Evaluate int_2^3 x^2 dx as the limit of a sum | vedclass

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Question

(D) We use the definition of the definite integral as the limit of a sum: $\int_a^b f(x) dx = \lim_{h 	o 0} h \sum_{r=0}^{n-1} f(a+rh)$,where $nh = b

Vedclass · Accepted Answer

$\frac{19}{3}$

Vedclass · Answer

(D) We use the definition of the definite integral as the limit of a sum: $\int_a^b f(x) dx = \lim_{h 	o 0} h \sum_{r=0}^{n-1} f(a+rh)$,where $nh = b-a$.Here,$a=2$,$b=3$,and $f(x)=x^2$. Thus,$nh = 3-2 = 1$.$I = \lim_{h 	o 0} h \sum_{r=0}^{n-1} (2+rh)^2 = \lim_{h 	o 0} h \sum_{r=0}^{n-1} (4 + 4rh + r^2h^2)$.$I = \lim_{h 	o 0} [4nh + 4h^2 \sum_{r=0}^{n-1} r + h^3 \sum_{r=0}^{n-1} r^2]$.Using the formulas $\sum_{r=0}^{n-1} r = \frac{(n-1)n}{2}$ and $\sum_{r=0}^{n-1} r^2 = \frac{(n-1)n(2n-1)}{6}$:$I = \lim_{h 	o 0} [4nh + 4h^2 \frac{(n-1)n}{2} + h^3 \frac{(n-1)n(2n-1)}{6}]$.Since $nh=1$,we have $I = \lim_{h 	o 0} [4(1) + 2(nh-h)(nh) + \frac{(nh-h)(nh)(2nh-h)}{6}]$.Substituting $nh=1$ and $h 	o 0$:$I = 4 + 2(1)(1) + \frac{(1)(1)(2)}{6} = 4 + 2 + \frac{1}{3} = 6 + \frac{1}{3} = \frac{19}{3}$.

Evaluate $\int_2^3 x^2 dx$ as the limit of a sum.

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