Given $f:[0,4]\rightarrow\mathbb{R}$ with the definition :
$f(x)=\begin{cases} 2x+3, & 0 \leq x<1 \\ 3, & x=1 \\ -x+1, & 1<x\leq 3 \\ 2, & 3<x\leq 4 \end{cases}$
If we're given the partition $P=\{0,1-h,1+h,3-h,3+h,4\}\subset[0,4]$ for any $h\in(0,\frac{1}{2})$, find the value of $U(f,P)-L(f,P)$.
Can someone please help me? I get stuck every time.
If $\mathcal P = \{x_0,\ldots,x_n\}$ is a partition then the upper and lower sums are \begin{align} U_f(\mathcal P) &= \sum_{j=0}^{n-1}\sup_{x\ \in\ [x_j,x_{j+1}]}f(x)(x_{j+1}-x_j)\\ L_f(\mathcal P) &= \sum_{j=0}^{n-1}\inf_{x\ \in\ [x_j,x_{j+1}]}f(x)(x_{j+1}-x_j). \end{align} So we compute \begin{align} U_f(\mathcal P) &= \sup_{x\in[0,1-h]}f(x)(1-h-0) + \sup_{x\in[1-h,1+h]}f(x)((1+h)-(1-h)) + \sup_{x\in[1+h,3-h]}f(x)(3-h - (1+h)) + \sup_{x\in[3-h,4]}f(x)(4 - (3-h)) \\ &= (2(1-h)+3)(1-h) + (2+3)(2h) + (-h)(2-2h) + 2(1-h)\\ &= 7 - h + 4h^2 \end{align} and \begin{align} L_f(\mathcal P) &= \inf_{x\in[0,1-h]}f(x)(1-h) + \inf_{x\in[1-h,1+h]}f(x)(2h) + \inf_{x\in[1+h,3-h]}f(x)(2h(h-1)) + \inf_{x\in[3-h,4]}f(x)(1-h) \\ &= 3(1-h) + (-h)(2h) + (-(2-h))(2h(h-1)) + (-2)(1-h)\\ &= 1+3h-8h^2+2h^3. \end{align} It follows that \begin{align} U_f(\mathcal P) - L_f(\mathcal P) &= 7 - h + 4h^2 - (1+3h-8h^2+2h^3)\\ &= 2(3-2h+6h^2-h^3). \end{align}