Question

Sumas de Riemann f(x) =2x^2-3x en el intervalo [0,4] alguien me podría ayudar a resolverlo se los agradecería bastante

Answer 1

$\mathrm{\large{Tema:Suma\ de\ Riemann}}$

Dada la función: $f(x)=2x^2-3x$

y el intervalo: $\left[ 0;4\right]$

$\mathrm{\large{Solución:}}$

se divide $\left[ 0;4\right]$ en $n$ sub-intervalos $\Delta x$

$\displaystyle{\Delta x=\frac{b-a}{n}}$

dónde:$a=0 \ \ \ \ b=4$

$\displaystyle{\Delta x=\frac{4-0}{n}\Rightarrow \Delta x=\frac{4}{n}}$

$\displaystyle{x_i=a+i\Delta x=0+\frac{4}{n}i\Rightarrow x_i=\frac{4}{n}i}$

$\mathrm{\large{La\ suma\ de\ Riemann}}$

$\displaystyle{\sum_{i=1}^n}f(x_i)\Delta x=\displaystyle{\sum_{i=1}^n}f\left(\frac{4}{n}i\right)\frac{4n}{n}$

$\mathrm{\large{Reemplazando\ x \ por\ x_i}}$

$\displaystyle{\sum_{i=1}^n}\left[2\left(\frac{4}{n}i\right)^2-3\left(\frac{4}{n}i\right)\right]\frac{4}{n}$

$\large{\frac{4}{n}\Rightarrow \mathrm{es\ constante}}$

$\frac{4}{n}\displaystyle{\sum_{i=1}^n}\frac{32}{n^2}i^2-\frac{12}{n}i$

$\displaystyle{\begin{cases}\frac{32}{n^2}\cr\cr \frac{12}{n}\end{cases}\Rightarrow \mathrm{\large{constantes}}}$



$\displaystyle{\frac{4}{n}\left(\frac{32}{n^2}\displaystyle{\sum_{i=1}^ni^2-\frac{12}{n}\sum_{i=1}^n}i\right)}$

$\displaystyle{\sum_{i=1}^n}i=\frac{n(n+1)}{2}\ \ \ \ y\ \ \ \displaystyle{\sum_{i=1}^n}i^2=\frac{n(n+1)(2n+1)}{6}$

$\mathrm{\large{Reemplazando}}$

$\displaystyle{\frac{4}{n}\left\lbrace\frac{32}{n^2}\left[\frac{n(n+1)(2n+1)}{6}\right]-\frac{12}{n}\left[\frac{n(n+1)}{2}\right]\right\rbrace}$

$\mathrm{\large{Desarrollando\ se \ llega:}}$

$\displaystyle{\sum_{i=1}^n}f(x_i)\Delta x=\frac{56n^2+120n+64}{3n^2}$

$\mathrm{\large{entonces:}}$
$\displaystyle{\lim_{n\to \infty}}\displaystyle{\sum_{i=1}^n}f(x_i)\Delta x=\displaystyle{\lim_{n\to \infty}}\frac{56n^2+12n+64}{3n^2}=\frac{n^2\left(56+\overbrace{\frac{120n}{n^2}}^0-\overbrace{\frac{64}{n^2}}^0\right)}{3n^2}=\frac{56}{3}$