Q.Evaluate $\int_0^2 (x^2+1)\,dx$ as the limit of a sum. OR Evaluate: $\displaystyle\int \dfrac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx$

The main part uses the definition $\int_a^b f(x)dx=\lim_{n\to\infty}h\sum_{r=0}^{n-1}f(a+rh)$; the OR part uses the substitution $x=\cos t$ then integration by parts.

Main question — limit of a sum

$f(x)=x^2+1$, $a=0$, $b=2$, $h=\dfrac{2}{n}$.

$\displaystyle\int_0^2 f(x)\,dx = \lim_{n\to\infty} h\sum_{r=0}^{n-1} f(rh) = \lim_{n\to\infty} h\sum_{r=0}^{n-1}\big[(rh)^2+1\big] = \lim_{n\to\infty}\Big[h^3\sum_{r=0}^{n-1}r^2 + nh\Big]$

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

$= \lim_{n\to\infty}\Big[\dfrac{8}{n^3}\cdot\dfrac{(n-1)n(2n-1)}{6} + 2\Big] = \lim_{n\to\infty}\Big[\dfrac{8(n-1)(2n-1)}{6n^2} + 2\Big]$

As $n\to\infty$, $\dfrac{(n-1)(2n-1)}{n^2}\to 2$, so the first term $\to\dfrac{8\times2}{6}=\dfrac{8}{3}$.

$\displaystyle\int_0^2(x^2+1)dx = \dfrac{8}{3}+2 = \dfrac{14}{3}$

OR — evaluate $\int\dfrac{x\cos^{-1}x}{\sqrt{1-x^2}}dx$

Let $t=\cos^{-1}x$, so $x=\cos t$, $dx=-\sin t\,dt$, $\sqrt{1-x^2}=\sin t$.

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