In a finite element analysis, I am evaluating the following integral:
$$\int_{0}^{h}\left ( 1-\frac{x}{h} \right )*\left ( x \right )dx$$
but I want to apply a transformation from x to integrate over a different coordinate system, xi, with a domain of xi= -1 to 1.
So, the interval in the x coordinate system ranges from 0 to h and I want to integrate over -1 to 1, so I derived the following transformation:
$$x(\xi)=\frac{h\xi+h}{2}$$
so, I apply change of variables to my original integral and obtain:
$$\int_{-1}^{1}(\frac{1-\xi}{2})(\frac{h\xi+h}{2})(\frac{2}{h})d\xi$$
However, I am not getting the same answers for both integrals.
Where did I go wrong?
If $x=\frac{(\xi+1)h}{2}$ then $\operatorname d x= \dfrac h 2 \operatorname d \xi$
$$\require{enclose}\begin{align}\int_{0}^{h} (1-\tfrac x {h})x \operatorname d x = & ~ \enclose{circle}[mathcolor="red"]{\color{black}{\tfrac {h}{2}}}\int_{-1}^1 (1- \tfrac{(\xi+1)}{2})\tfrac{(\xi+1)h}{2}\operatorname d \xi \\[1ex] = & ~ \tfrac {h^2}{8}\int_{-1}^1 (1-\xi)(\xi+1)\operatorname d \xi \\[1ex] = & ~ \tfrac {h^2}{8}\int_{-1}^1 (1-\xi^2)\operatorname d \xi \\[1ex] = & ~ \tfrac {h^2}{8}\left[\xi-\tfrac {\xi^3}3)\right]_{\xi=-1}^{\xi=1} \\[1ex] = & ~ \tfrac 1 6h^2\end{align}$$
To test$$\begin{align}\int_0^h (1-\tfrac xh)x\operatorname d x = & ~ \tfrac 1 h\left[\tfrac{hx^2}{2} -\tfrac {x^3} 3 \right]_{x=0}^{x=h} \\[1ex] = & ~ \tfrac 1 h(\tfrac{h^3}{2}-\tfrac {h^3}{3}) \\[1ex] = & ~ \tfrac 16 h^2\end{align}$$