Any chance I can hire a tutor to help me with my Calculus problems?

I hope that asking such a question is not against the rules here, and if so I apologize in advance!

I have a whole slew of homework problems having to do with antiderivatives, Computing sums using summation rules, and the like.....

Of course, I would expect someone who has a good reputation here and knows what they are talking about.

It is a total of 25 problems and I can provide them in any word format or pdf if desired, or probably other formats as well.

Thanks in advance!!

Re: Any chance I can hire a tutor to help me with my Calculus problems?

Bro, if you legit are trying your best to learn that stuff, you can post 25 problems here and people will try to help, and you will pretty much always get good replies, but if you are asking a guy to finish your problem set, that is not taken with good humor here.

Re: Any chance I can hire a tutor to help me with my Calculus problems?

Post them and people will guide you towards the right direction. Plus, that gives other people the opportunity to learn, too.

Re: Any chance I can hire a tutor to help me with my Calculus problems?

Quote:

Originally Posted by

**jakncoke** Bro, if you legit are trying your best to learn that stuff, you can post 25 problems here and people will try to help, and you will pretty much always get good replies, but if you are asking a guy to finish your problem set, that is not taken with good humor here.

I am not looking for someone to do my homework, that will not help me when it comes time to take a test. I need to actually understand how to do these problems. I am taking this class via distance learning and have had to try and teach myself, and it just isnt working out for me. So again, I am not simply wanting someone to answer the problems, but rather tutor me and help me to understand how to solve them.

1 Attachment(s)

Re: Any chance I can hire a tutor to help me with my Calculus problems?

So I am attaching the file and will just start with the first problem.

Find the general antiderivative.

4[(cosx)/(sin^{2}x)] dx

Im looking in my textbook but just am not getting it to sink in yet

Re: Any chance I can hire a tutor to help me with my Calculus problems?

I can offer you help through email or instant messaging sessions or something. I'd explain things in an intuitive way, and stress understanding rather than practice. If you are interested let me know.

Re: Any chance I can hire a tutor to help me with my Calculus problems?

Quote:

Originally Posted by

**SworD** I can offer you help through email or instant messaging sessions or something. I'd explain things in an intuitive way, and stress understanding rather than practice. If you are interested let me know.

I am definitely open to all the help I can get in understanding this stuff. If you do not mind me asking, what time zone are you in? I am in the USA on EST. Thanks in advance!

Re: Any chance I can hire a tutor to help me with my Calculus problems?

We are given:

$\displaystyle 4\int\frac{\cos(x)}{\sin^2(x)}\,dx$

The first thing I notice is that $\displaystyle \frac{d}{dx}(\sin(x))=\cos(x)$ so if we let:

$\displaystyle u=\sin(x)$ then we have $\displaystyle du=\cos(x)\,dx$ and we may write:

$\displaystyle 4\int\frac{1}{u^2}\,du=4\int u^{-2}\,du$

Can you proceed from here, and remember to back-substitute for $\displaystyle u$ when you find the anti-derivative.

Re: Any chance I can hire a tutor to help me with my Calculus problems?

Quote:

Originally Posted by

**JDS** I am definitely open to all the help I can get in understanding this stuff. If you do not mind me asking, what time zone are you in? I am in the USA on EST. Thanks in advance!

Same.

Re: Any chance I can hire a tutor to help me with my Calculus problems?

Quote:

Originally Posted by

**MarkFL2** We are given:

$\displaystyle 4\int\frac{\cos(x)}{\sin^2(x)}\,dx$

The first thing I notice is that $\displaystyle \frac{d}{dx}(\sin(x))=\cos(x)$ so if we let:

$\displaystyle u=\sin(x)$ then we have $\displaystyle du=\cos(x)\,dx$ and we may write:

$\displaystyle 4\int\frac{1}{u^2}\,du=4\int u^{-2}\,du$

Can you proceed from here, and remember to back-substitute for $\displaystyle u$ when you find the anti-derivative.

I am trying to do so now... thanks and bare with me if you can, lol

Re: Any chance I can hire a tutor to help me with my Calculus problems?

I am not sure what I am supposed to do with u^-2 du....but what about this....

=∫ 4 1/sinx * cosx/sinx dx

= ∫ 4 csc x cot x dx

= ∫4 -csc x + C

NOTE: it is almost 1 am here, and I gotta get up early so I am headed to bed for now. Thanks everyone for all your help and I will pick this back up tomorrow!

Re: Any chance I can hire a tutor to help me with my Calculus problems?

That will work as well, however you should write the result as:

$\displaystyle -4\csc(x)+C$

Once you integrate, you drop the integration symbol, as that operation has been completed.

Using the approach I showed you, the power rule for integration $\displaystyle \int u^n\,du=\frac{u^{n+1}}{n+1}+C$ gives us:

$\displaystyle 4\int u^{-2}\,du=-4u^{-1}+C=-4(\sin(x))^{-1}+C=-4\csc(x)+C$

Re: Any chance I can hire a tutor to help me with my Calculus problems?

Quote:

Originally Posted by

**JDS** I am definitely open to all the help I can get in understanding this stuff. If you do not mind me asking, what time zone are you in? I am in the USA on EST. Thanks in advance!

I am a maths tutor, but in Australia. Perhaps you would be interested in performing some tutoring via skype or facebook or some other means?

Re: Any chance I can hire a tutor to help me with my Calculus problems?

Since there are several posts relating to tutoring I'll mention my thoughts here, rather than over PM.

There is nothing against trying to find a tutor on the site, but I cannot allow it in a public forum. You are welcome to use the PM system to individually find possible tutors. (That is to say, don't PM thirteen members all at once.)

(whispers) Though you could make a post for tutors in the "Lounge" forum...shhh! I didn't say that. (Tmi)

-Dan

Re: Any chance I can hire a tutor to help me with my Calculus problems?

Next problem is:

$\displaystyle \int$ (4x - 2e^{x})dx

I am still trying to work this out.... I will post back with my progress