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 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
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...