## anonymous one year ago Calculus Help

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1. anonymous

Write $\lim_{n \rightarrow \infty}\sum_{k=1}^{n}(2+k*(5/n))^3 *5/n$ as a definite Integral

2. anonymous

Find $dx/dy \int\limits_{2}^{x^3}\ln(x^2)dx$

3. anonymous

Find the area bounded by the curves y^2= 2x+6 and x=y+1. Work must be an integral with one variable

4. anonymous

Write the integral in one variable to find the volume of the solid obtained by rotating the first quadrant region bounded by y=0.5x^2 and y=x about the line x=7.

5. anonymous

0.5x² = x 0.5x² - x = 0 x (0.5x - 1) = 0 x = 0, 2 Those are the x-coordinates of the two points of intersection of y = 00.5x² and y=x. The full points are (0,0) and (2,2). Revolving the region around a vertical axis like x = 3 is better suited to the cylindrical shell method so that you don't have to solve for y. You integrate the lateral surface area of a cylindrical shell having its height vertically within the bounded region. The radius is 3-x (because x is the distance from the y-axis to the outer edge of the shell, leaving you with 3-x as the distance from the outer edge of the shell to x=3). The height of the cylinder is upper curve minus lower curve = x - 0.5x². Therefore, volume = 2π∫rh dx = 2π∫ (3-x)(x - 0.5x²) dx, from x = 0 to 2 = 2π∫ (3x-x² - 1.5x² + 0.5x³) dx, from x = 0 to 2 = 2π∫ (3x - 2.5x² + 0.5x³) dx, from x = 0 to 2 = 2π [3x²/2 - 2.5x³/3 + 0.5x⁴/4], from x = 0 to 2 = 2π [6 - 20/3 + 2] = 2π [4/3] = 8π/3

6. anonymous

A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as -32 feet per second squared.

7. Michele_Laino

question #2: the integral is, integrating by parts, is: $\Large \begin{gathered} \int {dx\ln \left( {{x^2}} \right)} = x\ln \left( {{x^2}} \right) - \int {x\frac{{dx}}{{{x^2}}}} 2x = \hfill \\ \hfill \\ = x\ln \left( {{x^2}} \right) - 2x \hfill \\ \end{gathered}$

8. Michele_Laino

so the definite integral is: $\Large \left. {x\ln \left( {{x^2}} \right) - 2x} \right|_2^{{x^3}} = {x^3}\ln \left( {{x^6}} \right) - 2{x^3} - 2\ln 4 + 4$

9. anonymous

for iluvhomewurk's did they just confuse 3 with 7?

10. Michele_Laino

Question #3 we can rewrite your equations as follows: $\Large \left\{ \begin{gathered} x = \frac{{{y^2} - 6}}{2} \hfill \\ x = y + 1 \hfill \\ \end{gathered} \right.$

11. Michele_Laino

the intersection point are: $\begin{gathered} P = \left( { - 1, - 2} \right) \hfill \\ Q = \left( {5,4} \right) \hfill \\ \end{gathered}$

12. Michele_Laino

so the requested area, is given by the subsequent integral: $\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)}$

13. anonymous

are you still talking about question 3?

14. Michele_Laino

yes!

15. Michele_Laino

here is the next step: $\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {y - \frac{{{y^2}}}{2} + 4} \right|_{ - 2}^4 = 28$

16. Michele_Laino

|dw:1433964888030:dw|

17. anonymous

18. Michele_Laino

yes!

19. Michele_Laino

oops.. it is 18

20. Michele_Laino

sorry!

21. Michele_Laino

$\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {y - \frac{{{y^2}}}{2} + 4} \right|_{ - 2}^4 = 18$

22. anonymous

its okay Is the work that iluvhomework did correct?

23. Michele_Laino

24. anonymous

okay thank you. I think it is just its 7 instead of 3

25. anonymous

and i got 0 not 18

26. Michele_Laino

area=0?

27. anonymous

the step is to plug in 4 into equation then subtract the equation with -2 pulgged into it correct?

28. Michele_Laino

yes! correct!

29. anonymous

When i did that i got 0

30. Michele_Laino

I'm checking....

