Is there an easier way to solve these basic integration problems?

[ming2194]

Homework Statement

http://u1.imgupload.co.uk/1257897600/5a0e_image.bmp

Homework Equations

above

The Attempt at a Solution

1)
needs some hints here.
may i treat e^2x as (e^x)2 and then apply the formula to get the answers?

2)
http://u1.imgupload.co.uk/1257897600/bd3b_1.png

i didnt write a full attempt but it is the flow of my thought.
did my attempt correct? or in fact there are other easier way to get the answers?

3)
http://u1.imgupload.co.uk/1257897600/96d0_2.bmp
this one i just wonder whether my work is correct or not. (shown in pic)

thx a lots.

 

[Gib Z]

1) Assuming the formula is for the standard form \int \frac{1}{\sqrt{x^2-1}} dx, no you can not treat it that simply. You could try letting u=e^x.

2) Your method will work. I've tried some things and that seems to be the most elementary.

3) That is pretty much the correct evaluation of the "net area" but strictly the integral does not converge at all because the integrand is not well defined over the interval of integration.

 

[Mark44]

For 3, as Gib Z already noted, the integrand is undefined at a point in the interval. For this problem you need to split the integral into two improper integrals:
\lim_{a \rightarrow 1^-} \int_0^a \frac{dx}{x - 1}~+~\lim_{b \rightarrow 1^+} \int_b^3 \frac{dx}{x - 1}
Evaluate each integral and take the limit. There's no guarantee that either improper integral will converge, though.

 

[ming2194]

actually how can i know when the question is asking me using improper integral? or it's needed to decide by myself?

and for 2), i worked out the answer and found it's damn long and complex. so how can i check whether I am correct?

1)
http://u1.imgupload.co.uk/1257897600/492b_aaa.png
please judge

thx again.

 

[Mark44]

I think this is wrong, which you can confirm by differentiating your final result. How did you go from this integral:
\int \frac{du}{\sqrt{u^4 - 4u^2}}
to this?
ln(u^2 + \sqrt{u^4 - 4u^2}) + C

I would have left the integral like this:
\int \frac{du}{u\sqrt{u^2 - 4}}
and worked toward a trig substitution, with sec(\theta) = u/2.