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

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

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

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

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

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

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

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

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

hiralpatel121
 2 years ago
Best ResponseYou've already chosen the best response.0What do you do after you integrate and back substitute? I want to show all the steps.

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

hiralpatel121
 2 years ago
Best ResponseYou've already chosen the best response.0But the answer is cube root of 3x^2. You did something after you got cube root of u^2.

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2well 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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2That'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.

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2no, 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.

hiralpatel121
 2 years ago
Best ResponseYou've already chosen the best response.01/x  1/2x^2 + C = Answer?

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

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

hiralpatel121
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, 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?

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

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

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

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

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

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

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

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

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

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

Spacelimbus
 2 years ago
Best 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.

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2for 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.

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2the under bound is always the lower limits, meaning lower value, the upper bound is always the upper limit, meaning higher value.

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

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2So 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.

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

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2Both 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.

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2well 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.

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

hiralpatel121
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry, I'm so confused. This is due by 9:30 and I'm under stress. Ugh!

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

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

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2and 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.

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

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

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2no, 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)

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

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

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

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2exactly 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.

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

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

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

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

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

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

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

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

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2see \(\frac{\pi^2}{8}\) 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.

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

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

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

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

hiralpatel121
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, 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.

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

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

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

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2False: 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.

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

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

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

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

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

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.24 and 5 you answered perfectly already.

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

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

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2You 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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.2For 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.

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