Greetings everyone. For the following problem, I think I got the idea but I cannot find the mistake that I made. I know there is one because my answer is $\displaystyle k + \frac{5}{6}$ which is NOT

$\displaystyle k + \frac{11}{6}$. I am also trying to get this right before I sum each representative area between function and the floor function.

Thank you everyone.

My work

$\displaystyle \text{For }x\in (k,k-1)\text{ on }x\text{-axis, area = }\int_{k-1}^{k}{{{x}^{2}}+3x-1}\text{ }dx-1\cdot {{(k-1)}^{2}}+3(k-1)-1 $

$\displaystyle = \left[ \frac{{{x}^{3}}}{3}+\frac{3}{2}{{x}^{2}}-x \right]_{k-1}^{k}-1\cdot {{(k-1)}^{2}}+3(k-1)-1 $

$\displaystyle = \frac{{{k}^{3}}-{{k}^{3}}+3{{k}^{2}}-3k+1}{3}+\frac{3{{k}^{2}}-5{{k}^{2}}+10k-5}{2}-k+(k-1)-3\left( k-1 \right)+1 $

$\displaystyle = k+\frac{1}{3}-\frac{5}{2}+3=k+\frac{5}{6}

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