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 October 12th, 2012, 03:53 AM #1 Newbie   Joined: Apr 2012 Posts: 8 Thanks: 0 Changing index of an integral Hi, As source i have the gauss quadrature integral: $\int_{-1}^1\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2}\simeq[C_1 * f(t_1) + ... + C_n * f(t_n)]$ Now I've seen that, to be able to use that in the range between 0 and , where z is less or equal, i could shift the integral which results in $\int_0^z\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2}\simeq\frac{z}{2}[C_1 * f(\frac{z}{2} * t_1 + \frac{z}{2}) + ... + C_n * f(\frac{z}{2} * t_n + \frac{z}{2})]$ Now the integral always goes from 0 to z. What do i have to do when i want it to go from x to z something like this:$\int_{0.4}^{0.75}$ Do i have to transform the integral again, and how does that exactly work. Or do i have to do this: $\int_{0.4}^{0.75} - \int_{0}^{0.4}$ where i subtract the result of the first integral from the second. In the end this is to calculate the arc-lengths of the segments of a cubic bezier curve.
 October 14th, 2012, 03:58 PM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Changing index of an integral I'm not certain I understand what you are asking. If it helps, with $I= \int_{-1}^1 f(x)dx$ we have $\int_0^z f(x)dx= I- \int_{-1}^0 f(x)dx- \int_z^1 f(x)dx$

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