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## anonymous one year ago intergration problem

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1. anonymous

2. anonymous

my firsr step is antderv right

3. anonymous

i have an idea (although i totally suck at these)

4. anonymous

it will be easier to integrate in terms of $$y$$ not $$x$$ so lets solve each of these for $$x$$

5. anonymous

the first one is $$x=y+1$$ and the second is $$x=\frac{y^2-6}{2}$$

6. anonymous

we also need to find the limits of integration did you find where they intersect?

7. anonymous

oh and to answer your question, no, the first step is not to find the anti derivative, the first step is to find the thing you need to integrate, both the integrand and the limits of integration THEN you can take the anti - derivative to compute the integral

8. anonymous

how to do that

9. anonymous

set them equal and solve

10. anonymous

$y=x-1,y^2=2x+6$ replace the $$y$$ in the second equation by $$x-1$$ and solve $(x-1)^2=2x+6$ turns out you get an easy quadratic equation to solve (it factor, get integer solutions )

11. anonymous

that is if you want to do it by hand it if was me, i would cut to the chase http://www.wolframalpha.com/input/?i=%28x-1%29^2%3D2x%2B6

12. anonymous

k so i get x^2-4x+8

13. anonymous

you get $$x=-1$$ or $$x=5$$ the points where they intersect are $$(-1,-2)$$ and $$(5,4)$$

14. anonymous

hmm no i think if you do it by hand you get $x^2-2x+1=2x+6\\ x^2-4x-5=0$

15. anonymous

o wow yea yea

16. anonymous

factors as$(x-5)(x+1)=0\\ x=-1,x=5$

17. anonymous

ok then

18. anonymous

now this is a parabola that opens sideways so it will be easier to integrate wrt y not x

19. anonymous

here is a picture http://www.wolframalpha.com/input/?i=+y%3Dx+-+1%2C+y^2%3D2x%2B6

20. anonymous

so if you want to integrate wrt y, you need to write each of these as a function of $$y$$ i.e solve each for $$x$$

21. anonymous

first one is $x=y+1$ second is $x=\frac{y^2-6}{2}$

22. anonymous

so far so good?

23. anonymous

okey

24. anonymous

now if we turn it sideways we see the line is above the parabola the limits of integration in terms of y means we use the y values $\int_{-2}^5y+1-(\frac{y^2-6}{2})dy$

25. anonymous

do some algebra first maybe, then compute that integral

26. anonymous

i have to take the antiderv right

27. anonymous

i would do the algebra first then take anti derivatives

28. anonymous

oh u mean simplify

29. anonymous

oops $\int_{-2}^5\left(\frac{-y^2}{2}+y+4\right)dy$

30. xapproachesinfinity

seems good :)

31. anonymous

thanks !

32. anonymous

now take the anto right

33. anonymous

NOW take anti derivatives

34. anonymous

yeah i know you are dying to do it, do it now

35. anonymous

okey tge y2/2 is confusing

36. anonymous

would it be y3/6?

37. anonymous

no

38. xapproachesinfinity

yes! just don't forgot - sign

39. anonymous

$-\frac{y^2}{6}$ for the first term

40. anonymous

ok as @xapproachesinfinity said "yes" only "no"

41. anonymous

oh yea sorry

42. anonymous

the rest should be routine anti derivative is $-\frac{y^2}{6}+\frac{y^2}{2}+4y$ plug in 5, plug in -2 and subtract if you do it right, you will get 18i think

43. anonymous

so i get 12.5 after simliy

44. anonymous

damn typo $-\frac{y^3}{6}+\frac{y^2}{2}+4y$

45. anonymous

i think ur right one my firends got 18 two

46. anonymous

me, i like to know what the answer is before i begin http://www.wolframalpha.com/input/?i=+area+between+the+curves+y%3Dx+-+1%2C+y^2%3D2x%2B6

47. anonymous

ok so i have more intergral problems but just much eaiser i am just stuck since i am new to all this so could u help me more?

