Math 11 Final Review Packet for Remaining Material since Exam #3

Math 11 Final Review Packet for Remaining Material since Exam #3 Integration Compute each of the following integrals: 1. Z x(x2 + 1)14 dx = 1 2 Z u14 du = 1 2 1 u15 +C = (x + 1)15 + C 2 15 30 u = x2 + 1 du = 2xdx Here  1 2 du = xdx   Z 1 1 1 u2 + C = sin2 (4x) + C 2. sin(4x) cos(4x) dx = u du = 4 4 2 8  u = sin(4x)  du = cos(4x)4dx Here  1 4 du = cos(4x)dx Z x Z e − e−x 1 3. dx = du = ln |u| + C = ln |ex + e−x | + C x −x e +e u  u = ex + e−x Here du = ex − e−x dx Z 4. 5. 6. 7. 8. Z √ 3 1√ 2 3 2 1 + ln x dx = u du = u 2 + C = (1 + ln x) 2 + C x 3 3 ( u = 1 + ln x Here 1 du = dx x Z Z 1 1 dx = du = ln |u| + C = ln | ln(x + 1)| + C (x + 1) ln(x + 1) u   u = ln(x + 1) 1 Here dx  du = x+1 Z Z 1 sin x dx = − du = − ln |u| + C = − ln |7 + cos x| + C 7 + cos x u  u = 7 + cos x  du = − sin xdx Here  −du = sin xdx Z Z 6ex 1 dx = 6 dx = 6 ln |u| + C = 6 ln |ex + 7| + C x e +7 u  u = ex + 7 Here du = ex dx Z Z Or eln(sin x) eln(cos x+7) dx = Z sin x dx and this reduces to #6. cos x + 7 1 eln(sin x) dx = eln(sin x) − ln(cos x + 7) dx = ln(cos x + 7) e Z sin x = dx which still reduces to #6. cos x + 7 Z Z 9. Z 2 ln(ex ex e7 ) dx = Z ln ex 2 +x+7 dx = Z Z x2 + x + 7 dx = ln e  sin x cos x + 7 Z 1 1 1 1 1 dx = − du = − ln |u| + C = − ln |1 − 2x| + C 11. 1 − 2x 2 u 2 2  u = 1 − 2x  du = −2dx Here  1 − 2 du = dx Z 12. 3x+1 e   1 dx = 3 Z 1 1 eu du = eu + C = e3x+1 + C 3 3 u = 3x + 1 du = 3dx Here  1 3 du = dx 1 Z 1 1 1 1 e− x7 dx = eu du = eu + C = e− x7 + C 13. 8 x 7 7 7  1 u = − x7  du = 7x−8 dx Here  1 1 7 du = x8 dx Z Z Z 1 −x dx = e dx = − eu du = −eu + C = −e−x + C 14. ex  u = −x  du = −dx Here  −du = dx Z Z 8 1 1 1 u=8 1 1 1 dx = du = ln |u| = (ln 8 − ln 1) = ln 8 15. 7 u=1 u 7 7 7 0 7x + 1 1   u = 7x + 1  x = 0 =⇒ u = 1 du = 7dx and Here x = 1 =⇒ u = 8  1 7 du = dx Z 1 2 dx x3 x2 + + 7x + C 3 2 Or Z Z x3 x2 x2 x 7 + + 7x + C ln e + ln e + ln e dx = x2 + x + 7 dx = 3 2 Z Z 1 6x + 3 dx = 3 du = 3 ln |u| + C = 3 ln |x2 + x − 5| + C 10. x2 + x − 5 u  u = x2 + x − 5 Here du = 2x + 1dx Z  Z u=2 1 1 2 1 1 u−1 2 1 =− = − − (−1) = dx = du = 16. 2 2 x(ln x) u −1 u 2 2 u=1 e 1 1   x = e =⇒ u = ln e = 1 u = ln x and Here x = e2 =⇒ u = ln e2 = 2 du = x1 dx 17. Z e2 Z ln 7 9e2x dx = ln 4 Here 9 2   Z ln 49 ln 16 9 eu du = eu 2 u = 2x du = 2dx and  1 2 du = dx 2  ln 49 = ln 16  9 9  ln 49 297 9 e − eln 16 = (49 − 16) = (33) = 2 2 2 2 x = ln 4 =⇒ u = 2 ln 4 = ln 42 = ln 16 x = ln 7 =⇒ u = 2 ln 7 = ln 72 = ln 49 4 1 −2x ln 3 −2x −x + e dx = 4x − 4e − e = (4 ln 3 − 4e− ln 3 − ex 2 0 0 0 1 1 9 1 9 1 −2 ln 3 −1 −2 e ) − (0 − 4e0 − e0 ) = 4 ln 3 − 4eln 3 − eln 3 + = 4 ln 3 − (4)3−1 − 3−2 + = 2 2 2 2 2 2 9 4 1 9 24 1 81 1 1 1 56 + = 4 ln 3 − − + = 4 ln 3 − − + = 4 ln 3 − 18 4 ln 3 − (4) − 3 2 32 2 3 18 2 18 18 18 Z Z 2  1 wew 1  1 1 1 w2 w2 dw = 19. du = ln |u| + C = ln |17 + e | + C = ln 17 + e +C 2 u 2 2 2 17 + ew2  2  u = 17 + ew  2 Here du = ew (2w) dw   1 du = wew2 dw 2 18. 