Evaluate ∫limits_0^2 (x^2 + 3) dx

The correct answer is Option (3) → $\frac{26}{3}$

We know that $\int\limits_{a}^{b} f(x) dx = \lim\limits_{n \to \infty} h \sum\limits_{r=0}^{n-1} f(a + rh)$

For $I = \int\limits_{0}^{2} (x^2 + 3) dx$, we have $a = 0$ and $b = 2$

$I = \int\limits_{0}^{2} (x^2 + 3) dx$

Here, $a = 0$, $b = 2$ and $h = \frac{b-a}{n} = \frac{2-0}{n} = \frac{2}{n}$

$\Rightarrow nh = 2$

and $f(x) = (x^2 + 3)$

$∴I = \int\limits_{0}^{2} (x^2 + 3) dx = \lim\limits_{h \to 0} h \sum\limits_{r=0}^{n-1} f(a + rh)$

$= \lim\limits_{h \to 0} h \sum\limits_{r=0}^{n-1} f(rh)$

$= \lim\limits_{h \to 0} h \sum\limits_{r=0}^{n-1} (3 + r^2 h^2)$

$= \lim\limits_{h \to 0} h \left[ 3n + h^2 \left( \frac{(n-1)(n-1+1)(2n-2+1)}{6} \right) \right]$

$= \lim\limits_{h \to 0} \left[ 3nh + h^3 \left( \frac{(n-1)(n)(2n-1)}{6} \right) \right]$

$= \lim\limits_{h \to 0} \left[ 3nh + \frac{h^2}{6} (2n^3 - 3n^2 + n) \right]$

$= \lim\limits_{h \to 0} \left[ 3nh + \frac{2n^3 h^3 - 3n^2 h^2 \cdot h + nh \cdot h^2}{6} \right]$

$= \lim\limits_{h \to 0} \left[ 3 \cdot 2 + \frac{2 \cdot 2^3 - 3 \cdot 2^2 \cdot h + 2 \cdot h^2}{6} \right] = 6 + \frac{16}{6} = \frac{26}{3}$