Answer:
Given definite integral as a limit of Riemann sums is:
![\lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Csum%5E%7Bn%7D%20_%7Bi%3D1%7D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
Step-by-step explanation:
Given definite integral is:
Substituting (2) in above
![f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=f%28x_%7Bi%7D%29%3D%5Cfrac%7B1%7D%7B2%7D%284%2B%5Cfrac%7B3%7D%7Bn%7Di%29%2B%284%2B%5Cfrac%7B3%7D%7Bn%7Di%29%5E%7B3%7D%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%282%2B%5Cfrac%7B3%7D%7B2n%7Di%29%2B%2864%2B%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B3%2816%29%5Cfrac%7B3%7D%7Bn%7Di%2B3%284%29%5Cfrac%7B9%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%29%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B108%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B3%7D%7B2n%7Di%2B%5Cfrac%7B144%7D%7Bn%7Di%2B66%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B108%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B291%7D%7B2n%7Di%2B66%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
Riemann sum is:
![= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Csum%5E%7Bn%7D%20_%7Bi%3D1%7D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
Answer:
Yi Xing invented the astronomical clock and introduced some new methods of interpolation in mathematics.
Step-by-step explanation:
Yi Xing was both an astronomer and a mathematician during the era. He invented the astronomical clock which was more accurate than the initial water and Sun's clock in use.
Furthermore, Yi Xing also discovered some new methods of interpolation in mathematics of which the meaning and interpretation became controversial. Interpolation is a method majorly in mathematics that can be used to estimate a value of a function from its discrete values. It involves first order differences and second order differences.
Also, Yi Xing was able to design a calendar in A.D. 727.
<h2>Question:</h2>
Find x to the nearest degree
<h2>
Answer:</h2>
a. 45
