## hiralpatel121 3 years ago How can I find the indefinite integral of the attached problem?

1. hiralpatel121

2. Spacelimbus

a substitution will do, let =3x

3. Spacelimbus

let u=3x*

4. hiralpatel121

When I solved it, I got:

5. Spacelimbus

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

6. Spacelimbus

can you go from here?

7. hiralpatel121

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

8. Spacelimbus

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

9. hiralpatel121

10. Spacelimbus

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

11. hiralpatel121

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

12. Spacelimbus

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

13. hiralpatel121

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

14. Spacelimbus

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

Oh okay, can you help me with one more?

16. Spacelimbus

sure

17. hiralpatel121

18. Spacelimbus

Do you know basic integration of exponents yes?

19. hiralpatel121

Some of it.

20. Spacelimbus

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

21. Spacelimbus

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

22. hiralpatel121

Can we do it step by step?

23. Spacelimbus

yes one second

24. hiralpatel121

Okay

25. Spacelimbus

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

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

27. hiralpatel121

Okay, I got that part.

28. Spacelimbus

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

29. hiralpatel121

Wait, I mean -1/2x^2

30. Spacelimbus

yes you did it right

31. hiralpatel121

It's the first or second one?

32. Spacelimbus

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

33. hiralpatel121

Okay, what is the step after that?

34. Spacelimbus

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

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

36. Spacelimbus

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

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

38. Spacelimbus

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

39. Spacelimbus

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

40. hiralpatel121

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

oops, you are absolutely right, I messed it up

42. Spacelimbus

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

43. Spacelimbus

well done! (-: you discovered my mistake.

44. hiralpatel121

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

45. Spacelimbus

I believe so. just post them.

46. hiralpatel121

47. Spacelimbus

you know how to setup this integral?

48. hiralpatel121

No

49. hiralpatel121

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

50. Spacelimbus

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

51. hiralpatel121

No

52. Spacelimbus

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

53. Spacelimbus

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

$\large \int_a^b f(x) dx$ 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

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

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

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

58. hiralpatel121

Okay

59. Spacelimbus

So do you have the picture open?

60. hiralpatel121

Yes, I have it open

61. Spacelimbus

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

0 to pi/2

63. Spacelimbus

exactly!

64. Spacelimbus

and the function is y=x+cosx right?

65. hiralpatel121

Yes

66. Spacelimbus

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

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

Okay, that's all there is to it?

69. Spacelimbus

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

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

71. hiralpatel121

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

72. Spacelimbus

ohhh, I see.

73. hiralpatel121

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

74. Spacelimbus

first you need to compute the antiderivate.

75. Spacelimbus

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

76. hiralpatel121

x^2/2

77. Spacelimbus

exactly!

78. Spacelimbus

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

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

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

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

82. Spacelimbus

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

Wouldn't it be negative sin(x)?

84. Spacelimbus

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

85. Spacelimbus

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

86. hiralpatel121

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

87. Spacelimbus

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

I get sin(pi/2) = 1

89. Spacelimbus

that's true!

90. Spacelimbus

so the entire area is?

91. hiralpatel121

-1 ?

92. Spacelimbus

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

93. hiralpatel121

So it's 1.

94. Spacelimbus

yes, but that's not the entire area

95. Spacelimbus

Remember what we got for the first integral?

96. Spacelimbus

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

97. hiralpatel121

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

98. Spacelimbus

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

99. hiralpatel121

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

100. Spacelimbus

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

Yes so how would I write the final answer?

102. Spacelimbus

just like I did, up there

103. Spacelimbus

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

104. Spacelimbus

because that is the area.

105. hiralpatel121

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

we can give it a try.

107. hiralpatel121

108. Spacelimbus

109. hiralpatel121

How do I explain it in sentences though?

110. Spacelimbus

111. Spacelimbus

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

If I use partial integration, how would it look?

113. Spacelimbus

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

114. hiralpatel121

Okay

115. hiralpatel121

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

116. Spacelimbus

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

117. hiralpatel121

That's all I have to say for that?

118. Spacelimbus

totally

119. Spacelimbus

for 3.

120. hiralpatel121

Okay, great, 4 and 5 left.

121. Spacelimbus

122. Spacelimbus

there is nothing I would add.

123. hiralpatel121

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

124. Spacelimbus

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

125. Spacelimbus

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

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

Okay, thank you so so much!!

128. Spacelimbus

welcome