20. Z ln 3  1 2+ x e dx = Z ln 3 4+ ln 3  1  2 2 1  2 ln 3 1 = e − e2 ln 2 = eln(3 ) − eln(2 ) e2x dx = e2x 2 2 2 ln 2 ln 2  1 5 1  ln 9 e − eln 4 = (9 − 4) = = 2 2 2 Z ln 3 OR we can use u-substitution to finish the integral. Z ln 3 Z  9−4 1 ln 9 u 1 ln 9 1  ln 9 5 e2x dx = e du = eu e − eln 4 = = = 2 2 2 2 2 ln 2 ln 4 ln 4   u = 2x  x = ln 2 =⇒ u = 2 ln 2 = ln 22 = ln 4 du = 2 dx and Here x = ln 3 =⇒ u = 2 ln 3 = ln 32 = ln 9  1 du = dx 2 Z 2 (ln(1 + e−x )) u2 e−x ln(1 + e−x ) dx = − u du = − + C = − +C 21. 1 + e−x 2 2  u = ln(1 + e−x )    1    du = (e−x )(−1) dx −x 1 + e Here    −x    −du = e dx 1 + e−x Z e4 Z u=4 Z u=4 u=4 √ √ 1 1 1 1 √ √ du = =2 4−2 1= 2 22. u− 2 du = 2u 2 dx = u x ln x e u=1 u=1 u=1 3 Z Here 23. Z 24. Z ( 3x (e u = ln x 1 du = dx x −7x 2 +e ) dx = and Z  x = e =⇒ u = ln e = 1 x = e4 =⇒ u = ln(e4 ) = 4 e6x + 2e−4x + e−14x dx = 1 6x 1 −4x 1 e − e − e−14x + C 6 2 14 Z 1 1 1 1 1 dx = du = ln |u| + C = ln |2x − 1| + C 2x − 1 2 u 2 2  u = 2x − 1   du = 2 dx Here 1   du = dx 2 Derivatives/Tangent Lines 25. Find the equation of the tangent line to the curve y = (x + 2)e−x at the point (0, 2). y ′ = (x + 2)e−x (−1) + e−x (1) = e−x (1 − (x + 2)) = e−x (1 − x − 2) = e−x (−1 − x) Then, the slope at x = 0 is given by y ′ (0) = e0 (−1 − 0) = −1. As a result, the equation of the tangent line is y − 2 = (−1)(x − 0) or y = −x + 2 . 26. Find the equation of the tangent line to the curve y = ln(xe−3x ) at the point (1, −3). First we can simplify the function y = ln(xe−3x ) = ln x + ln e−3x = ln x − 3x. 1 Then y ′ = − 3. x The slope at x = 1 is given by y ′ (1) = 1 − 3 = −2. As a result, the equation of the tangent line is y + 3 = −2(x − 1) or y = −2x − 1 . 27. Let y = ln x , find f ′ (1). 1 + x2 Using the Quotient Rule, y′
(1 + x2 ) x1 − ln x(2x) = = (1 + x2 )2 Then the slope at x = 1 is given by y ′ (1) = 1 x + x − 2x ln x . (1 + x2 )2 1 1 + 1 − 2 ln 1 = . 4 2 28. Let f (x) = x ln x with x > 0. Where is f (x) concave up? 1 First f ′ (x) = x + ln x(1) = 1 + ln x. x 1 Second f ′′ (x) = . Notice that the second derivative is always positive for x > 0. So f is x concave up for all x > 0 which is the entire domain of the original function f . 4 29. Let x2 ey = ln(xy). Find dy . dx
d d x2 ey = (ln(xy)). Implicit differentiation yields dx dx   dy 1 1 dy 1 dy Then x2 ey + ey 2x = +y = + x dx xy dx y dx x 1 dy 1 dy − = − 2xey Simplifying yields x2 ey dx y dx x 1 − 2xey dy x = Finally, 1 dx x2 ey − y 30. Find all local maximum and minimum value(s) of the function f (x) = (x2 − 7)e−x . Compute f ′ (x) = (x2 − 7)e−x (−1) + e−x (2x) = e−x (−x2 √ + 7 + 2x). Setting f ′ (x) = 0, we need 2 to solve −x + 2x + 7 = 0 which has solutions x = 1√ ± 2 2. Sign-testing √ √around