Defined integral $\{x^2-1\}$ from to $2$ to $\pi$.

Question

So I separate this integral from $2$ to $\sqrt5$, from $\sqrt5$ to $\sqrt6, \ldots$ from $3$ to $\pi$. Then I made a substitution of $x^2-1=t$. $dx=\frac{dt}{2 \sqrt{t+1}}$. So the length of every interval is 1.

So I can make the sum. But I have no I idea how to count it. Or I don’t need to change integral to sum? $\{x\}=x-[x]$ or I need to use this?

Answer 1

\begin{align} &\int_2^\pi \{ x^2-1\}\, dx \\ &=\int_2^\pi ( x^2-1) - \lfloor x^2-1\rfloor\, dx \\ &=\int_2^\pi ( x^2-1)\, dx - \int_2^\pi \lfloor x^2-1\rfloor\, dx \\ &=\int_2^\pi ( x^2-1)\, dx -\left( \int_3^\pi \lfloor x^2-1\rfloor\, dx + \sum_{k=4}^8\int_{\sqrt{k}}^{\sqrt{k+1}} \lfloor x^2-1\rfloor\, dx \right)\\ &=\int_2^\pi ( x^2-1)\, dx -\left( \int_3^\pi \lfloor x^2-1\rfloor\, dx + \sum_{k=4}^8\int_{\sqrt{k}}^{\sqrt{k+1}} \lfloor x^2-1\rfloor\, dx \right)\\ &=\int_2^\pi ( x^2-1)\, dx -\left( \int_3^\pi (8)\, dx + \sum_{k=4}^8\int_{\sqrt{k}}^{\sqrt{k+1}} (k-1)\, dx \right)\\ \end{align}

Can you complete this?

Edit:

Let $k$ be a positive integer, then for $\sqrt{k} \le x < \sqrt{k+1}$, we have $k \le x^2 < k+1$ and hence we have $k-1 \le x^2-1 < k$, hence $\lfloor x^2-1\rfloor=k-1$

Answer 2

To evaluate the integral $$\int_2^\pi \{ x^2-1\}\ dx$$

We partition the interval $[2,\pi ]$ into $$\{2, \sqrt 5, \sqrt 6, \sqrt 7, \sqrt 8, 3, \pi\}$$

The function $\{ x^2-1\}$ on each partition subinterval in a linear function ranging from $0$ to $1$

On the first interval $[2,\sqrt 5]$ the function is$$ y_1=\frac {x-2}{\sqrt 5 -2}$$ On the second interval $[\sqrt 5,\sqrt 6 ]$ the function is$$ y_2=\frac {x-\sqrt 5}{\sqrt 6 -\sqrt 5}$$ and so forth. Thus our integral is

$$ \int_2^\pi \{ x^2-1\}\ dx = \int _2^{\sqrt 5}y_1dx +\int _{\sqrt 5}^{\sqrt 6}y_2dx+...+\int _{3}^{\pi }y_6dx \approx 0.554179673$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{2}^{\pi}\braces{x^{2} - 2}\,\dd x & = \int_{2}^{\pi}\pars{x^{2} - 2 - \left\lfloor x^{2} - 2\right\rfloor}\dd x \\[5mm] & = {1 \over 3}\,\pi^{3} - 2\pi + {4 \over 3} - {1 \over 2}\int_{2}^{\pi^{2} - 2} {\left\lfloor x\right\rfloor \over \root{x + 2}}\dd x \\[5mm] & = {1 \over 3}\,\pi^{3} - 2\pi + {4 \over 3} - {1 \over 2}\sum_{k = 2}^{6}\int_{k}^{k + 1} {k \over \root{x + 2}}\dd x - {1 \over 2}\int_{7}^{\pi^{2} - 2} {7 \over \root{x + 2}}\dd x \\[5mm] & = {1 \over 3}\,\pi^{3} - 2\pi + {4 \over 3} - \sum_{k = 2}^{6}k\pars{\root{3 + k} - \root{k + 2}} + 7\pars{3 - \pi} \\[5mm] & = \bbx{\pi^{3} -27\pi + 3\root{7} + 3\root{6} + 3\root{5} + 6\root{2} + 25 \over 3} \approx 0.5542 \end{align}