The only issue with self-studying is how am I going to know that my work is correct without some feedback from a grader? I appreciate your advice though. I just don't see it being very practical in my case.
I really feel dissapointed in myself that I didn't perform as well as I wanted last semester. I took Modern Algebra I and Geometry. The Geometry class covered Euclidean and non-Euclidean geometries. I bombed the final but earned an overall of a B+ because of a 90-something percentile homework...
Thanks for taking the time in relating your thoughts about the subjects. I wanted to get through all the hard courses this junior year and then have an easier senior year, but I guess that would be rushing and not learning.
Hi PF members,
I've been thinking about taking Modern Algebra I and Real Analysis I in the same semester, but I'm having doubts if this is a realistic schedule. A semester is only four and a half months long and I don't believe that I can fully comprehend the subjects I'm studying for that...
I believe the answer is simply T = 780 N.
T - mg = m(0)
T = mg = 780 N
They didn't give you any value for the acceleration which if they did would've changed alot.
The algebra that I did from expanding it like you suggested was:
(a^3x + 3a^2x b^x + 3a^x b^2x + b^3x)/(a^x b^x)
= (a^2 /b)^x + 3a^x + 3b^x + (b^2/a)^x
But then what do I do with this strange exponent? I don't know how to integrate an expression like b^x
Homework Statement
Integrate (a^x + b^x)^3 /((a^x)*(b^x))
The Attempt at a Solution
I have only 2 semesters of calculus under my belt yet nothing in my experience has taught me to do anything like this
Hi again
Alright, kinda tired at this point (its 2am here) but I'm willing to show all steps
For the shell the method the way I thought about it was folding it into a crown in "3 d." The circumference is just 2*pi*y times height x thickness dy which in your terms would be 2*pi*y*(2 - x)...
Homework Statement
The function is y = 2 - x. The region is bounded by x = 2 and x = 4. Calculate its volume by the shell method by rotating it by the x axis.
The Attempt at a Solution
This problem has been consuming my mind. I calculated it by the disc method and shell method but I...
I'm not sure where to begin. First off I don't know what its really asking. I have an idea that the stone at the end of point A will fling at a higher range than point B, but other than that I don't see how to get there
So I think the appropriate eq is xf = x0 + v0t --> 1.20 + 1.50cos(30...
Homework Statement
A stone at the end of a sling is whirled in a vertical circle of radius 1.20 m at a constant speed v0 = 1.50 m/s. The center of the sling is 1.50 m above the ground. What is the range of the stone if it is released when the sling is inclined at 30 degrees with the...
Oh, sorry about that, here it is:
http://integrals.wolfram.com/index.jsp?expr=sec%5E3%28x%29&random=false
tanh^-1(tan(x/2)) + 1/(4 - 4sin(x)) - 1/(4(cos(x/2) + sin(x/2))^2)
That method leads to (1/2)sec(x)tan(x) + (1/2)ln(sec(x) + tan(x)) + C. I am interested in getting an inverse hyperbolic function as displayed on Wolfram.
I do not know how inverse hyperbolic functions are related to integrals. The only success I've had was integrating sec(x) into...
This is not homework, but I'm just wondering, how do you integrate this deceptive looking integrand to get what Wolfram has?
I don't get why the answer has an inverse hyperbolic function. Please teach me!