Question

How can I find the indefinite integral of the attached problem?

Answer 1

a substitution will do, let =3x

Answer 2

let u=3x*

Answer 3

When I solved it, I got:

Answer 4

\[ u = 3x \longrightarrow \frac{du}{dx}=3
\\
\frac{2}{3}\cdot \frac{du}{dx}=2 \]

\[ \large \frac{2}{3}\int u^{-\frac{1}{3}}du\]

Answer 5

can you go from here?

Answer 6

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

Answer 7

you integrate and backsubstitute. you will get:

\[ \sqrt[3]{u}^2\]

Answer 8

So that's the answer?

Answer 9

the answer is

\[ \sqrt[3]{3x}^2 \]

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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.

Answer 14

Oh okay, can you help me with one more?

Answer 15

sure

Answer 16

Do you know basic integration of exponents yes?

Answer 17

Some of it.

Answer 18

because you can write this integral like this:

\[ \int (x^{-2} -x^{-3})dx\]

Answer 19

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

Answer 20

Can we do it step by step?

Answer 21

yes one second

Answer 22

Okay

Answer 23

so you can actually split this integral up if that makes it easier for you

\[ \int x^{-2} dx- \int x^{-3}dx \]

Answer 24

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

Answer 25

Okay, I got that part.

Answer 26

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

Answer 27

Wait, I mean -1/2x^2

Answer 28

yes you did it right

Answer 29

It's the first or second one?

Answer 30

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

Answer 31

Okay, what is the step after that?

Answer 32

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.

Answer 33

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

Answer 34

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.

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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?

Answer 39

oops, you are absolutely right, I messed it up

Answer 40

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

Answer 41

well done! (-: you discovered my mistake.

Answer 42

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

Answer 43

I believe so. just post them.

Answer 44

you know how to setup this integral?

Answer 45

No

Answer 46

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

Answer 47

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

Answer 48

No

Answer 49

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

Answer 50

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

Answer 51

\[ \large \int_a^b f(x) dx \] 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.

Answer 52

a and b are the boundaries of your integral, more correctly it should be written like this

\[ \int_{x=a}^{x=b} f(x)dx\]

Answer 53

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.

Answer 54

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

Answer 55

Okay

Answer 56

So do you have the picture open?

Answer 57

Yes, I have it open

Answer 58

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.

Answer 59

0 to pi/2

Answer 60

exactly!

Answer 61

and the function is y=x+cosx right?

Answer 62

Yes

Answer 63

so this means it looks like that :

\[ \large \int_0^{\frac{\pi}{2}}(x + \cos x)dx =\int_0^{\frac{\pi}{2}}xdx + \int_0^{\frac{\pi}{2}}\cos(x)dx \]

Answer 64

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.

Answer 65

Okay, that's all there is to it?

Answer 66

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.

Answer 67

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

Answer 68

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

Answer 69

ohhh, I see.

Answer 70

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

Answer 71

first you need to compute the antiderivate.

Answer 72

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

Answer 73

x^2/2

Answer 74

exactly!

Answer 75

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.

Answer 76

Usually you see this written like that

\[ \left. \frac{x^2}{2} \right|_{x=0}^{x=\frac{\pi}{2}}\ = \frac{\frac{\pi ^2}{4}}{2} - \frac{0^2}{2} = \frac{\pi^2}{8} \]

Answer 77

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

Answer 78

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

Answer 79

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)

Answer 80

Wouldn't it be negative sin(x)?

Answer 81

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

Answer 82

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

Answer 83

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

Answer 84

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.

Answer 85

I get sin(pi/2) = 1

Answer 86

that's true!

Answer 87

so the entire area is?

Answer 88

-1 ?

Answer 89

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

Answer 90

So it's 1.

Answer 91

yes, but that's not the entire area

Answer 92

Remember what we got for the first integral?

Answer 93

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