Evaluación de integrales, sumas de Riemann

La suma Riemann será

$$\begin{align}&S=\lim_{n \to \infty}\sum_{i=0}^{n-1}(5x_i-6)\Delta x\\ &\\ &Donde\\ &\\ &\Delta x = \frac{5-2}{n}=\frac 3n\\ &\\ &x_i= 2+\frac{3i}{n}\\ &\\ &\\ &\\ &S =\lim_{n \to \infty}\sum_{i=0}^{n-1}\left[5\left(  2+\frac{3i}{n}\right)-6 \right] \frac 3n=\\ &\\ &\frac{3}{n}\lim_{n \to \infty} \sum_{i=0}^{n-1}\left(4+\frac{15i}{n}  \right)=\\ &\\ &\frac{3}{n}(4n)+\frac {45}{n^2} \lim_{n\to\infty}\sum_{i=0}^{n-1}i=\\ &\\ &12+ \lim_{n\to\infty}\frac {45n(n-1)}{2n^2}=\\ &\\ &\\ &12+ \frac{45}{2}= \frac{24+45}{2}=\frac{69}{2}\end{align}$$

Y la integral definida será:

$$\begin{align}&\int_2^5 (5x-6)dx=\\ &\\ &\left[\frac{5x^2}{2}-6x  \right]_2^5=\\ &\\ &\\ &\frac{125}{2}-30-10+12=\\ &\\ &\\ &\frac {125}{2}-28 =\frac{125-56}{2}=\frac{69}{2}\\ &\\ &\end{align}$$

Y ambas coinciden como tiene que ser. Si no habríamos contradicho el teorema fundamental del cálculo.