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(C) Let $f(x) = 2+x^2$. By the definition of a definite integral as the limit of a sum, $\int_a^b f(x) dx = \lim_{n \rightarrow \infty} h \sum_{r=1}^{

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$\lim _{n \rightarrow \infty} \frac{1}{n} \left[6n + \frac{1^2+2^2+\ldots+(3n)^2}{n^2} \right]$

Vedclass · Answer

(C) Let $f(x) = 2+x^2$. By the definition of a definite integral as the limit of a sum, $\int_a^b f(x) dx = \lim_{n \rightarrow \infty} h \sum_{r=1}^{n} f(a+rh)$, where $h = \frac{b-a}{n}$. Here, $a=0, b=3$, so $h = \frac{3-0}{n} = \frac{3}{n}$. Thus, $\int_0^3 (2+x^2) dx = \lim_{n \rightarrow \infty} \frac{3}{n} \sum_{r=1}^{n} f\left(0 + r \cdot \frac{3}{n}\right) = \lim_{n \rightarrow \infty} \frac{3}{n} \sum_{r=1}^{n} \left(2 + \frac{9r^2}{n^2}\right)$. Alternatively, using the form $\int_0^b f(x) dx = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{bn} f\left(\frac{r}{n}\right)$, we have: $\int_0^3 (2+x^2) dx = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{3n} \left(2 + \left(\frac{r}{n}\right)^2\right)$ $= \lim_{n \rightarrow \infty} \frac{1}{n} \left[ \sum_{r=1}^{3n} 2 + \sum_{r=1}^{3n} \frac{r^2}{n^2} \right]$ $= \lim_{n \rightarrow \infty} \frac{1}{n} \left[ 2(3n) + \frac{1^2+2^2+\ldots+(3n)^2}{n^2} \right]$ $= \lim_{n \rightarrow \infty} \frac{1}{n} \left[ 6n + \frac{1^2+2^2+\ldots+(3n)^2}{n^2} \right]$.

$\int_0^3 (2+x^2) dx = $

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