anonymous
  • anonymous
How can I find the indefinite integral of the attached problem?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
a substitution will do, let =3x
anonymous
  • anonymous
let u=3x*

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

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