31. Michele_Laino

you are right, I gave you the integrand function, please here is the integral: $\Large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {\frac{{{y^2}}}{2} - \frac{{{y^3}}}{6} + 4y} \right|_{ - 2}^4 = 18$

32. Michele_Laino

$\large \int_{ - 2}^4 {dy\left( {y + 1 - \frac{{{y^2} - 6}}{2}} \right)} = \left. {\frac{{{y^2}}}{2} - \frac{{{y^3}}}{6} + 4y} \right|_{ - 2}^4 = 18$

33. Michele_Laino

question #3 I make this traslation: $\Large \left\{ \begin{gathered} x = X + 7 \hfill \\ y = Y \hfill \\ \end{gathered} \right.$ where X, and Y are the new coordinates

34. Michele_Laino

as you can see my new reference system has the Y-axis along the axis of rotation

35. Michele_Laino

|dw:1433966377727:dw|

36. Michele_Laino

so the equations of your lines, becomes: $\Large \begin{gathered} y = \frac{{{x^2}}}{2} \to Y = \frac{{{{\left( {X + 7} \right)}^2}}}{2} \hfill \\ \hfill \\ y = x \to Y = X + 7 \hfill \\ \end{gathered}$

37. anonymous

what problem is this for? the one that ilovehomework started?

38. Michele_Laino

here is our situation: |dw:1433966625656:dw|

39. Michele_Laino

yes!

40. anonymous

Im confused on what work i am supposed to show

41. anonymous

is ilovehomeworks work not correct?

42. Michele_Laino

I give my solution, after that you can choose the solution which you prefer

43. Michele_Laino

I got a different result

44. anonymous

okay then please outline step by step on how you did the problem. Tnak you

45. anonymous

okay then please outline step by step on how you did the problem. Tnak you

46. Michele_Laino

ok! Now our volume is given integrating the area of annulus like below, along the Y-axis: |dw:1433967160201:dw|

47. anonymous

okay how do you show that mathematically.

48. Michele_Laino

here is the integral which gives the requested volume: $\Large \int_0^2 {dY\left\{ {{{\left( { - 7 + Y} \right)}^2} - {{\left( { - 7 + \sqrt {2Y} } \right)}^2}} \right\}} \pi$

49. anonymous

so thats my first step to write down?

50. Michele_Laino

your first step are the equations of traslation, I think

51. anonymous

okay

52. Michele_Laino

now, developing that integral, we get: $\Large \begin{gathered} \int_0^2 {dY\left\{ {{{\left( { - 7 + Y} \right)}^2} - {{\left( { - 7 + \sqrt {2Y} } \right)}^2}} \right\}} \pi = \hfill \\ \hfill \\ = \pi \left( {\left. {\frac{{{Y^3}}}{3} - 16\frac{{{Y^2}}}{2} + 14\sqrt 2 \times \frac{{2{Y^{3/2}}}}{3}} \right|_0^2} \right) \hfill \\ \end{gathered}$

53. Michele_Laino

and the final result is: $\Large V = 8\pi$

54. anonymous

A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as -32 feet per second squared.

55. Michele_Laino

here we have to use the subsequent vector equation: $\Large {\mathbf{OP}}\left( t \right) = {\mathbf{O}}{{\mathbf{P}}_{\mathbf{0}}} + {{\mathbf{v}}_{\mathbf{0}}}t + \frac{1}{2}{\mathbf{g}}{t^2}$

56. Michele_Laino

|dw:1433968218028:dw|

57. anonymous

okay

58. Michele_Laino

furthermore, we have to consider this one: $\Large {\mathbf{v}}\left( t \right) = {{\mathbf{v}}_{\mathbf{0}}} + {\mathbf{g}}t$ |dw:1433968355520:dw|

59. anonymous

so we use both equations?

60. Michele_Laino

yes!

61. Michele_Laino

now I rewrite both equations using z-components: $\Large \left\{ \begin{gathered} z\left( t \right) = {z_0} + {v_0}t - \frac{1}{2}g{t^2} \hfill \\ {v_z}\left( t \right) = {v_0} - gt \hfill \\ \end{gathered} \right.$

62. Michele_Laino

when the ball reaches its maximum height, then its velocity is zero, so we can write: $\Large 0 = {v_0} - g\tau$ where \tau is the time at which our ball reaches its maximum height

63. Michele_Laino

so: $\Large \tau = \frac{{{v_0}}}{g}$

64. anonymous

so t=40/-32?