48. anonymous

ok sure why not

49. anonymous

if i can i kind of suck at these though

50. anonymous

ok give me 5 minutes i am finishing somthing

51. anonymous

kk

52. anonymous

lets see

53. anonymous

|dw:1433296429808:dw|

54. anonymous

i did everything but getting the answer wrong

55. anonymous

here is my work just before the solution

56. anonymous

2/3x^3/2 + x +lnx , am i right

57. anonymous

looks like you have an extra term there

58. anonymous

$\frac{\sqrt{x}+1}{x}=x^{-\frac{1}{2}}+\frac{1}{x}$

59. anonymous

not sure where the x came from

60. anonymous

how is squrtx , x ^-1/2? half should be positve right

61. anonymous

also note that $\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$ not $$\sqrt{x}$$

62. anonymous

or if you like $\frac{\sqrt{x}}{x}=\frac{x^{\frac{1}{2}}}{x}=x^{\frac{1}{2}-1}=x^{-\frac{1}{2}}$

63. anonymous

you got that?

64. anonymous

yes

65. anonymous

wait i thiught it was +1 when taking anti

66. anonymous

ok now take the anti derivative

67. anonymous

yes we didn't do that yet

68. anonymous

then why did u -1?

69. anonymous

as you put it, we 'simplified' first to get $x^{-\frac{1}{2}}+\frac{1}{x}$ now go from there

70. anonymous

ok lets go slow

71. anonymous

first off you are given $\frac{\sqrt{x}+1}{x}$ right?

72. anonymous

yes

73. anonymous

then BEFORE we find the anti derivative we are going to divide each term in the numerator by x that is not integrating, that is dividing

74. anonymous

$\frac{\sqrt{x}+1}{x}=\frac{\sqrt{x}}{x}+\frac{1}{x}$ right?

75. anonymous

yes

76. anonymous

we still have not taken the anti derivative yet the anti derivative of $$\frac{1}{x}$$ is what you said $$\ln(x)$$

77. anonymous

but in order to take the anti derivative of $\frac{\sqrt{x}}{x}$ we need to write it as x to some power

78. anonymous

ohh k

79. anonymous

then we can use the power rule (backwards) for finding it as a power it is $\frac{\sqrt{x}}{x}=x^{-\frac{1}{2}}$ that is not finding the anti derivative, that is just writing it in exponential form

80. anonymous

yea i think that my only mistake

81. anonymous

THEN you can add one to the exponent etc

82. anonymous

you should get $2x^{\frac{1}{2}}$ or if you prefer $2\sqrt{x}$

83. anonymous

questions or clear?

84. anonymous

k i am going to finish this uo

85. anonymous

kk

86. anonymous

i am getting 6.38 , u ?

87. anonymous

@satellite73 the answer on my book is like 2+ln4

88. anonymous

@satellite73 u their ?

89. anonymous

plug in 4, plug in 1 subtract

90. anonymous

if the book as $$\ln(4)$$ as part of the answer, do not use a calculator you may have got the same answer but as a decimal

91. anonymous

$F(x)=2\sqrt{x}+\ln(x)$

92. anonymous

$F(4)=2\sqrt4 +\ln(4)=4+\ln(4)$

93. anonymous

$F(1)=2\sqrt 1+\ln(1)=2$

94. anonymous

i did 1^1/2 i got 1

95. anonymous

did you forget the 2 out front?

96. anonymous

2-1=1

97. anonymous

omg yea

98. anonymous

$\int x^{-\frac{1}{2}}dx=2\sqrt{x}$

99. anonymous

damn silly mistakes errrr

100. anonymous

lol there are lots of ways to make a mistake!

101. anonymous

serioulsy the only reasons i lose mrks in calc

102. anonymous

but i am making miskates here so i wont makw them on the test

103. anonymous

good!

104. anonymous

thats the number 2 reason for doing homework

105. anonymous

do u know questions realted to area under graph where u do the same thing just now draw

106. anonymous

go ahead i will look

107. anonymous

"fnid teh aera fo teh reiogn beteewn the gienv curesv, setch of the reiogn

108. anonymous

y=x^2 +3 and y= x+a from 2 to 4

109. anonymous

i mean x+1

110. anonymous

what i did was made them equal to each other then anti

111. anonymous

no not in this case

112. anonymous

you are given the limits of integration, you don't have to find them

113. anonymous

the only thing you need to know is that the parabola lies above the line

114. anonymous

so $\int_2^2\left(x^2+3-(x+1)\right)dx$

115. anonymous

simplify then anti ?

