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help evaluating integral........

T

theMan

Diamond Member
ok, so, i can't figure this out at all...

definite integral of [1 + (4/9)x^(-2/3)]^(1/2) dx

i just dont know where to start, any hints?
 
so how do i do it?

ok, here is the original problem:

Find the EXACT length of the curve

y=x^(2/3) for x is between x=0 and x=2
 
this should be in the form of (a^2 + u^2)^1/2

so a = 1; u = 2/3[1/x]^(1/3)

an integral table says that it should come out to be...
1/2[u*(u^2+a^2)^(1/2) + a^2 *ln|u+ (u^2+a^2)^(1/2)|]+c
 
ok, i just used my calculator, and found the arc length to be 2.5999521037. but, i really need to actually do the problem.
 
Originally posted by: reverend boltron
this should be in the form of (a^2 + u^2)^1/2

so a = 1; u = 2/3[1/x]^(1/3)

an integral table says that it should come out to be...
1/2[u*(u^2+a^2)^(1/2) + a^2 *ln|u+ (u^2+a^2)^(1/2)|]+c
Click to expand...

uhh, for one thing, i don't have an integral table, but i looked one up, and i am still confused.
 
Originally posted by: theman
i dont think i learned trig substitution yet, i hope, because i have never seen it before.
Click to expand...

in that case, it sucks to be you and you need to use integration by substitution😛
 
Originally posted by: theman
so how do i do it?

ok, here is the original problem:

Find the EXACT length of the curve

y=x^(2/3) for x is between x=0 and x=2
Click to expand...

can anyone confirm that the integral i got for solving this is correct, because it seems like this is way over my head.
 
Okay, I figured it out. It's a little tricky because you use both integration by parts and substitution.

Since I don't want to deal with constants, I'll drop the 4/9 in my calculations.

Start by doing integration by parts with u=(1+x^(-2/3))^(1/2) and dv=dx.

Then du=(-5/6)x^(-5/3)(1+x^(-2/3))^(-1/2)dx and v=x.

Now our original integral is equal to uv-int vdu, so I'll concentrate on the integral of vdu.

-vdu=(5/6)x^(-2/3)(1+x^(-2/3))^(-1/2)dx

This looks bad, but put all negative powers in the denominator and distribute half of the x power (x^(1/3)) into the square root. This multiplies the terms inside the square root by x^(2/3) and we are left with:

(5/6)1/(x^(1/3)(x^(2/3)+1)^1/2))

Do a substitution of u=1+x^(2/3). Then du=x^(-1/3), which is exactly the power of x we have outside the square root.

To finish, you just have int(u^(-1/2))du, which is an easy power rule, and then back substitution for x.

Hope this helps!
 
Originally posted by: Gibson486
the derivative of

x^(2/3)

is not

(4/9)x^(-2/3)
Click to expand...

its the derivative squared, its in the formula for finding arc length,

definite integral of (1+(dy/dx)^2)^(1/2)
 
You can try the Integrator website =P. Just looking at these numbers, there's trig substitution written all over it. If you haven't learned it yet, maybe they want you to derive trig substitution? Heh =P Good luck!
 
I refer to these for my students as "tricky trig substitutions" (vs trig substitutions such as making 2sinxcosx = sin2x, etc.)
 
anyway, its a long time later, but i can explain how i did it with just plain u substitution... for those of you who have been anxiously waiting. 🙂

so, first, i write it as:

sqrt[1+ 4/9x^(2/3)]

which then with a common denominator =

sqrt[(9x^(2/3)+4)/9x^(2/3)]

the denominator can be taken out of the square root,

3x^(-1/3) * sqrt[9x^(2/3)+4]

then, just use easy u substitution

u = 9x^(2/3)+4
du = 6x^(-1/3)

easy. why couldn't i have thought of that before?
 
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