Calc help(please)

[parwana]

When particle is at distance meters from the origin, there is a force of Newtons (N) pulling it toward the origin. How much work is needed to move the particle from the position to the position ?



and




What is integral of 5^ radical x ?

[AKG]

That's a very strangely worded question. Let's call the distance from the origin, d, the force, F. Now what on earth do you mean "from the position to the position?" What position to what position? Very strange?!...

\int 5^{\root n\of x}dx

Let u = x^{1/n}, therefore du = \frac{1}{n}x^{1/n - 1}dx.

dx = \frac{ndu}{x^{1/n - 1}}

n\int \frac{5^udu}{u^{1 - n}}

Hmmm... maybe this is a start, I don't know where to go with this.

EDIT: Look up integral tables, see if they have ways to solve things like \int u5^udu, you might be able to solve this for the first few n (n = 1, 2, 3), and if you find a pattern, you might be able to use induction to prove it generally.

[Zurtex]

Your integral is way beyond me. But http://integrals.wolfram.com/index.en.cgi says:

\int 5^{x^{\frac{1}{n}}}dx = -nx \left(-x^{\frac{1}{n}} \right)^{-n} \Gamma \left( n, -x^{\frac{1}{n}} \ln 5 \right) \left(\ln 5\right)^{-n} + C

[AKG]

\int 5^{x^{\frac{1}{n}}}dx = -nx \left(-x^{\frac{1}{n}} \right)^{-n} \Gamma \left( n, -x^{\frac{1}{n}} \ln 5 \right) \left(\ln 5\right)^{-n} + CHah! Yes, I knew I was close :wink:.

[µ³]

Wouldn't the constant force towards the origin indicate a vector field of constant magnitude with all vectors pointing toward the origin? So the work from any two positions would just be the line integral for the straight line between any two points over this vector field.

[HallsofIvy]

Wouldn't the constant force towards the origin indicate a vector field of constant magnitude with all vectors pointing toward the origin? So the work from any two positions would just be the line integral for the straight line between any two points over this vector field.

The line integral [b]of the component of force parallel to the straight line[b] over the line segment.