Recent content by wvcaudill2
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Basic limit involving absolute value
Im still not seeing the big picture here. Can you write out exactly what this problem should look like with all the steps at least partially worked out?- wvcaudill2
- Post #13
- Forum: Calculus and Beyond Homework Help
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Basic limit involving absolute value
Ok, I see what you are saying now, but when I graphed the parabola, I also saw that coming from the right-hand side, the values of x were also positive up until x=1.- wvcaudill2
- Post #11
- Forum: Calculus and Beyond Homework Help
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Basic limit involving absolute value
Im still not sure how it is you are coming to the statement that "if x comes from the left, then this expression is positive." Can you explain how you are getting this? Here is what I have now:- wvcaudill2
- Post #9
- Forum: Calculus and Beyond Homework Help
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Basic limit involving absolute value
I do not see how they should be switched. Can you explain your process in more detail?- wvcaudill2
- Post #5
- Forum: Calculus and Beyond Homework Help
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Basic limit involving absolute value
Ok, from the left-hand side of the limit, I got -2, and from the right-hand side I got 2, and when x-1=0 I got "does not exist." Is this right?- wvcaudill2
- Post #3
- Forum: Calculus and Beyond Homework Help
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Basic limit involving absolute value
Homework Statement lim as x\rightarrow-1 = \frac{|x^{2}-1|}{x^{2}+x} Homework Equations N/a The Attempt at a Solution Tried to write this as a piecewise function, but I got lost.- wvcaudill2
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- Absolute Absolute value Limit Value
- Replies: 12
- Forum: Calculus and Beyond Homework Help
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Integration by parts & inv. trig fxn
Oh, ok. Thanks again!- wvcaudill2
- Post #5
- Forum: Calculus and Beyond Homework Help
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Integration by parts & inv. trig fxn
I still don't see how this gets (1/8), and even then, what is the purpose of doing this, and where does the (2x) come from?- wvcaudill2
- Post #3
- Forum: Calculus and Beyond Homework Help
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Integration by parts & inv. trig fxn
Homework Statement \int xarcsin2xdx 2. The attempt at a solution Can someone explain to me what is happening at step 2? I understand how the integration by parts was done, but where does the (1/8) or (2x) come from?- wvcaudill2
- Thread
- Integration Integration by parts parts Trig
- Replies: 4
- Forum: Calculus and Beyond Homework Help
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Applying Integration by Parts & Trig Substitution
Okay, I got it now. Thanks for all your help!- wvcaudill2
- Post #13
- Forum: Calculus and Beyond Homework Help
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Applying Integration by Parts & Trig Substitution
Okay, that's what I thought you wanted me to do. I am now left with this: (7/4)\int sec^{3}Θ = (3/4)secΘtanΘ - (3/4)ln\left|secΘ + tanΘ\right| + C Now I need to manipulate the fraction so as to get (4/3)\int sec^{3}Θ How do I do this?- wvcaudill2
- Post #11
- Forum: Calculus and Beyond Homework Help
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Applying Integration by Parts & Trig Substitution
I am still not seeing it. Currently, I have \int sec^{3}Θ = (3/4)(secΘtanΘ - \int tan^{2}ΘsecΘ)- wvcaudill2
- Post #9
- Forum: Calculus and Beyond Homework Help
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Applying Integration by Parts & Trig Substitution
Is there anyone that can help me finish this problem?- wvcaudill2
- Post #7
- Forum: Calculus and Beyond Homework Help
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Applying Integration by Parts & Trig Substitution
Alright, I still seem to have a problem. I now have (3/4)(secθtanθ)-\int tan^{2}θsecθ Where do I go from here?- wvcaudill2
- Post #6
- Forum: Calculus and Beyond Homework Help