I was doing some riemann sum excercises with a couple of friends to study for an exam but i got confused with the next sum.
I would appreciate to know the sum and the process behind the result.
Compute the Riemann sum $\displaystyle \sum_{{k}={1}}^{{n}}{{f}({x}_{k}^\ast)}∆x_{k} $ for the given partition $${f}\left({x}\right)={x}^{x2}+{1},\ \left[{1},{3}\right]$$
three subintervals, $${x}_{0}={1},\ {x}_{1}=\frac{{3}}{{2}},\ {x}_{2}=\frac{{5}}{{2}},\ {x}_{3}={3}; {x}_{1}^\ast=\frac{{5}}{{4}},\ {x}_{2}^\ast=\frac{{7}}{{4}},\ {x}_{3}^\ast={3} $$
The sum you’re looking for is the result of calculating
$(x_1-x_0)\cdot f(x_1^*)+(x_2-x_1)\cdot f(x_2^*)+(x_3-x_2)\cdot f(x_3^*)$
Looks like you have a mistake on f(x).