Donde $\qquad f(x_i)=\bigg(1 + \bigg(\frac1n\bigg)i\bigg)^2+5 \qquad$ entonces,

$\begin{aligned} \int_{1}^{2} (x^2+5)\, dx &= \lim_{n \to{}\infty}{\sum_{i=1}^n \bigg(6 + (\frac2n)i + (\frac{1}{n^2})i^2\bigg) .\bigg(\frac1n\bigg)} \\ &= \lim_{n \to{}\infty} \bigg(\frac1n\bigg){\sum_{i=1}^n \bigg(6 + (\frac2n)i + (\frac{1}{n^2})i^2\bigg)} \\ &= \lim_{n \to{}\infty} \bigg(\frac1n\bigg)\bigg[{\sum_{i=1}^n 6} + \frac2n{\sum_{i=1}^n i} + \frac{1}{n^2}{\sum_{i=1}^n i^2}\bigg] \\ &= \lim_{n \to{}\infty} \bigg(\frac1n\bigg)\bigg[6n + \frac{2n(n+1)}{2n} + \frac{n(n+1)(2n+1)}{6n^2}\bigg] \\ &= \lim_{n \to{}\infty} \bigg[6 + \bigg(1+ \frac1n\bigg) + \frac16 \bigg(1+ \frac1n\bigg)\bigg(2+ \frac1n\bigg)\bigg] \\ &= 6+1+\frac13 \\ \end{aligned}$

por tanto, $\displaystyle\qquad \int_{1}^{2} (x^2+5)\, dx =\frac{22}{3}$

Ejercicio 3.

Se tiene que, $\displaystyle\qquad \triangle x = \frac{4-1}{n}=\frac3n\qquad$ y $\displaystyle\qquad x_i = 1 + \bigg(\frac3n\bigg)i$

Donde $\qquad f(x_i)=\bigg(1 + \bigg(\frac3n\bigg)i\bigg)^2-5 \qquad$ entonces,

$\begin{aligned} \int_{1}^{4} (x^2-5)\, dx &= \lim_{n \to{}\infty}{\sum_{i=1}^n \bigg(-4 + (\frac6n)i + (\frac{9}{n^2})i^2\bigg) .\bigg(\frac3n\bigg)} \\ &= \lim_{n \to{}\infty} \bigg(\frac3n\bigg){\sum_{i=1}^n \bigg(-4 + (\frac6n)i + (\frac{9}{n^2})i^2\bigg)} \\ &= \lim_{n \to{}\infty} \bigg(\frac3n\bigg)\bigg[{\sum_{i=1}^n (-4)} + \frac6n{\sum_{i=1}^n i} + \frac{9}{n^2}{\sum_{i=1}^n i^2}\bigg] \\ &= \lim_{n \to{}\infty} \bigg(\frac3n\bigg)\bigg[-4n + \frac{6n(n+1)}{2n} + \frac{9n(n+1)(2n+1)}{6n^2}\bigg] \\ &= \lim_{n \to{}\infty} \bigg[-12 + 9\bigg(1+ \frac1n\bigg) + \frac92 \bigg(1+ \frac1n\bigg)\bigg(2+ \frac1n\bigg)\bigg] \\ &= -12+9+9 \\ \end{aligned}$

por tanto, $\displaystyle\qquad \int_{1}^{4} (x^2-5)\, dx =6$

Ejercicio 4.

Se tiene que, $\displaystyle\qquad \triangle x = \frac{2-1}{n}=\frac1n\qquad$ y $\displaystyle\qquad x_i = 1 + \bigg(\frac1n\bigg)i$

Donde $\qquad f(x_i)=\bigg(1 + \bigg(\frac1n\bigg)i\bigg)^2-5 \qquad$ entonces,

$\begin{aligned} \int_{1}^{2} (x^2-5)\, dx &= \lim_{n \to{}\infty}{\sum_{i=1}^n \bigg(-4 + (\frac2n)i + (\frac{1}{n^2})i^2\bigg) .\bigg(\frac1n\bigg)} \\ &= \lim_{n \to{}\infty} \bigg(\frac1n\bigg){\sum_{i=1}^n \bigg(-4 + (\frac2n)i + (\frac{1}{n^2})i^2\bigg)} \\ &= \lim_{n \to{}\infty} \bigg(\frac1n\bigg)\bigg[{\sum_{i=1}^n (-4)} + \frac2n{\sum_{i=1}^n i} + \frac{1}{n^2}{\sum_{i=1}^n i^2}\bigg] \\ &= \lim_{n \to{}\infty} \bigg(\frac1n\bigg)\bigg[-4n + \frac{2n(n+1)}{2n} + \frac{n(n+1)(2n+1)}{6n^2}\bigg] \\ &= \lim_{n \to{}\infty} \bigg[-4 + \bigg(1+ \frac1n\bigg) + \frac16 \bigg(1+ \frac1n\bigg)\bigg(2+ \frac1n\bigg)\bigg] \\ &= -4+1+\frac13 \\ \end{aligned}$

por tanto, $\displaystyle\qquad \int_{1}^{2} (x^2-5)\, dx =-\frac{8}{3}$

Ejercicio 5.

Se tiene que, $\displaystyle\qquad \triangle x = \frac{3-0}{n}=\frac3n\qquad$ y $\displaystyle\qquad x_i = 0 + \bigg(\frac3n\bigg)i$

Donde $\qquad f(x_i)=\bigg(\frac3n i\bigg)^2+5 \qquad$ entonces,

$\begin{aligned} \int_{0}^{3} (x^2+5)\, dx &= \lim_{n \to{}\infty}{\sum_{i=1}^n \bigg(5+\frac{9}{n^2}i^2\bigg) .\bigg(\frac3n\bigg)} \\ &= \lim_{n \to{}\infty} \bigg(\frac3n\bigg){\sum_{i=1}^n \bigg(5 + \frac{9}{n^2}i^2\bigg)} \\ &= \lim_{n \to{}\infty} \bigg(\frac3n\bigg)\bigg[{\sum_{i=1}^n 5} + \frac{9}{n^2}{\sum_{i=1}^n i^2}\bigg] \\ &= \lim_{n \to{}\infty} \bigg(\frac3n\bigg)\bigg[5n + \frac{9n(n+1)(2n+1)}{6n^2}\bigg] \\ &= \lim_{n \to{}\infty} \bigg[15 + \frac92 \bigg(1+ \frac1n\bigg)\bigg(2+ \frac1n\bigg)\bigg] \\ &= 15+9 \\ \end{aligned}$

por tanto, $\displaystyle\qquad \int_{0}^{3} (x^2+5)\, dx =24$

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