x x = ex→∞ = ex→∞   1
 · e x − x12 +  4 x 4 x2  − x12        1  1 4 1  1   1  2 x x + (−x ) lim · e · e lim  − (1) − 4  1 1 4 4 x→∞ x→∞ x2 x2 ex − ex − x x =e =e = e−3    3 x    3 ln(ln x) lim = ex→∞ x lim ln (ln x) = ex→∞ x→∞ x→∞     3 1  ln x  x    ∞  lim    3 3 ln(ln x) ∞ x→∞ 1 lim lim L’H x = e = ex→∞ x ln x = e0 = 1 ex→∞ 3 3. (a) ∞0 lim (ln x) x = lim e (c) [30 Points] Z  3 ln (ln x) x Compute the following definite integral. Please simplify your answer. ln 7 ln 7 x sinh x dx == x cosh x 0 0 − Z ln 7 ln 7 ln 7 cosh x dx = x cosh x 0 0 − sinh x = 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 2 0 50 7 = ln 7 2 ! 48 7 − 2 u=x ! = ln 7  25 7  −  24 7  dv = sinh xdx I.B.P. du = dx v = cosh x √ √ 3 3 √ x 3 3 1 x 3 1 1 √ + (b) dx = arcsin arctan + 6 3 3 3 3 36 − x2 9 + x2 3 √ ! √ !     3 3 1 1 3 3 3 3 = arcsin − arcsin + arctan − arctan 6 6 3 3 3 3 √ !   √  1 3 1 1 3 − arctan (1) = arcsin − arcsin + arctan 2 2 3 3 Z 3 π π 1 π  1 π  π π π 7π π 12π 6π 4π 3π − = − + − − + = − + − = 3 6 3 3 3 4 3 6 9 12 36 36 36 36 36 = Z (c) e 1 1 dx = arctan(ln x) x[1 + (ln x)2 ] = arctan(1) − arctan(0) = e 1 = arctan(ln e) − arctan(ln 1) π π −0= 4 4 OR you can use a standard u-substitution, being careful to change the limits of integration or mark u = ln x x = 1 ⇒ u = ln(1) = 0 them as x-limits. du = Z e 1 1 dx = x[1 + (ln x)2 ] OR Z e 1 4. 1 dx = x[1 + (ln x)2 ] [30 Points] Z Z 1 dx x 1 0 x = e ⇒ u = ln(e) = 1 1 du = arctan u 1 + u2 x=e x=1 1 0 1 du = arctan u 1 + u2 = arctan(1) − arctan(0) = x=e π π −0= 4 4 e = arctan(ln x) = ... 1 x=1 Compute the following indefinite integral. Z 1 x2 x2 √ arcsin x − dx = (a) x arcsin x dx = 2 2 1 − x2 Z Z x2 x2 1 1 sin2 θ √ = ·cos θ dθ = arcsin x− arcsin x− 2 2 2 2 cos2 θ Z 3 sin2 θ p · cos θ dθ 1 − sin2 θ Z x2 1 sin2 θ ·cos θ dθ = arcsin x− sin2 θdθ cos θ 2 2 1 x2 arcsin x − 2 2 Z x2 1 = arcsin x − 2 2 Z x2 1 1 − cos(2θ) dθ = arcsin x − 1 − cos(2θ)dθ 2 2 4   1 1 x2 1 1 x2 θ − sin(2θ) + C = arcsin x − arcsin x − θ + sin(2θ) + C = 2 4 2 2 4 8 = Z x2 x2 1 1 1 1 p 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 ex (e2x + 4) 7 2 dx = Z x 1 7 2 (u2 du = Z 1 (4 tan2 θ 7 2 · 2 sec2 θ dθ + 4) + 4) Z Z 1 1 1 2 2 √ · 2 sec2 θ dθ · 2 sec θ dθ = 7 · 2 sec θ dθ = 7 2 θ)7 2 (2 sec θ) 2 ( 4 sec (4 sec θ) Z Z Z 1 1 1 sec2 θ 1 dθ = 6 dθ = cos5 θ dθ 26 sec7 θ 2 sec5 θ 64 Z Z Z 1 1 1 4 2 2 cos θ cos θ dθ = (cos θ) cos θ dθ = (1 − sin2 θ)2 cos θ dθ 64 