## hiralpatel121 Group Title How can I find the indefinite integral of the attached problem? 2 years ago 2 years ago

1. hiralpatel121 Group Title

2. Spacelimbus Group Title

a substitution will do, let =3x

3. Spacelimbus Group Title

let u=3x*

4. hiralpatel121 Group Title

When I solved it, I got:

5. Spacelimbus Group Title

$u = 3x \longrightarrow \frac{du}{dx}=3 \\ \frac{2}{3}\cdot \frac{du}{dx}=2$ $\large \frac{2}{3}\int u^{-\frac{1}{3}}du$

6. Spacelimbus Group Title

can you go from here?

7. hiralpatel121 Group Title

I don't know what to do after this..

8. Spacelimbus Group Title

you integrate and backsubstitute. you will get: $\sqrt[3]{u}^2$

9. hiralpatel121 Group Title

So that's the answer?

10. Spacelimbus Group Title

the answer is $\sqrt[3]{3x}^2$

11. hiralpatel121 Group Title

What do you do after you integrate and back substitute? I want to show all the steps.

12. Spacelimbus Group Title

Then you are done. That's the indefinite integral.

13. hiralpatel121 Group Title

But the answer is cube root of 3x^2. You did something after you got cube root of u^2.

14. Spacelimbus Group Title

well at the beginning I said substitute u=3x, and you want your result to be in forms of x right? so you have to back-substitute, where there is an u, you put a 3x instead.

15. hiralpatel121 Group Title

Oh okay, can you help me with one more?

16. Spacelimbus Group Title

sure

17. hiralpatel121 Group Title

18. Spacelimbus Group Title

Do you know basic integration of exponents yes?

19. hiralpatel121 Group Title

Some of it.

20. Spacelimbus Group Title

because you can write this integral like this: $\int (x^{-2} -x^{-3})dx$

21. Spacelimbus Group Title

And then you just integrate normally, increase the exponent by one and divide by the new exponent.

22. hiralpatel121 Group Title

Can we do it step by step?

23. Spacelimbus Group Title

yes one second

24. hiralpatel121 Group Title

Okay

25. Spacelimbus Group Title

so you can actually split this integral up if that makes it easier for you $\int x^{-2} dx- \int x^{-3}dx$

26. Spacelimbus Group Title

if you integrate the first term you should get $- \frac{1}{x}$

27. hiralpatel121 Group Title

Okay, I got that part.

28. Spacelimbus Group Title

think you can figure out the second one? it works exactly the same.

29. hiralpatel121 Group Title

Wait, I mean -1/2x^2

30. Spacelimbus Group Title

yes you did it right

31. hiralpatel121 Group Title

It's the first or second one?

32. Spacelimbus Group Title

$\frac{x^{-3+1}}{-3+1} = -\frac{1}{2x^2}$

33. hiralpatel121 Group Title

Okay, what is the step after that?

34. Spacelimbus Group Title

That's it, you have taken the indefinite integral so you can just write it down. usually you add a + C at the end for indefinite integrals to show that an arbitrary constant could have been lost.

35. hiralpatel121 Group Title

So it would be: -1/2x^2+C?

36. Spacelimbus Group Title

no, both of them, you have taken two indefinite integrals remember? Look at the formula I have posted above, we did split them up, you have derived the first one and the second one, both are part of your solution.

37. hiralpatel121 Group Title

-1/x - 1/2x^2 + C = Answer?

38. Spacelimbus Group Title

$\int x^{-2} dx- \int x^{-3}dx = -\frac{1}{x} + \frac{1}{x^2}+C$

39. Spacelimbus Group Title

almost, the minus becomes a plus, because there is a minus in front of the integral. See it?

40. hiralpatel121 Group Title

Yes, but wasn't the second part -1/ 2x^2? Then in the answer, why did you leave out the 2 in front of the x^2?

41. Spacelimbus Group Title

oops, you are absolutely right, I messed it up

42. Spacelimbus Group Title

$\int x^{-2} dx- \int x^{-3}dx = -\frac{1}{x} + \frac{1}{2x^2}+C$

43. Spacelimbus Group Title

well done! (-: you discovered my mistake.

