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hiralpatel121
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How can I find the indefinite integral of the attached problem?
 2 years ago
 2 years ago
hiralpatel121 Group Title
How can I find the indefinite integral of the attached problem?
 2 years ago
 2 years ago

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Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
a substitution will do, let =3x
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
let u=3x*
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
When I solved it, I got:
 2 years ago

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

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
can you go from here?
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
I don't know what to do after this..
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
you integrate and backsubstitute. you will get: \[ \sqrt[3]{u}^2\]
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
So that's the answer?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
the answer is \[ \sqrt[3]{3x}^2 \]
 2 years ago

hiralpatel121 Group TitleBest 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.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
Then you are done. That's the indefinite integral.
 2 years ago

hiralpatel121 Group TitleBest 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.
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Oh okay, can you help me with one more?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
Do you know basic integration of exponents yes?
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Some of it.
 2 years ago

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

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Can we do it step by step?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
yes one second
 2 years ago

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

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
if you integrate the first term you should get \[ \frac{1}{x}\]
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Okay, I got that part.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
think you can figure out the second one? it works exactly the same.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Wait, I mean 1/2x^2
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
yes you did it right
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
It's the first or second one?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
\[ \frac{x^{3+1}}{3+1} = \frac{1}{2x^2}\]
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Okay, what is the step after that?
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
So it would be: 1/2x^2+C?
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
1/x  1/2x^2 + C = Answer?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
\[ \int x^{2} dx \int x^{3}dx = \frac{1}{x} + \frac{1}{x^2}+C \]
 2 years ago

Spacelimbus Group TitleBest 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?
 2 years ago

hiralpatel121 Group TitleBest 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?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
oops, you are absolutely right, I messed it up
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
\[ \int x^{2} dx \int x^{3}dx = \frac{1}{x} + \frac{1}{2x^2}+C \]
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
well done! (: you discovered my mistake.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Yay! I need help with a few more if you have time?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
I believe so. just post them.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
you know how to setup this integral?
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
I'm not good with trigs..at all :/
 2 years ago

Spacelimbus Group TitleBest 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 ?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
The integral calculates the Area below a graph, do you know that?
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

Spacelimbus Group TitleBest 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 \[ \int_{x=a}^{x=b} f(x)dx\]
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
So do you have the picture open?
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Yes, I have it open
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
0 to pi/2
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
exactly!
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
and the function is y=x+cosx right?
 2 years ago

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

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Okay, that's all there is to it?
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest 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!
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
ohhh, I see.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
So, now I plug in pi/2 and then plug in 0?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
first you need to compute the antiderivate.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
have you done that? can you tell me what the integral of x is ?
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
x^2/2
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
exactly!
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

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

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
That's the difficult part. Wouldn't I need a unit circle?
 2 years ago

Spacelimbus Group TitleBest 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)
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Wouldn't it be negative sin(x)?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
no the derivative of cos(x) is sin(x)
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
but we are doing integration , that's the other way around.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Okay, now I just plug in pi/2 into sin x?
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
I get sin(pi/2) = 1
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
that's true!
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
so the entire area is?
 2 years ago

Spacelimbus Group TitleBest 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
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
So it's 1.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
yes, but that's not the entire area
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
Remember what we got for the first integral?
 2 years ago

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

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Now I have to subtract pi/8+1 by 1?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
no, that's the solution already up there, I did underline it.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Ohhh so the entire area is pi/8 +1.
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Yes so how would I write the final answer?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
just like I did, up there
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
\[ A = \underline{\underline{\frac{\pi^2}{8}+1}}\]
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
because that is the area.
 2 years ago

hiralpatel121 Group TitleBest 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.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
we can give it a try.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
one you answered right
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
How do I explain it in sentences though?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
two you answered right two, but your reasoning isn't quite correct
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
If I use partial integration, how would it look?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
\[ \large\int f'(x)g(x)dx= f(x)g(x) \int f(x)g'(x) dx \]
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
For 35, I just don't know how to word it.
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
That's all I have to say for that?
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
totally
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Okay, great, 4 and 5 left.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
4 and 5 you answered perfectly already.
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
there is nothing I would add.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
But I have to rephrase it all into sentence form and I don't know how.
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

Spacelimbus Group TitleBest 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
 2 years ago

Spacelimbus Group TitleBest 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.
 2 years ago

hiralpatel121 Group TitleBest ResponseYou've already chosen the best response.0
Okay, thank you so so much!!
 2 years ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
welcome
 2 years ago
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