these points √ √ −(1+2 2) (2 + 4 2)) and a local min yields a local max at (1 + 2 2, f (1 + 2 2)) = (1 + 2 2, e √ √ √ √ √ at (1 − 2 2, f (1 − 2 2)) = (1 − 2 2, e−(1−2 2) (2 − 4 2)). (Recall that e−x 6= 0) 31. Compute the derivatives of the following functions. (Hint: You may want to simplify first.) (a) f (x) = ln(5xe−5x ) = ln(5x) + ln(e−5x ) = ln(5x) − 5x 1 1 Then f ′ (x) = (5) − 5 = −5 5x x ln(x2 + x) − ln x “ (b) f (x) = e = (x2 + x)x−1 = ” 2 2 (−1) ) = eln(x + x) e(− ln x) = eln(x + x) e(ln x (x2 + x) =x+1 x Then f ′ (x) = 1 .  x  √  
1  xe (c) f (x) = ln √ = ln (xex ) − ln e7x = ln x + ln (ex ) − ln e7x 2 e7x
1 1 = ln x + x − ln e7x = ln x + x − (7x) 2 2 1 7 +1− . Then f ′ (x) = x 2 32. Let f (x) = xcos x . Compute f ′ (x). Set y = xcos x Then ln y = ln (xcos x ) = cos x ln x. Implicit differentation yields 1 dy 1 = cos x + ln x(− sin x). y dx x  cos x  dy =y − ln x sin x = xcos x Solve for dx x 5 cos x x
− ln x sin x 33. Let f (x) = (tan x)x . Compute f ′ (x). Set y = (tan x)x Then ln y = ln ((tan x)x ) = x ln(tan x). Implicit differentation yields  
1 1 dy 2 =x sec x + ln(tan x)(1). y dx tan x   2     sec x dy sec2 x =y x + ln(tan x) = (tan x)x xtan Solve for + ln(tan x) x dx tan x 34. Let f (x) = x4 e−x . For this function, discuss domain, vertical and horizontal asymptote(s), interval(s) of increase or decrease, local extreme value(s), concavity, and inflection point(s). Then use this information to present a detailed and labelled sketch of the curve. Take my word that lim f (x) = 0 and lim f (x) = +∞ x→∞ x→−∞ • Domain= R. Note that f (x) = x4 ex and ex is never zero in the denominator. • It has no vertical asymptotes. • There is a horizontal asymptote for this f at y = 0 because lim f (x) = 0. x→∞ • First Derivative Information We compute f ′ (x) = x4 e−x (−1) + e−x (4x3 ) = e−x (4x3 − x4 ) = e−x x3 (4 − x) and set it equal to 0 and solve for x to find critical numbers. The critical points occur where f ′ is undefined (never here) or zero. The latter happens when x = 0 or x = 4. Using sign testing/analysis for f ′ , f′ ⊖  ⊕ t t ⊖ - 0 4 f ց local ր ց local min max So f is increasing on the interval (0, 4); and f is decreasing on (−∞, 0) and (4, ∞). Moreover, f has a local max at x = 4 with f (4) = 256e−4 , and a local min at x = 0 with f (0) = 0. • Second Derivative Information Recall, f ′ (x) = e−x (4x3 − x4 ). Next f ′′ = e−x (12x2 −4x3 )−e−x (4x3 −x4 ) = e−x (12x2 −4x3 −4x3 +x4 ) = e−x x2 (12−8x+x2 ) = e−x x2 (x − 2)(x − 6). Setting f ′′ = 0 we solve for our possible inflection points x = 0, x = 2, or x = 6. Using sign testing/analysis for f ′′ , f ′′  f ⊕ ⊕ t t S 0S 2 6 infl. point ⊖ t T 6 infl. point ⊕ - S So f is concave down on the interval (2, 6) and concave up on (−∞, 2) and (6, ∞), with inflection points at x = 2 and x = 6. • Piece the first and second derivative information together t  f f f t t t - 0 2 4 6 ց ր ր ց ց S S T T S      local infl. local infl. min pt. max pt. Areas between Curves and Volumes of Revolution SEE ME FOR THE SKECTHES OF THE REGIONS IF YOU HAVE ANY QUESTIONS! 