65. Michele_Laino

no, it is \tau = 40/32=...

66. Michele_Laino

g is the magnitude of gravity: g=9.81 m/sec^2=32 feet/sec^2

67. anonymous

wont it be -32?

68. Michele_Laino

no, since I wrote my scalar equations using the magnitudes or length of the respective vectors the maximum height is then (substituting \tau into the first scalar equation): $\Large \begin{gathered} z\left( \tau \right) = {z_0} + {v_0}\tau - \frac{1}{2}g{\tau ^2} = h + \frac{{v_0^2}}{g} - \frac{1}{2}\frac{{v_0^2}}{g} = \hfill \\ \hfill \\ = h + \frac{{v_0^2}}{{2g}} \hfill \\ \end{gathered}$

69. Michele_Laino

where h=500 feet

70. Michele_Laino

next, in order to find the requested velocity, we can apply the formula of Torricelli, like below: $\Large v = \sqrt {2gz\left( \tau \right)} = \sqrt {2gh + v_0^2}$

71. Michele_Laino

substituting your data we find: $\Large \begin{gathered} v = \sqrt {2gz\left( \tau \right)} = \sqrt {2gh + v_0^2} = \sqrt {2 \times 32 \times 500 + {{40}^2}} = \hfill \\ = 183.3\;{\text{feet/sec}} \hfill \\ \end{gathered}$

72. Michele_Laino

$\large \begin{gathered} v = \sqrt {2gz\left( \tau \right)} = \sqrt {2gh + v_0^2} = \sqrt {2 \times 32 \times 500 + {{40}^2}} = \hfill \\ = 183.3\;{\text{feet/sec}} \hfill \\ \end{gathered}$

73. anonymous

74. Michele_Laino

yes!

75. anonymous

Thank you so much. Could you help me with about 3 more?

76. Michele_Laino

ok!

77. anonymous

Use the midpoint rule with n=4 to approximate the area of the region bounded by y=x^3 and y=x

78. Michele_Laino

79. anonymous

okay thank you

80. Michele_Laino

the intersection points between those equations are: $\begin{gathered} P = \left( { - 1, - 1} \right) \hfill \\ Q = \left( {1,1} \right) \hfill \\ O = \left( {0,0} \right) \hfill \\ \end{gathered}$

81. Michele_Laino

|dw:1433970078818:dw|

82. anonymous

ok i see the intersection points

83. Michele_Laino

so, it is suffice to compute the bounded area which lies into the first quadrant

84. anonymous

yes I agree with that

85. Michele_Laino

|dw:1433970288453:dw|

86. Michele_Laino

the area under the cubic parabola, is given by the subsequent formula: $\Large A = \frac{{\left( {1 - 0} \right)}}{2}\left( {f\left( {1/4} \right) + f\left( {3/4} \right)} \right)$ where f(x) = x^3

87. Michele_Laino

so we get: $\Large \begin{gathered} A = \frac{{\left( {1 - 0} \right)}}{2}\left( {f\left( {1/4} \right) + f\left( {3/4} \right)} \right) = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {\frac{1}{{64}} + \frac{{27}}{{64}}} \right\} = \frac{{14}}{{64}} \hfill \\ \end{gathered}$

88. anonymous

simplidy it to 7/32?

89. Michele_Laino

|dw:1433970540095:dw| the area of bounded region is: $\Large {A_B} = \frac{1}{2} - \frac{{14}}{{64}} = \frac{7}{{16}}$

90. anonymous

oh okay

91. Michele_Laino

so the requested area is: $\Large 2{A_B} = \frac{7}{8}$

92. anonymous

who do you format it like that?

93. anonymous

why

94. Michele_Laino

since our drawing is symmetric with respect to the y-axis: |dw:1433970857126:dw|

95. anonymous

wouldnt one of the areas be negative because it is below the x axis?

