$$\int_c xyds,$$
$$C: x=t^2, y=2t, 0 \le t \le 1$$
Evaluate the line integral, where C is the given curve.

Question

experimentx · Answer

$$ ds = \sqrt{ \left ( dx \over dt \right )^2 + \left ( dy \over dt \right )^2} dt$$
change everything into t's and integrate

anonymous · Answer

ok

anonymous · Answer

$$\int_c 2t^3 \sqrt{4t^2 +4} dt$$
$$4\int_c t^3 \sqrt{t^2 +1} dt$$

experimentx · Answer

anonymous · Answer

$$t=sec\theta$$
$$dt=sec\theta tan\theta d\theta$$
$$4\int_c sec^4 \theta tan^2 \theta d\theta$$

experimentx · Answer

t = tan 	heta

anonymous · Answer

I sit easier that way?

anonymous · Answer

*is

experimentx · Answer

i think integration by parts is also easier
u = t^2
v = t sqrt(t^2 + 1)

experimentx · Answer

anonymous · Answer

Wouldn't trig substitution be easier than the above link?

experimentx · Answer

oh .. you have to evaluate 
\int tan^3 x sec ^3 x dx

experimentx · Answer

and i'm not fond of high power of trigs ...

anonymous · Answer

$$4\int_c sec^4 \theta tan^2 \theta d\theta$$
$$4\int_c (tan^2 \theta +1)tan^2 \theta d\theta$$

anonymous · Answer

$$4\int_c (tan^4 \theta +tan^2 \theta) d\theta$$

anonymous · Answer

$$4\int_c tan^4 \theta d \theta +4\int_ctan^2 \theta d\theta$$

anonymous · Answer

Does this look right so far?

anonymous · Answer

anonymous · Answer

integration by parts I mean

experimentx · Answer

yep ... i would try that.

anonymous · Answer

I finally got:
$$\frac 4 5 (t^2+1)^{\frac 52}-\frac 4 3 (t^2+1)^{\frac 32}$$
is the the given curve?

anonymous · Answer

*this

experimentx · Answer

anonymous · Answer

yep, so this "c" by the integral is specific for curvature?

experimentx · Answer

hmm 
$$ \int_0^1$$

anonymous · Answer

Thanks!

experimentx · Answer