PLEASE HELP: what is the length of the curve?

Question

length of the curve from x=1 to x=3, y= ((x^3)/3) + (x/4)

I get to a point where I have to take the integral of a square root, but the inside function does not have the derivative in the integral so I am stuck and can't go any further.

y= ((x^3)/3) +(x/4)

y'= x^2 +1/4

(y')^2 = (x^4)+ (1/16)+ ((x^2)/2)

surface area = integral from 1 to 3 of sqrt(1+ (y')^2) dx

= (1/4))* sqrt ((16 + 16x^4 +8x^2))dx

then i get stuck... Am i doing anything wrong?

Answer 1

You have to compute Int (1^3) sqrt((1 + (dy/dx)^2) dx.

dy/dx = 2x + 1/4
1 + (dy/dx)^2 = 1 + 4x^2 + x + 1/16 = 4x^2 + x + 17/16.

This 2nd degree polynomial has discriminant 1 - 17 = -16 <0, and can be written as 4[(x +1/8)^2 + 1/4]. Integrating sqrt(4[(x +1/8)^2 + 1/4]) from 1 to 3 is the same as integrating sqrt(4[(x^2 + 1/4]) = 2sqrt(x^2 + 1/4) from 1 -1/8 to 3 - 3/8
To integrate this last integrand, put x = sinh(t), so that dx = coosh (t) dt. You'll get the integral of cosh^2(t) which is easy.

Answer 2

The arclength of the curve is
∫(1..3) √(1 + (dy/dx)²
dy/dx = x² +1/4
(dy/dx)² = x^4 + x²/2+ 1/16
1 + (dy/dx)²= x^4 + x²/2 + 17/16.
Unfortunately, this is not a square
so you have to integrate the square root of a 4th
degree polynomial. This is an elliptic integral,
so you have to evaluate it numerically. Simpson's rule
would work fine.

Update: Your integrand is wrong. It should be
1/4* √(16x^4 + 8x^2 + 17). You cannot
evaluate this integral using antiderivatives, as I
pointed out above.

Now: I found this problem in my old calculus book.
It looks like you miscopied the problem.
The problem in the book had 1/(4x), not x/4
and this time 1 + (dy/dx)² turns out to be a
square, so you can calculate the integral.
Let y = x^3/3 + 1/(4x)
then dy /dx = x² - 1/4x²
and (dy/dx)² = x^4 -1/2 + 1/16*x*4
1+ (dy/dx)² = x^4 + 1/2 + 1/16*x^4 = (x²+1/4x²)².
So we have to evaluate
∫(1..3) (x² + 1/(4x²) dx
and you can do the rest from here!

Answer 3

i don't know the equation off the top of my head. show us your integral and we'll try to solve that