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How can I find the indefinite integral of the attached problem?
 one year ago
 one year ago
How can I find the indefinite integral of the attached problem?
 one year ago
 one year ago

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SpacelimbusBest ResponseYou've already chosen the best response.2
a substitution will do, let =3x
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
When I solved it, I got:
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
\[ u = 3x \longrightarrow \frac{du}{dx}=3 \\ \frac{2}{3}\cdot \frac{du}{dx}=2 \] \[ \large \frac{2}{3}\int u^{\frac{1}{3}}du\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
can you go from here?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
I don't know what to do after this..
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
you integrate and backsubstitute. you will get: \[ \sqrt[3]{u}^2\]
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
So that's the answer?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
the answer is \[ \sqrt[3]{3x}^2 \]
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
What do you do after you integrate and back substitute? I want to show all the steps.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
Then you are done. That's the indefinite integral.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
But the answer is cube root of 3x^2. You did something after you got cube root of u^2.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.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 backsubstitute, where there is an u, you put a 3x instead.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Oh okay, can you help me with one more?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
Do you know basic integration of exponents yes?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
because you can write this integral like this: \[ \int (x^{2} x^{3})dx\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
And then you just integrate normally, increase the exponent by one and divide by the new exponent.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Can we do it step by step?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
so you can actually split this integral up if that makes it easier for you \[ \int x^{2} dx \int x^{3}dx \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
if you integrate the first term you should get \[ \frac{1}{x}\]
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Okay, I got that part.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
think you can figure out the second one? it works exactly the same.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Wait, I mean 1/2x^2
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
yes you did it right
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
It's the first or second one?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
\[ \frac{x^{3+1}}{3+1} = \frac{1}{2x^2}\]
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Okay, what is the step after that?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
So it would be: 1/2x^2+C?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
1/x  1/2x^2 + C = Answer?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
\[ \int x^{2} dx \int x^{3}dx = \frac{1}{x} + \frac{1}{x^2}+C \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
almost, the minus becomes a plus, because there is a minus in front of the integral. See it?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
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?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
oops, you are absolutely right, I messed it up
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
\[ \int x^{2} dx \int x^{3}dx = \frac{1}{x} + \frac{1}{2x^2}+C \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
well done! (: you discovered my mistake.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Yay! I need help with a few more if you have time?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
I believe so. just post them.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
you know how to setup this integral?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
I'm not good with trigs..at all :/
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
hmm well I more meant, do you know what you have to do in order to calculate the area of this ?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
The integral calculates the Area below a graph, do you know that?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
So whenever you want to calculate the area underneath a graph of a function, of any kind, you want to setup an integral.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
\[ \large \int_a^b f(x) dx \] 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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
a and b are the boundaries of your integral, more correctly it should be written like this \[ \int_{x=a}^{x=b} f(x)dx\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
the under bound is always the lower limits, meaning lower value, the upper bound is always the upper limit, meaning higher value.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
So do you have the picture open?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Yes, I have it open
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
and the function is y=x+cosx right?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
so this means it looks like that : \[ \large \int_0^{\frac{\pi}{2}}(x + \cos x)dx =\int_0^{\frac{\pi}{2}}xdx + \int_0^{\frac{\pi}{2}}\cos(x)dx \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Okay, that's all there is to it?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
just like you did before, take the antiderivative, only this time you have values to plug in and see what happens.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Sorry, I'm so confused. This is due by 9:30 and I'm under stress. Ugh!
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
So, now I plug in pi/2 and then plug in 0?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
first you need to compute the antiderivate.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
have you done that? can you tell me what the integral of x is ?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
Usually you see this written like that \[ \left. \frac{x^2}{2} \right_{x=0}^{x=\frac{\pi}{2}}\ = \frac{\frac{\pi ^2}{4}}{2}  \frac{0^2}{2} = \frac{\pi^2}{8} \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
but that's only the first step, now you need to do the exact same with the cos(x) function, your second integral.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
That's the difficult part. Wouldn't I need a unit circle?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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)
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Wouldn't it be negative sin(x)?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
no the derivative of cos(x) is sin(x)
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
but we are doing integration , that's the other way around.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Okay, now I just plug in pi/2 into sin x?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
I get sin(pi/2) = 1
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SpacelimbusBest ResponseYou've already chosen the best response.2
so the entire area is?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
no you add them both together now, just like we did before with the indefinite integrals, now you add the values together
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
yes, but that's not the entire area
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
Remember what we got for the first integral?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
\[\large \int_0^{\frac{\pi}{2}}(x + \cos x)dx =\int_0^{\frac{\pi}{2}}xdx + \int_0^{\frac{\pi}{2}}\cos(x)dx \\ = \underline{\underline{\frac{\pi^2}{8}+1}}\]
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Now I have to subtract pi/8+1 by 1?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
no, that's the solution already up there, I did underline it.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Ohhh so the entire area is pi/8 +1.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Yes so how would I write the final answer?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
just like I did, up there
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
\[ A = \underline{\underline{\frac{\pi^2}{8}+1}}\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
because that is the area.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
we can give it a try.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
one you answered right
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
How do I explain it in sentences though?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
two you answered right two, but your reasoning isn't quite correct
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
If I use partial integration, how would it look?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
\[ \large\int f'(x)g(x)dx= f(x)g(x) \int f(x)g'(x) dx \]
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hiralpatel121Best ResponseYou've already chosen the best response.0
For 35, I just don't know how to word it.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
false because y= ln e =1 is a number, the derivative of a number is always zero.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
That's all I have to say for that?
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Okay, great, 4 and 5 left.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
4 and 5 you answered perfectly already.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
there is nothing I would add.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
But I have to rephrase it all into sentence form and I don't know how.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
ah yeh? Well, this is math, sometimes showing the example is the best way to make a phrase.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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 xlnxx + C
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.2
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.
 one year ago

hiralpatel121Best ResponseYou've already chosen the best response.0
Okay, thank you so so much!!
 one year ago
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