64 64 Z Z 1 1 (1 − w2 )2 dw = 1 − 2w2 + w4 dw 64 64     1 2w3 w5 1 2 sin3 θ sin5 θ + + w− +C = sin θ − +C 64 3 5 64 3 5 ! u5 1 2u3 u √ − 3 + 5 + C 64 u2 + 4 3(u2 + 4) 2 5(u2 + 4) 2 Z 1 = 64 √ ex e2x + 4 − 2e3x 3 3(e2x + 4) 2 + e5x 5 5(e2x + 4) 2 ! +C u = ex Standard u substitution to simplify at the start: du = ex dx 4 √ u = 2 tan θ u2 + 4 u θ Trig. Substitute du = 2 sec2 θdθ 2 Standard w substitution for odd trig. integral Z cos5 θ dθ technique: w = sin θ dw = cos θ dθ (c) Z ln(x2 + 1) dx = x ln(x2 + 1) − 2 Z x2 dx +1 x2 x2 + 1 − 1 dx = x ln(x2 + 1) − 2 = + 1) − 2 x2 + 1  Z Z 1 2 dx = x ln(x + 1) − 2 1 dx − x2 + 1 Z x ln(x2 Z x2 + 1 dx − x2 + 1 Z 1 dx x2 + 1  = 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 x2 + 1 NOTE: OR you can use a trig substitution to finish Z x2 dx. x2 + 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 √ √ √ 2 √ √ √  √ √ 1 + x e 1+ x − 3 1 + x e 1+ x + 5 1 + x e 1+ x − 5e 1+ x + C =4 = 4 ((∗) − (∗∗)) = 4 5 = 4e √ √ 1+ x √ √ h 1+ x (1   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 √ + √ 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 Here 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 = w 3 ew = w 3 ew − 3w2 ew − 3w2 ew +6 +6 Z Z 3 w 2 w  we dw = w e − 3w e + 6 we − w w Z w e dw  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 u=w v= du = 2wdw v = ew ew dv = ew dw Third I.B.P. du = dw v = ew ***************************************************************************** Z Z w w (**)Aside: we dw = we − ew dw = wew − ew + C u=w dv = ew dw I.B.P. du = dw v = ew OPTIONAL BONUS #2 2. Z ln(x − 1) √ dx = x Z Compute the following indefinite integral. √ Z Z ln(( x)2 − 1) √ dx = 2 ln(u2 − 1) dx = 2 ln((u − 1)(u + 1)) du x 6 =2 Z ln(u − 1) + ln(u + 1) du = 2 ((u − 1) ln(u − 1) − (u − 1) + (u + 1) ln(u + 1) − (u + 1)) + C √ √ √ √ √ √ = 2 (( x − 1) ln( x − 1) − ( x − 1) + ( x + 1) ln( x + 1) − ( x + 1)) + C √ √ √ √ √ √ = 2 (( x − 1) ln( x − 1) − x + 1 + ( x + 1) ln( x + 1) − x − 1) + C √ √ √ √ √ = 2 (( x − 1) ln( x − 1) − 2 x + ( x + 1) ln( x + 1)) + C Here u = du = 2du = √ x 1 √ dx 2 x 1 √ dx x √ √ √ Z Z ln(x − 1) ln(( x)2 − 1) ln(( x − 1)( x + 1)) √ √ √ OR recognize dx = dx = dx x x x √ √ √ √ Z Z Z ln( x − 1) ln( x + 1) ln( x − 1) + ln( x + 1) √ √ √ dx = dx + dx = x x x Z = . . . do subsitution on both pieces, and then I.B.P. √ √ √ √ √ √ = 2 (( x − 1) ln( x − 1) − ( x − 1) + ( x + 1) ln( x + 1) − ( x + 1)) + C 7