Q.Evaluate ∫ (1 to 3) (x² + 4) dx as limit of a sum.

Building the Riemann sum with h=(3−1)/n and taking n→∞ evaluates the integral to 50/3.

Using $\displaystyle\int_a^b f(x)\,dx = \lim_{n\to\infty} h\sum_{r=0}^{n-1} f(a+rh)$, $h=\dfrac{b-a}{n}$

Here $a=1,b=3,f(x)=x^2+4$, so $h=\dfrac2n$.

$f(1+rh) = (1+rh)^2+4 = 1+2rh+r^2h^2+4 = 5+2rh+r^2h^2$

Sum $= h\displaystyle\sum_{r=0}^{n-1}\left(5+2rh+r^2h^2\right) = 5hn + 2h^2\sum r + h^3\sum r^2$

Using $\displaystyle\sum_{r=0}^{n-1} r = \dfrac{n(n-1)}{2}$, $\displaystyle\sum_{r=0}^{n-1} r^2=\dfrac{n(n-1)(2n-1)}{6}$:

$= 5hn + h^2n(n-1) + \dfrac{h^3n(n-1)(2n-1)}{6}$

With $h=2/n$:

$5hn = 10$

$h^2n(n-1) = \dfrac{4}{n^2}\cdot n(n-1) = 4\left(1-\dfrac1n\right) \to 4$ as $n\to\infty$

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