Given $f(x) = 1+x^2, \alpha(x) = x^3, x \in [-1,1], P = \{-1,\frac{-1}{2},0,\frac{1}{2},1\}$. Find $U(P,f,\alpha)$ and $L(P,f,\alpha)$.
Ok, for this problem, we have $\alpha (x) = x^3$. And, I am confused where should I apply $\alpha(x) = x^3$. For the upper bound, I have use this: $\sum_{i = 1}^{n} f(xi)\Delta x$ and for lower bound I've used this: $\sum_{i = 1}^{n} (f(xi)-\Delta x)\cdot \Delta x$. I know when $\alpha(x) = x, \Delta \alpha(x) = \Delta(x)$. But, now that I have $\alpha(x) = x^3, \Delta \alpha(x) \neq \Delta x$. So, is $\Delta \alpha (x) = (1)^3 -(-1)^3 = 2$ in this case? And, what would be the formula for $\alpha (x)$? Any help? Thanks.
We have that given a partition $P=\{-1,\frac{-1}{2},0,\frac{1}{2},1\}$ of $[-1,1]$ with $\alpha(x)=x^3$, then \begin{align*} U(P,f,\alpha)=\sum_{i=1}^nM_i\cdot\Delta\alpha_i \end{align*} where $M_i=\sup_{x_{i-1}\le x\le x_i}f(x)$ and $\Delta\alpha_i=\alpha(x_{i})-\alpha(x_{i-1})$. Now it's a matter of just plugging in.
I will work through the computation for you so you can see.
\begin{align*} \sum_{i=1}^4M_i\cdot\Delta\alpha_i&=f(-1)\left(\frac{7}{8}\right)+f\left(\frac{-1}{2}\right)\left(\frac{1}{8}\right)+f\left(\frac{1}{2}\right)\left(\frac{1}{8}\right)+f\left(1\right)\left(\frac{7}{8}\right)\\ &=2\cdot\frac{7}{8}+\frac{5}{4}\cdot\frac{1}{8}+\frac{5}{4}\cdot\frac{1}{8}+2\cdot\frac{7}{8}\\ &=\frac{61}{16}. \end{align*}