96. anonymous

so it would be 0?

97. Michele_Laino

no, since an area is the measure of the subset of the x,y-plane, and the measure of a set, as it is defined by mathematicians, is always positive

98. anonymous

okay so the fonal answer would be 7/8?

99. Michele_Laino

yes!

100. anonymous

okay heres the next one

101. Michele_Laino

ok!

102. anonymous

I have completed it i just wanna make sure i did it correct.

103. anonymous

Trash is delivered to a dumpsite between the hours of midnight(t=0) and 6am (t=6). Selected Values of the rates of deleivery are shown in the table below with t measured in hours and r(t) measured in tons per hours. Use 4 trapezoids to estimate total amount of trash. Round to nearest ton

104. anonymous

t 0 1 3 5 6 R(t) 4 2 3 2 1

105. anonymous

Lsum= (1)(4)+(2)(2)+(2)(3)+(1)(2)=4+4+6+2= 16 Rsum= (1)(2)+(2)(3)+(2)(2)+(1)(1)= 2+6+4+1= 13 Tsum= ½(Rsum+Lsum)= ½(16+13)=1/2(29)= 14.5 15 Tons

106. anonymous

I think i did it correctly

107. Michele_Laino

108. anonymous

Lsum= (1)(4)+(2)(2)+(2)(3)+(1)(2)=4+4+6+2= 16 Rsum= (1)(2)+(2)(3)+(2)(2)+(1)(1)= 2+6+4+1= 13 Tsum= ½(Rsum+Lsum)= ½(16+13)=1/2(29)= 14.5 15 Tons

109. anonymous

sorry dont know why it sent again

110. Michele_Laino

Lsum is right for me whereas I write Rsum like this: Rsum= (2)(2)+(2)(3)+(2)(2)+(1)(1)= 4+6+4+1= 15

111. anonymous

okat thank you for al this help just two more!

112. Michele_Laino

ok!

113. anonymous

Show that the function $F(x)= \int\limits_{x}^{3x}1/t dt$ is constant on the interval(0, infty)

114. Michele_Laino

it is simple, since we have: $\Large \int_x^{3x} {\frac{{dt}}{t}} = \ln \left( {3x} \right) - \ln \left( x \right) = \ln \left( {\frac{{3x}}{x}} \right) = \ln 3$

115. anonymous

how did you get ln(3x) and ln(x) from the integral?

116. Michele_Laino

yes! since we can write: $\Large \int {\frac{{dt}}{t}} = \ln t + k$

117. anonymous

oh okay!

118. anonymous

and since it comes out to be a number that means it is constant?

119. Michele_Laino

correct!

120. anonymous

perfect! last question

121. Michele_Laino

ok!

122. anonymous

write$\lim_{n \rightarrow \infty} \sum_{k=1}^{n} (2+k*5/n)^3*5/n$ as a definite integral

123. Michele_Laino

I'm pondering...

124. anonymous

Thank you

125. Michele_Laino

what is k

126. anonymous

constant

127. Michele_Laino

is it a parameter?

128. Michele_Laino

ok!

129. anonymous

I think i got it

130. anonymous

131. anonymous

does that look correct?

132. Michele_Laino

here is my answer: I write the sum as integral, like below: $\Large \int_0^n {dk} {\left( {2 + \frac{{5k}}{n}} \right)^3}\frac{5}{n}$

133. anonymous

Thank you for all your help i am finished

134. Michele_Laino

which is equal to: $\Large \int_0^n {dk} {\left( {2 + \frac{{5k}}{n}} \right)^3}\frac{5}{n} = \left. {{{\left( {2 + \frac{{5k}}{n}} \right)}^4}\frac{1}{4}} \right|_0^n$

135. Michele_Laino

and finally, I get: $\Large \begin{gathered} \int_0^n {dk} {\left( {2 + \frac{{5k}}{n}} \right)^3}\frac{5}{n} = \left. {{{\left( {2 + \frac{{5k}}{n}} \right)}^4}\frac{1}{4}} \right|_0^n = \hfill \\ \hfill \\ = \frac{{{7^4} - {2^4}}}{4} \hfill \\ \end{gathered}$