# Use the Composite Trapezoidal, Simpson's and Midpoint Rules to approximate the integral int_{1}^{2} x ln(x)dx, n=4 (For the Midpoint Rule use n + 2 subintervals.)

Question
Confidence intervals
Use the Composite Trapezoidal, Simpson's and Midpoint Rules to approximate the integral
$$\int_{1}^{2}\ x\ \ln(x)dx,\ n=4$$
(For the Midpoint Rule use $$n\ +\ 2$$ subintervals.)

2020-12-13
a) Trapezoidal Rule:
The interval [1, 2 can be partitioned into 4 sub-intervals, $$P=\left\{1,\ \frac{5}{4},\ \frac{3}{2},\ \frac{7}{4},\ 2\right\},$$ where:
$$\Delta\ x=\frac{b\ -\ a}{4}=\frac{1}{4}.$$
$$\int_{1}^{2}\ x\ln(x)dx=\frac{1}{4}\left\{f(1)\ +\ 2f\left(\frac{5}{4}\right)\ +\ 2f\left(\frac{3}{2}\right)\ +\ 2f\left(\frac{7}{4}\right)\ +\ f(2)\right\}$$
$$\frac{1}{4}\left\{(\ln(1))\ +\ 2\left(\frac{5}{4}\ \ln\left(\frac{5}{4}\right)\right)\ +\ 2\left(\frac{3}{2}\ln\left(\frac{3}{2}\right)\right)\ +\ 2\left(\frac{7}{4}\ln\left(\frac{7}{4}\right)\right)\ +\ 2(2\ln(2))\right\}$$
$$\frac{1}{4} \{0\ +\ 0.55788589\ +\ 1.2163953\ +\ 1.9586553\ +\ 2.7725887\} = 1.6263813$$
b) Simpson's Rule:
The interval [1, 2] can be partitioned $$\Delta\ x = \frac{b\ -\ a}{n} = \frac{1}{4}.$$
The partition set $$P = \left\{1,\ \frac{5}{4},\ \frac{3}{2},\ \frac{7}{4},\ 2\right\}.$$
Therefore,
$$\int_{1}^{2}\ x\ln(x)dx=\frac{\Delta\ x}{3}\ \{f(x_{0}\ +\ 4f(x_{1})\ +\ 2f(x_{2})\ +\ \cdots\ +\ f(x_{n})\}$$
$$=\frac{1}{12}\ \left\{f(1)\ +\ 4f\left(\frac{5}{4}\right)\ +\ 2f\left(\frac{3}{2}\right)\ +\ 4f\left(\frac{7}{4}\right)\ +\ f(2)\right\}$$
$$=\frac{1}{12}\ \left\{(\ln)\ +\ 4\left(\frac{5}{4}\ \ln\left(\frac{5}{4}\right)\right)\ +\ 2\left(\frac{3}{2}\ \ln\left(\frac{3}{2}\right)\right)\ +\ 4\left(\frac{7}{4}\ \ln\left(\frac{7}{4}\right)\right)\ +\ 2\ \ln(2)\right\}$$
$$\frac{1}{12} \{0\ +\ 1.1157178\ +\ 1.2163953\ +\ 3.9173105\ +\ 1.3862943\}$$
$$= 0.6363098$$
c) Midpoint Rule:
The interval [1, 2] can be partitioned $$\Delta\ x = \frac{b\ -\ a}{n\ +\ 2} = \frac{1}{6}.$$
The partition set $$P = \left\{1,\ \frac{7}{6},\ \frac{4}{3},\ \frac{3}{2},\ \frac{5}{3},\ \frac{11}{6},\ 2\right\}.$$
Therefore,
$$\int_{1}^{2}\ x\ln(x)dx=\ \Delta\ x\ \sum_{i=1}^{7}f(m_{i})$$
$$\frac{1}{6}\ \left\{(\ln(1))\ +\ \left(\frac{7}{6}\ \ln\left(\frac{7}{6}\right)\right)\ +\ \left(\frac{4}{3}\ \ln\left(\frac{4}{3}\right)\right)\ +\ \left(\frac{3}{2}\ \ln\left(\frac{3}{2}\right)\right)\ +\ \ \left(\frac{5}{3}\ \ln\left(\frac{5}{3}\right)\right)\ +\ \left(\frac{11}{6}\ \ln\left(\frac{11}{6}\right)\right)\ +\ 2\ln(2)\right\}$$
$$= 0.75342259$$

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