Q.If $\int f(x)dx=g(x)+c$, then $\int f(x)g'(x)dx$ is

Divide $[1,3]$ into $n$ equal parts of width $h=2/n$, form the Riemann sum $h\sum_{r=0}^{n-1}(1+rh)^2$, and take the limit as $n\to\infty$ to get $26/3$.

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

Here $a=1,\ b=3,\ f(x)=x^2$, so $h=\dfrac{2}{n}$.

$\displaystyle\int_1^3 x^2\,dx = \lim_{n\to\infty} h\sum_{r=0}^{n-1}(1+rh)^2 = \lim_{n\to\infty} h\sum_{r=0}^{n-1}\big[1+2rh+r^2h^2\big]$

$= \lim_{n\to\infty}\left[hn + 2h^2\sum_{r=0}^{n-1}r + h^3\sum_{r=0}^{n-1}r^2\right]$

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

Term 1: $hn = \dfrac{2}{n}\cdot n = 2$

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

Term 3: $h^3\cdot\dfrac{(n-1)n(2n-1)}{6} = \dfrac{8}{n^3}\cdot\dfrac{n(n-1)(2n-1)}{6} = \dfrac{8}{6}\cdot\dfrac{(n-1)(2n-1)}{n^2} = \dfrac{4}{3}\left(2-\dfrac{3}{n}+\dfrac{1}{n^2}\right) \to \dfrac{4}{3}\times 2 = \dfrac{8}{3}$ as $n\to\infty$

Adding the limits:

$\displaystyle\int_1^3 x^2\,dx = 2+4+\dfrac{8}{3} = 6+\dfrac{8}{3} = \dfrac{26}{3}$

(This matches the direct evaluation $\left[\dfrac{x^3}{3}\right]_1^3=\dfrac{27}{3}-\dfrac{1}{3}=\dfrac{26}{3}$.)