116. anonymous

no first algebra, then anti derivative

117. anonymous

so its x2 -x +2

118. anonymous

so after antoi i get the wuation 1/3x3 - 1/2x2 +2x , am i right

119. anonymous

yes looks good

120. anonymous

you see the difference in that and the previous one is that you were asked for the area between the curves then you had to find where they intersect in this case you are given the limits of integration

121. anonymous

ok

122. anonymous

then what do we do

123. anonymous

plug in 4, plug in 2, subtract

124. anonymous

that is what you always do $F(x)=\frac{x^3}{3}-\frac{x^2}{2}+2x$ compute $F(4)-F(2)$

125. anonymous

what do u get , i get 12.05

126. anonymous

idk i didn't do it

127. anonymous

ok what do now

128. anonymous

i will check your answer

129. anonymous

i got $$\frac{50}{3}$$ hmmm

130. anonymous

oh ur right maybe i did wrong sub

131. anonymous

$F(4)=\frac{4^3}{3}-\frac{4^2}{2}+2\times 4$ $F(2)=\frac{2^3}{3}+\frac{2^2}{4}+2\times 2$need a calculator or sommat

132. anonymous

i think i need to stop subbing EVERYRthing into calc

133. anonymous

ok so what now

134. anonymous

how can i draw the graph from this info

135. anonymous

now you are done go have a beer

136. anonymous

if you have to graph, graph the line and the parabola, that is all

137. anonymous

both are real easy to graph

138. anonymous

sooo??|dw:1433301650060:dw|

139. anonymous

|dw:1433301637176:dw|

140. anonymous

yeah

141. anonymous

on the back of my book its like |dw:1433301723404:dw|

142. anonymous

why is that area shaded

143. anonymous

because that is what you were asked for the area bounded by the cuve $$y=x^2+3$$ the line $$y=x+1$$ between the vertical lines $$x=2$$ and $$x=4$$

144. anonymous

so all the work we did had nothing to do with drawing the graph?

145. anonymous

we found the area of that region, but no, finding the area is not the same as graphing the region, they are two different things

146. anonymous

how long r u gonna stay here?

147. anonymous

lol not much longer i hope you got more?

148. anonymous

i wanna do more probs , i might get stuck again

149. anonymous

their is a question

150. anonymous

find the area under the given curve from a to b

151. anonymous

y=x2 +1 from 0 to 4

152. anonymous

that is real real easy

153. anonymous

set up the integral and do it

154. anonymous

we can graph it even before we solve it right ? and whats the difference

155. anonymous

why bother, you know what $$x^2+1$$ looks like right? all you have to compute here is $\int_0^4(x^2+1)dx$

156. anonymous

idk the answer on the back has a graph

157. anonymous

that is because... they have a graph in the back the graphics jack up the price of the book

158. anonymous

probs

159. anonymous

that integral is real real easy right?

160. anonymous

yea

161. anonymous

lots easier than the first two for sure

162. anonymous

their are different chapters on my homwork but all the same thing

163. anonymous

one says "area under curve " and the other "area between curves"

164. anonymous

under the curve means below the curve and above the x axis between the curve means between the curves the first one you just use the limits of integration given you the second you need the upper curve minus the lower curve

165. anonymous

|dw:1433302834136:dw|

166. anonymous

didnt we do upper minus bottom in both of them

167. anonymous

not for the last one no

168. anonymous

so the last one i asked was area under the curve

169. anonymous

i toke the anti of x2 + 1

170. anonymous

i got x3/3 + x

171. anonymous

then i did F(4)-F(0)

172. anonymous

yes

173. anonymous

final soltuion 76/3 which is correct is that top minus bottom ?

174. anonymous

$F(0)=0$ in this case it should be easier than most

175. anonymous

$\frac{4^3}{3}+4$ is all

176. anonymous

u said area under the curve is NOT top minus bottom but thats what i just did ?

177. anonymous

slow

178. anonymous

it is always $F(b)-F(a)$

179. anonymous

but if it is the area between two curves you have to subtract the lower one from the upper one BEFORE you integrate

180. anonymous

so in your example it was only one curve $$y=x^2+1$$ from $$0$$ to $$4$$ you do $\int_0^2 (x^2+1)dx$

181. anonymous

ohh u mean we have to simplify then anti but in area under their is no simplfying to do just straight

182. anonymous

but if it was between to curves, you have to subtract first, then integrate

183. anonymous

yes right, what you said

184. anonymous

now i have a question

185. anonymous

do you really love homework?

186. anonymous

according to os you posted this question two hours ago! so i guess you must

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