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

Mathematics
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a substitution will do, let =3x
let u=3x*

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Other answers:

When I solved it, I got:
\[ u = 3x \longrightarrow \frac{du}{dx}=3 \\ \frac{2}{3}\cdot \frac{du}{dx}=2 \] \[ \large \frac{2}{3}\int u^{-\frac{1}{3}}du\]
can you go from here?
I don't know what to do after this..
you integrate and backsubstitute. you will get: \[ \sqrt[3]{u}^2\]
So that's the answer?
the answer is \[ \sqrt[3]{3x}^2 \]
What do you do after you integrate and back substitute? I want to show all the steps.
Then you are done. That's the indefinite integral.
But the answer is cube root of 3x^2. You did something after you got cube root of u^2.
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.
Oh okay, can you help me with one more?
sure
Do you know basic integration of exponents yes?
Some of it.
because you can write this integral like this: \[ \int (x^{-2} -x^{-3})dx\]
And then you just integrate normally, increase the exponent by one and divide by the new exponent.
Can we do it step by step?
yes one second
Okay
so you can actually split this integral up if that makes it easier for you \[ \int x^{-2} dx- \int x^{-3}dx \]
if you integrate the first term you should get \[- \frac{1}{x}\]
Okay, I got that part.
think you can figure out the second one? it works exactly the same.
Wait, I mean -1/2x^2
yes you did it right
It's the first or second one?
\[ \frac{x^{-3+1}}{-3+1} = -\frac{1}{2x^2}\]
Okay, what is the step after that?
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.
So it would be: -1/2x^2+C?
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.
-1/x - 1/2x^2 + C = Answer?
\[ \int x^{-2} dx- \int x^{-3}dx = -\frac{1}{x} + \frac{1}{x^2}+C \]
almost, the minus becomes a plus, because there is a minus in front of the integral. See it?
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?
oops, you are absolutely right, I messed it up
\[ \int x^{-2} dx- \int x^{-3}dx = -\frac{1}{x} + \frac{1}{2x^2}+C \]
well done! (-: you discovered my mistake.
Yay! I need help with a few more if you have time?
I believe so. just post them.
you know how to setup this integral?
No
I'm not good with trigs..at all :/
hmm well I more meant, do you know what you have to do in order to calculate the area of this ?
No
The integral calculates the Area below a graph, do you know that?
So whenever you want to calculate the area underneath a graph of a function, of any kind, you want to setup an integral.
\[ \large \int_a^b f(x) dx \] 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.
a and b are the boundaries of your integral, more correctly it should be written like this \[ \int_{x=a}^{x=b} f(x)dx\]
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.
the under bound is always the lower limits, meaning lower value, the upper bound is always the upper limit, meaning higher value.
Okay
So do you have the picture open?
Yes, I have it open
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.
0 to pi/2
exactly!
and the function is y=x+cosx right?
Yes
so this means it looks like that : \[ \large \int_0^{\frac{\pi}{2}}(x + \cos x)dx =\int_0^{\frac{\pi}{2}}xdx + \int_0^{\frac{\pi}{2}}\cos(x)dx \]
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.
Okay, that's all there is to it?
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.
just like you did before, take the antiderivative, only this time you have values to plug in and see what happens.
Sorry, I'm so confused. This is due by 9:30 and I'm under stress. Ugh!
ohhh, I see.
So, now I plug in pi/2 and then plug in 0?
first you need to compute the antiderivate.
have you done that? can you tell me what the integral of x is ?
x^2/2
exactly!
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.
Usually you see this written like that \[ \left. \frac{x^2}{2} \right|_{x=0}^{x=\frac{\pi}{2}}\ = \frac{\frac{\pi ^2}{4}}{2} - \frac{0^2}{2} = \frac{\pi^2}{8} \]
but that's only the first step, now you need to do the exact same with the cos(x) function, your second integral.
That's the difficult part. Wouldn't I need a unit circle?
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)
Wouldn't it be negative sin(x)?
no the derivative of cos(x) is -sin(x)
but we are doing integration , that's the other way around.
Okay, now I just plug in pi/2 into sin x?
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.
I get sin(pi/2) = 1
that's true!
so the entire area is?
-1 ?
no you add them both together now, just like we did before with the indefinite integrals, now you add the values together
So it's 1.
yes, but that's not the entire area
Remember what we got for the first integral?
\[\large \int_0^{\frac{\pi}{2}}(x + \cos x)dx =\int_0^{\frac{\pi}{2}}xdx + \int_0^{\frac{\pi}{2}}\cos(x)dx \\ = \underline{\underline{\frac{\pi^2}{8}+1}}\]
Now I have to subtract pi/8+1 by 1?
no, that's the solution already up there, I did underline it.
Ohhh so the entire area is pi/8 +1.
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.
Yes so how would I write the final answer?
just like I did, up there
\[ A = \underline{\underline{\frac{\pi^2}{8}+1}}\]
because that is the area.
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.
we can give it a try.
1 Attachment
one you answered right
How do I explain it in sentences though?
two you answered right two, but your reasoning isn't quite correct
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.
If I use partial integration, how would it look?
\[ \large\int f'(x)g(x)dx= f(x)g(x)- \int f(x)g'(x) dx \]
Okay
For 3-5, I just don't know how to word it.
false because y= ln e =1 is a number, the derivative of a number is always zero.
That's all I have to say for that?
totally
for 3.
Okay, great, 4 and 5 left.
4 and 5 you answered perfectly already.
there is nothing I would add.
But I have to rephrase it all into sentence form and I don't know how.
ah yeh? Well, this is math, sometimes showing the example is the best way to make a phrase.
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
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.
Okay, thank you so so much!!
welcome

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