35. Consider the region in the plane bounded by the curves y = ex+1 , y = e2x , and the y-axis. (a). Find the area of this region. First, the two curves intersect when ex+1 = e2x or when ln ex+1 = ln e2x so that x + 1 = 2x =⇒ x = 1. The other boundary line is the y-axis, which is x = 0. Between x = 0 and x = 1, the curve y = ex+1 lies above the curve y = e2x . 7 Then Area= Z 1 x+1 e 0 2x −e dx = e x+1 1 − e2x 2 1   e2 1 0 1 e2 1 − e − e = − e1 + =e − 2 2 2 2 2 0 (b). Rotate this region about the x-axis. What is the volume of the resulting solid? Z 1  Z 1
2
2  π ex+1 − e2x dx π((outer radius)2 − (inner radius)2 ) dx = Volume= 0 = Z 1 0 = π
π e2x+2 − e4x dx = π  e4 e2 1 − + 4 2 4  e2x+2 2 − e4x 4  1 =π 0  0 e4 e4 − 2 4  −  e2 e0 − 2 4   36. Consider the region enclosed by y = e−x , y = ex , and x = 2 and rotate it about the x–axis. What is the volume of the resulting solid? Z 2 Z 2 
2  2 2 Volume= π((outer radius) − (inner radius) ) dx = π (ex )2 − e−x dx 0 = Z 2 π e 0 = π 2x  0 −2x −e
dx = π  e2x 2 − e−2x −2  2 =π 0  e4 e−4 + 2 2  −  e0 e0 + 2 2   e4 1 + 4 −1 2 2e 1 37. Consider the region enclosed by y = , y = 0, x = 1 and x = 3 and rotate it about the x x–axis. What is the volume of the resulting solid? Z 3  2 Z 3 Z 3 1 x−1 3 π 3 2 π π(f (x)) dx = Volume= πx−2 dx = π dx = =− x −1 1 x 1 1 1 1 2π π π  π =− +π = =− − − 3 1 3 3 38. Find the area enclosed by y = ex , y = e3x and x = 1. The curves y = ex and y = e3x intersect when ex = e3x or when ln(ex ) = ln(e3x ) =⇒ x = 3x =⇒ x = 0.     Z 1 1 1 3x 1 0 1 3 3x x x 1 0 e − e dx = e − e Finally, the area A = e −e − e −e = = 3 3 3 0 0     1 3 1 1 2 e − e1 − − 1 = e3 − e + 3 3 3 3 Properties of ex and ln x 39. Simplify each of the following 8 (a) ln(eln e ) = ln(e1 ) = 1 (b) ln ln or ln(e(ln e) ) = (ln e)(ln e) = (1)(1) = 1 1 = ln ln e−1 = ln |−1| = ln 1 = 0 e 40. Solve each of the the following equations for x: (a) ln(ln x) = 1 First eln(ln x) = e1 =⇒ ln x = e. Then, eln x = ee =⇒ x = ee . (b) ln(x2 ) = 2 + ln x Note that x > 0 is required here. 2 First eln(x ) = e2+ln x = e2 eln x = e2 x =⇒ x2 = e2 x. Then, x2 − e2 x = 0 =⇒ x(x − e2 ) = 0 =⇒ x = 0 or x = e2 . Since we must have x > 0 we can ingnore the solution x = 0, so x = e2 . (c) e3x−4 = 7
ln 7 + 4 First ln e3x−4 = ln 7 =⇒ 3x − 4 = ln 7 =⇒ x = . 3 41. Decide whether each statement is True or False. Explain why or why not. 2 (a) (ex )2 = ex 2 False, because (ex )2 = e2x 6= ex . “piggy-back exponents multiply” (b) ln 5 − ln 3 = ln 2   5 6= ln 2. False, because ln 5 − ln 3 = ln 3 (c) (ln x)(ln x) = ln(x2 ) False because(ln x)(ln x) = (ln x)2 6= 2 ln x which equals ln(x2 ). 9