Respuesta :

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Splitting up the interval [0, 6] into 6 subintervals means we have

[tex][0,1]\cup[1,2]\cup[2,3]\cup\cdots\cup[5,6][/tex]

and the respective midpoints are [tex]\dfrac12,\dfrac32,\dfrac52,\ldots,\dfrac{11}2[/tex]. We can write these sequentially as [tex]{x_i}^*=\dfrac{2i+1}2[/tex] where [tex]0\le i\le5[/tex].

So the integral is approximately

[tex]\displaystyle\int_0^6x^2\,\mathrm dx\approx\sum_{i=0}^5({x_i}^*)^2\Delta x_i=\frac{6-0}6\sum_{i=0}^5({x_i}^*)^2=\sum_{i=0}^5\left(\frac{2i+1}2\right)^2[/tex]

Recall that

[tex]\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6[/tex]
[tex]\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2[/tex]
[tex]\displaystyle\sum_{i=1}^n1=n[/tex]

so our sum becomes

[tex]\displaystyle\sum_{i=0}^5\left(\frac{2i+1}2\right)^2=\sum_{i=0}^5\left(i^2+i+\frac14\right)[/tex]
[tex]=\displaystyle\frac{5(6)(11)}6+\frac{5(6)}2+\frac54=\frac{143}2[/tex]

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