Question

The value of the integral, ${\int }_1^3\left[x^2-2x-2\right]dx$ where $\left[x\right]$ denotes the greatest integer less than or equal to $x$ , is :

• A. $–4$
• B. $–5$
• C. $-\sqrt{2}-\sqrt{3}-1$
• D. $-\sqrt{2}-\sqrt{3}+1$

Answer 1

The correct option is C

$-\sqrt{2}-\sqrt{3}-1$

Explanation for the correct option:

Find the value of the give integral.

Given: ${\int }_1^3\left[x^2-2x-2\right]dx$

${\int }_1^3\left[x^2-2x-2\right]dx={\int }_1^3\left[x^2-2x+1-1-2\right]dx$

$$\begin{gathered} ={\int }_1^3\left[{\left(x-1\right)}^2-3\right]dx \\ ={\int }_1^3\left[{\left(x-1\right)}^2\right]dx-3{\int }_1^{3}dx \end{gathered}$$

Let, $x-1=t$

$\Rightarrow dx=dt$

So, for $x=1$, $t=0$ and for $x=3$, $t=2$

$={\int }_0^2\left[t^2\right]dt-3{\int }_1^{3}dt$

$={\int }_0^{1}0dt+{\int }_1^{\sqrt{2}}1dt+{\int }_{\sqrt{2}}^{\sqrt{3}}2dt+{\int }_{\sqrt{3}}^{2}3dt-3{\left[t\right]}_1^{3}dt$ [By using the property of $\left[x\right]$]

$$\begin{gathered} =0+{\left[t\right]}_1^{\sqrt{2}}+2{\left[t\right]}_{\sqrt{2}}^{\sqrt{3}}+3{\left[t\right]}_{\sqrt{3}}^2-3{\left[t\right]}_1^3 \\ =0+\left[\sqrt{2}-1\right]+2\left[\sqrt{3}-\sqrt{2}\right]+3\left[2-\sqrt{3}\right]-3\left[3-1\right] \\ =\sqrt{2}-1+2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3}-6 \\ =\sqrt{2}-2\sqrt{2}+2\sqrt{3}-3\sqrt{3}-1 \\ =-\sqrt{2}-\sqrt{3}-1 \end{gathered}$$

Therefore, the value of the integral, ${\int }_1^3\left[x^2-2x-2\right]dx$ where $\left[x\right]$ denotes the greatest integer less than or equal to $x$ , is $-\sqrt{2}-\sqrt{3}-1$

Hence, option (C) is the correct option.