MA005: Calculus I

Practice 1:

(a) $\sum_{\mathrm{k}=1}^{5} \mathrm{k}^{3}=1+8+27+64+125$.

(b) $\sum_{j=2}^{7}(-1)^{j} \frac{1}{j}=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}$.

(c) $\sum_{\mathrm{m}=0}^{4}(2 \mathrm{~m}+1)=1+3+5+7+9$.

Practice 2:

(a) $\sum_{\mathrm{k}=2}^{5} \mathrm{~g}(\mathrm{k})=\mathrm{g}(2)+\mathrm{g}(3)+\mathrm{g}(4)+\mathrm{g}(5)=1+(-2)+3+5=7$.

(b) $\sum_{j=1}^{4} h(j)=h(1)+h(2)+h(3)+h(4)=3+3+3+3=12$.

(c) $\sum_{\mathrm{k}=3}^{5}(\mathrm{~g}(\mathrm{k})+\mathrm{f}(\mathrm{k}-1))=(\mathrm{g}(3)+\mathrm{f}(2))+(\mathrm{g}(4)+\mathrm{f}(3))+(\mathrm{g}(5)+\mathrm{f}(4))=(-2+3)+(3+1)+(5+0)=10$.

Practice 3: For $\mathrm{g}(x)=1 / x, \sum_{\mathrm{k}=2}^{4} \mathrm{~g}(\mathrm{k})=\mathrm{g}(2)+\mathrm{g}(3)+\mathrm{g}(4)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}$ and $\sum_{\mathrm{k}=1}^{3} \mathrm{~g}(\mathrm{k}+1)=\mathrm{g}(2)+\mathrm{g}(3)+\mathrm{g}(4)=\frac{13}{12}$.

Practice 4: Rectangular areas = $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{25}{12}=\sum_{j=1}^{4} \frac{1}{j}$.

Practice 5: $\mathrm{f}\left(\mathrm{x}_{0}\right)\left(\mathrm{x}_{1}-\mathrm{x}_{0}\right)+\mathrm{f}\left(\mathrm{x}_{1}\right)\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)+\mathrm{f}\left(\mathrm{x}_{2}\right)\left(\mathrm{x}_{3}-\mathrm{x}_{2}\right)$
$=\sum_{j=1}^{3} f\left(x_{j-1}\right)\left(x_{j}-x_{j-1}\right) \text { or } \sum_{k=0}^{2} f\left(x_{k}\right)\left(x_{k+1}-x_{k}\right)$

Practice 6: Interval is $[3, 8]$. 6 subintervals. mesh = length of longest subinterval = 1.2.
$\mathrm{x}_{2}=4.8 \text { and } \Delta \mathrm{x}_{2}=\mathrm{x}_{2}-\mathrm{x}_{1}=4.8-3.8=1$.

Practice 7: $\mathrm{RS}=(3)\left(\frac{1}{3}\right)+(1)\left(\frac{1}{4}\right)=1.25$.

Practice 8: smallest RS = $(3)\left(\frac{1}{4}\right)+(1)\left(\frac{1}{5}\right)=0.95$     largest RS = $(3)(1)+(1)\left(\frac{1}{4}\right)=3.25$.

Practice 9: $\mathrm{RS}=(0)\left(\frac{\pi}{4}\right)+(1)\left(\frac{\pi}{4}\right)+(1)\left(\frac{\pi}{2}\right) \approx 2.356$.