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Ejercicio 1.

Se tiene que, △x=4−1n=3n\displaystyle\qquad \triangle x = \frac{4-1}{n}=\frac3n\qquad y xi=1+(3n)i\displaystyle\qquad x_i = 1 + \bigg(\frac3n\bigg)i

Donde f(xi)=(1+(3n)i)2+5\qquad f(x_i)=\bigg(1 + \bigg(\frac3n\bigg)i\bigg)^2+5 \qquad entonces,

∫14(x2+5) dx=lim⁡n→∞∑i=1n(6+(6n)i+(9n2)i2).(3n)=lim⁡n→∞(3n)∑i=1n(6+(6n)i+(9n2)i2)=lim⁡n→∞(3n)[∑i=1n6+6n∑i=1ni+9n2∑i=1ni2]=lim⁡n→∞(3n)[6n+6n(n+1)2n+9n(n+1)(2n+1)6n2]=lim⁡n→∞[18+9(1+1n)+92(1+1n)(2+1n)]=18+9+9\begin{aligned} \int_{1}^{4} (x^2+5)\, dx &= \lim_{n \to{}\infty}{\sum_{i=1}^n \bigg(6 + (\frac6n)i + (\frac{9}{n^2})i^2\bigg) .\bigg(\frac3n\bigg)} \\ &= \lim_{n \to{}\infty} \bigg(\frac3n\bigg){\sum_{i=1}^n \bigg(6 + (\frac6n)i + (\frac{9}{n^2})i^2\bigg)} \\ &= \lim_{n \to{}\infty} \bigg(\frac3n\bigg)\bigg[{\sum_{i=1}^n 6} + \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[6n + \frac{6n(n+1)}{2n} + \frac{9n(n+1)(2n+1)}{6n^2}\bigg] \\ &= \lim_{n \to{}\infty} \bigg[18 + 9\bigg(1+ \frac1n\bigg) + \frac92 \bigg(1+ \frac1n\bigg)\bigg(2+ \frac1n\bigg)\bigg] \\ &= 18+9+9 \\ \end{aligned}

por tanto, ∫14(x2+5) dx=36\displaystyle\qquad \int_{1}^{4} (x^2+5)\, dx =36

Ejercicio 2.

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

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

∫12(x2+5) dx=lim⁡n→∞∑i=1n(6+(2n)i+(1n2)i2).(1n)=lim⁡n→∞(1n)∑i=1n(6+(2n)i+(1n2)i2)=lim⁡n→∞(1n)[∑i=1n6+2n∑i=1ni+1n2∑i=1ni2]=lim⁡n→∞(1n)[6n+2n(n+1)2n+n(n+1)(2n+1)6n2]=lim⁡n→∞[6+(1+1n)+16(1+1n)(2+1n)]=6+1+13\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, ∫12(x2+5) dx=223\displaystyle\qquad \int_{1}^{2} (x^2+5)\, dx =\frac{22}{3}

Ejercicio 3.

Se tiene que, △x=4−1n=3n\displaystyle\qquad \triangle x = \frac{4-1}{n}=\frac3n\qquad y xi=1+(3n)i\displaystyle\qquad x_i = 1 + \bigg(\frac3n\bigg)i

Donde f(xi)=(1+(3n)i)2−5\qquad f(x_i)=\bigg(1 + \bigg(\frac3n\bigg)i\bigg)^2-5 \qquad entonces,

∫14(x2−5) dx=lim⁡n→∞∑i=1n(−4+(6n)i+(9n2)i2).(3n)=lim⁡n→∞(3n)∑i=1n(−4+(6n)i+(9n2)i2)=lim⁡n→∞(3n)[∑i=1n(−4)+6n∑i=1ni+9n2∑i=1ni2]=lim⁡n→∞(3n)[−4n+6n(n+1)2n+9n(n+1)(2n+1)6n2]=lim⁡n→∞[−12+9(1+1n)+92(1+1n)(2+1n)]=−12+9+9\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, ∫14(x2−5) dx=6\displaystyle\qquad \int_{1}^{4} (x^2-5)\, dx =6

Ejercicio 4.

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

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

∫12(x2−5) dx=lim⁡n→∞∑i=1n(−4+(2n)i+(1n2)i2).(1n)=lim⁡n→∞(1n)∑i=1n(−4+(2n)i+(1n2)i2)=lim⁡n→∞(1n)[∑i=1n(−4)+2n∑i=1ni+1n2∑i=1ni2]=lim⁡n→∞(1n)[−4n+2n(n+1)2n+n(n+1)(2n+1)6n2]=lim⁡n→∞[−4+(1+1n)+16(1+1n)(2+1n)]=−4+1+13\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, ∫12(x2−5) dx=−83\displaystyle\qquad \int_{1}^{2} (x^2-5)\, dx =-\frac{8}{3}

Ejercicio 5.

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

Donde f(xi)=(3ni)2+5\qquad f(x_i)=\bigg(\frac3n i\bigg)^2+5 \qquad entonces,

∫03(x2+5) dx=lim⁡n→∞∑i=1n(5+9n2i2).(3n)=lim⁡n→∞(3n)∑i=1n(5+9n2i2)=lim⁡n→∞(3n)[∑i=1n5+9n2∑i=1ni2]=lim⁡n→∞(3n)[5n+9n(n+1)(2n+1)6n2]=lim⁡n→∞[15+92(1+1n)(2+1n)]=15+9\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, ∫03(x2+5) dx=24\displaystyle\qquad \int_{0}^{3} (x^2+5)\, dx =24

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