44. hiralpatel121 Group Title

Yay! I need help with a few more if you have time?

45. Spacelimbus Group Title

I believe so. just post them.

46. hiralpatel121 Group Title

47. Spacelimbus Group Title

you know how to setup this integral?

48. hiralpatel121 Group Title

No

49. hiralpatel121 Group Title

I'm not good with trigs..at all :/

50. Spacelimbus Group Title

hmm well I more meant, do you know what you have to do in order to calculate the area of this ?

51. hiralpatel121 Group Title

No

52. Spacelimbus Group Title

The integral calculates the Area below a graph, do you know that?

53. Spacelimbus Group Title

So whenever you want to calculate the area underneath a graph of a function, of any kind, you want to setup an integral.

54. Spacelimbus Group Title

$\large \int_a^b f(x) dx$ this is the general setup of an integral, it is how it looks like, y=f(x) is the function you want to integrate, and the dx just tells you in respect to what you have to integrate it with.

55. Spacelimbus Group Title

a and b are the boundaries of your integral, more correctly it should be written like this $\int_{x=a}^{x=b} f(x)dx$

56. Spacelimbus Group Title

for a < b, that means you integrate a function, starting from the point a, and you integrate it's entire area until you hit the point b.

57. Spacelimbus Group Title

the under bound is always the lower limits, meaning lower value, the upper bound is always the upper limit, meaning higher value.

58. hiralpatel121 Group Title

Okay

59. Spacelimbus Group Title

So do you have the picture open?

60. hiralpatel121 Group Title

Yes, I have it open

61. Spacelimbus Group Title

So can you tell me what the boundaries (upper and lower) for this integral are?, from where to where does it run? Look at the area you want to integrate.

62. hiralpatel121 Group Title

0 to pi/2

63. Spacelimbus Group Title

exactly!

64. Spacelimbus Group Title

and the function is y=x+cosx right?

65. hiralpatel121 Group Title

Yes

66. Spacelimbus Group Title

so this means it looks like that : $\large \int_0^{\frac{\pi}{2}}(x + \cos x)dx =\int_0^{\frac{\pi}{2}}xdx + \int_0^{\frac{\pi}{2}}\cos(x)dx$

67. Spacelimbus Group Title

Both are easy integrals for themselves, when you look at them individually, keep in mind, the integrals turns a function into it's antiderivative, that means if you derive an integrated function, you should get back to the original function.

68. hiralpatel121 Group Title

Okay, that's all there is to it?

69. Spacelimbus Group Title

well now you have to evaluate it, you want to determine the area of the given region, that means you want to compute the integral and evaluate it.

70. Spacelimbus Group Title

just like you did before, take the antiderivative, only this time you have values to plug in and see what happens.

71. hiralpatel121 Group Title

Sorry, I'm so confused. This is due by 9:30 and I'm under stress. Ugh!

72. Spacelimbus Group Title

ohhh, I see.

73. hiralpatel121 Group Title

So, now I plug in pi/2 and then plug in 0?

74. Spacelimbus Group Title

first you need to compute the antiderivate.

75. Spacelimbus Group Title

have you done that? can you tell me what the integral of x is ?

76. hiralpatel121 Group Title

x^2/2

77. Spacelimbus Group Title

exactly!

78. Spacelimbus Group Title

and see, there you plug in the values now, it always works like this, plug in the higher value, then subtract from that the value that you get when you plug in the lower value.

79. Spacelimbus Group Title

Usually you see this written like that $\left. \frac{x^2}{2} \right|_{x=0}^{x=\frac{\pi}{2}}\ = \frac{\frac{\pi ^2}{4}}{2} - \frac{0^2}{2} = \frac{\pi^2}{8}$

80. Spacelimbus Group Title

but that's only the first step, now you need to do the exact same with the cos(x) function, your second integral.

81. hiralpatel121 Group Title

That's the difficult part. Wouldn't I need a unit circle?

82. Spacelimbus Group Title

no, you need to know the trig functions and their derivative by heart the derivative of sin(x) = cos(x) therefore the antiderivative of cos(x) = sin(x)

83. hiralpatel121 Group Title

Wouldn't it be negative sin(x)?

