Recent content by LinkMage

I think the answer is +\infty because a number divided by 0 tends to infinity.

Sorry, I don't understand what you mean. I arranged the function in the first step to get a "0/0" indetermination and applied L'Hopital to that. Can you show me what you would do please?

You mean that: \lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{-\frac{1}{2s (\frac{1}{2} ln (s) - \frac{1}{8})^2}} = 0 And that the problem ends there?

I have to solve this: \lim_{\substack{s\rightarrow 0^+}} s^4 (\frac{1}{2} ln (s) - \frac{1}{8}) Here is what I did so far: \lim_{\substack{s\rightarrow 0^+}} \frac{s^4}{\frac{1}{\frac{1}{2} ln (s) - \frac{1}{8}} = = \lim_{\substack{s\rightarrow 0^+}}...

Well, that's how they taught it. I'm sure I would be taken marks off if I didn't do it like that. BTW, what calculator do you have? I want one like that. :P

And then: \lim_{\substack{s\rightarrow 0^+}} ${\displaystyle x^4(\frac {1} {2} ln (x) - \frac {1} {8})|_{s}^{1}}$ = \lim_{\substack{s\rightarrow 0^+}} {[1^4(\frac {1} {2} ln (1) - \frac {1} {8})] - {[s^4(\frac {1} {2} ln (s) - \frac {1} {8})] = -\frac {1} {8} - 0 = -\frac {1} {8}

Here is what I did (if I remember correctly): \int {2x^3ln(x)} dx u = ln (x) v' = 2x^3dx u' = \frac {1} {x} v = \frac {1} {2} x^4 \int {2x^3ln(x)} dx = {\frac {1} {2} x^4} ln (x) - \int {{\frac {1} {x}}{\frac {1} {2} x^4}} dx = {\frac {1} {2} x^4} ln (x)...

I had an exam today and this was one of the problems: \int_{0}^{1} {2x^3 ln (x)} dx My answer was -1/8. Is this correct or not?

\int {\frac {1} {\sqrt{-x}}} dx = \int {\frac {1} {(-x)^{1/2}} dx = \int {(-x)^{-1/2}} dx = \frac {(-x)^{-1/2+1}} {-1/2+1} = \frac {(-x)^{1/2}} {1/2} = 2 \sqrt {-x}

How do I do this? \int {\frac {1} {(\sqrt {-x})}} dx I got 2 \sqrt {-x} but my teacher got -2 \sqrt {-x} and I don't know how he got there.

Is it possible to integrate \int {cos (x^2)} dx ? If so, which method do I have to use?

You have to remember this equation a = \frac{Vf - Vo}{\Delta t} where Vf = final speed, Vo = initial speed, a = acceleration, \Delta t = time variation. Maybe Vf and Vo are Sf and So, I don't know.

You mean I just replace values for x until I get that the equation equals 0. That's what I did first and got aproximately -2.001. I was just wondering if I could solve the equation instead of replacing numbers.

I don't know how to solve x for things like e^x=20. The problem I have to solve is x^3+e^(2x)+8=0 Can anyone help, please?

I finally did it. Thanks a lot Viet Dao. I would never have guessed I had to do that.