Recent content by renyikouniao

Yes,and this is where I am stuck..

A surface is obtained by rotating around the x-axis the arc over the integral(-1,0.5) of an ellipse given by: x^2+4y^2=1 What is its surface area? Here's my solution: I use the equation: S=integral( upper bound: a lower bound: b ) 2(pi)y*[1+(f'(x))^2]^0.5 dx Since x^2+4y^2=1...

find the length of a curve given by: f(x)=integral(upper bound: x lower bound: 0) (cos^2(x)+4cos(x)+1)^0.5 dx Here's my solution: I use the equation L=integral ( upper bound: a lower bound: b ) sqrt[1+(f'(x))^2] f'(x)=(cos^2(x)+4cos(x)+1)^0.5 1+[f'(x)]^2=2+cos^(x)+4cos(x)=[cos(x)+2]^2-2...

Since the integral(upper bound: 1, lower bound: 0) 1/x is divergent, so the definite integral can't be evaluated? Same as the second part, integral (upper bound:1, lower bound: 0) 1/(x-1) is divergent also.

1) integral (upper bound:1, lower bound:0) (x^2+1)/(x^3+x^2+4x) dx 2) integral (upper bound:1, lower bound:0) (x^4+x^2+1)/(x^3+x^2+x-3) dx Now I know how to use Partial Fractions,My question is: 1) For the first part ln(x) is not defined at 0 ¼ʃ1/x dx + ¼ʃ(3x-1)/(x²+x+4) dx = ¼ ln|x| +...

Can I state:if the integral (upper bound: a lower bound: b )is convergent/divergent. then the negative integral (upper bound: a lower bound: b )is convergent/divergent.For ex:If I prove 1/(x^2-1) is convergent,then I can say 1/(1-x^2) is convergent too.

Use the comparison test to find out whether or not the following improper integral exist(converge)? integral(upper bound:infinity lower bound:2) 1/(1-x^2) dx Here's my solution for 3),but I think something went wrong For all x>=2 0<=-(2-2x)<=-(1-x^2) that means: 0<=-1/(1-x^2)<=-1/(2-2x)...

thank you, can you help me with 2 and 3?

Thank you very much!:D How about I think it this way: (e^x-1/In(x))/(1/sinhx) (e^x-1/In(x)) this whole thing approches negative infinity and 1/sinhx approches negative infinity as x approches 1

Thank you very much for you patients(flower)(flower)(flower)

Thank you;),but what about the x on the top,should I rewrite x=2u?But if I do so,I can't use integral 1/((1-x^2)^0.5) right? Integrate x/((4-x^4)^0.5)

Thank you for the reponse. For part 2. (e^x-1)/Inx is one part,sinhx is another part.They both go to infinity as x approches 1.

Do you have any suggestions on how to solve this?:confused:

Question:Determine 1.Lim(x->infinity)e^(-x)coshx 2.Lim(x->1)[(e^x-1)/In(x)]sinhx For 1.=Lim(x->infinity)e^(-x)/(1/coshx) so the top and bottem both go to 0.then apply the rule. Repeat this procedure 3 times the top and bottem still both go to 0,so I am stuck...

No,we can't.That's the problem