84. Spacelimbus Group Title

no the derivative of cos(x) is -sin(x)

85. Spacelimbus Group Title

but we are doing integration , that's the other way around.

86. hiralpatel121 Group Title

Okay, now I just plug in pi/2 into sin x?

87. Spacelimbus Group Title

exactly and then you subtract from that sin(0) , but sin(0) =0, so you don't have to worry about that, just to empathize the order.

88. hiralpatel121 Group Title

I get sin(pi/2) = 1

89. Spacelimbus Group Title

that's true!

90. Spacelimbus Group Title

so the entire area is?

91. hiralpatel121 Group Title

-1 ?

92. Spacelimbus Group Title

no you add them both together now, just like we did before with the indefinite integrals, now you add the values together

93. hiralpatel121 Group Title

So it's 1.

94. Spacelimbus Group Title

yes, but that's not the entire area

95. Spacelimbus Group Title

Remember what we got for the first integral?

96. Spacelimbus Group Title

$\large \int_0^{\frac{\pi}{2}}(x + \cos x)dx =\int_0^{\frac{\pi}{2}}xdx + \int_0^{\frac{\pi}{2}}\cos(x)dx \\ = \underline{\underline{\frac{\pi^2}{8}+1}}$

97. hiralpatel121 Group Title

Now I have to subtract pi/8+1 by 1?

98. Spacelimbus Group Title

no, that's the solution already up there, I did underline it.

99. hiralpatel121 Group Title

Ohhh so the entire area is pi/8 +1.

100. Spacelimbus Group Title

see $$\frac{\pi^2}{8}$$ is what we got for the first integral, when we evaluated it, remember? x^2/2, you did that yourself, now the next integral - it has a plus there, you evaluated that integral and you said the area is 1 so the entire area is both summed together.

101. hiralpatel121 Group Title

Yes so how would I write the final answer?

102. Spacelimbus Group Title

just like I did, up there

103. Spacelimbus Group Title

$A = \underline{\underline{\frac{\pi^2}{8}+1}}$

104. Spacelimbus Group Title

because that is the area.

105. hiralpatel121 Group Title

Okay, thank you so much! Are you good with true and false questions? I have the answers already but I have to write sentences and I need help with that.

106. Spacelimbus Group Title

we can give it a try.

107. hiralpatel121 Group Title

108. Spacelimbus Group Title

one you answered right

109. hiralpatel121 Group Title

How do I explain it in sentences though?

110. Spacelimbus Group Title

two you answered right two, but your reasoning isn't quite correct

111. Spacelimbus Group Title

False: The integral of two functions cannot be computed by taking the integral of each function individually and multiplying the results. The correct method is using partial integration.

112. hiralpatel121 Group Title

If I use partial integration, how would it look?

113. Spacelimbus Group Title

$\large\int f'(x)g(x)dx= f(x)g(x)- \int f(x)g'(x) dx$

114. hiralpatel121 Group Title

Okay

115. hiralpatel121 Group Title

For 3-5, I just don't know how to word it.

116. Spacelimbus Group Title

false because y= ln e =1 is a number, the derivative of a number is always zero.

117. hiralpatel121 Group Title

That's all I have to say for that?

118. Spacelimbus Group Title

totally

119. Spacelimbus Group Title

for 3.

120. hiralpatel121 Group Title

Okay, great, 4 and 5 left.

121. Spacelimbus Group Title

4 and 5 you answered perfectly already.

122. Spacelimbus Group Title

there is nothing I would add.

123. hiralpatel121 Group Title

But I have to rephrase it all into sentence form and I don't know how.

124. Spacelimbus Group Title

ah yeh? Well, this is math, sometimes showing the example is the best way to make a phrase.

125. Spacelimbus Group Title

You can always make sentences though of course, for 4 you can write. False, if you derive the natural log of x you get the inverse function, when you integrate it you again have to use the technic of partial integration which leads to xlnx-x + C

126. Spacelimbus Group Title

For 5. There is a function which is equal to it's derivative, this function is e^x, it's slope at any given point of x is equal to it's function valued at given point.

127. hiralpatel121 Group Title

Okay, thank you so so much!!

128. Spacelimbus Group Title

welcome