Q.Find the value of $\int_a^b x^2\, dx$ with the help of definite integral as the limit of a sum.

Sum $h\sum(a+rh)^2$ using the formulas for $\sum r$ and $\sum r^2$, then let $h\to0$: the limit is $\dfrac{b^3-a^3}{3}$.

Concept. The definite integral is the limit of a Riemann sum: with $h=\dfrac{b-a}{n}$,

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

Set up with $f(x)=x^2$:

$\int_a^b x^2\,dx=\lim_{h\to0}h\sum_{r=0}^{n-1}(a+rh)^2=\lim_{h\to0}h\sum_{r=0}^{n-1}\big(a^2+2arh+r^2h^2\big).$

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

$=\lim_{h\to0}h\left[na^2+2ah\cdot\frac{n(n-1)}{2}+h^2\cdot\frac{(n-1)n(2n-1)}{6}\right].$

Since $nh=b-a$, write $nh=b-a$ and let $h\to0$ (so $n\to\infty$, $(n-1)h\to b-a$, $(2n-1)h\to2(b-a)$): …