Question

If $I_n={\int }_{-n}^n\left(\right\{x+1\left\}\right\{x^2+2\}+\{x^2+3\left\}\right\{x^2+4\left\}\right)dx$, (where $.$ denotes the fractional part), then $I_1$ is equal to

• A. $-\frac{1}{3}$
• B. $-\frac{2}{3}$
• C. $\frac{1}{3}$
• D. none of these

Answer 1

The correct option is C $\frac{1}{3}$

$I_n={\int }_{-n}^n(\{x+1\}\{x^2+2\}+\{x^2+3\}\{x^2+4\})dx{I}_n={\int }_{-n}^n((x+1-[x+1])(x^2+2-[x^2+2])+(x^3+3-[x^3+3]))(x^4+4-[x^4-4])dx{I}_1={\int }_{-}^1((x+1-[x+1])(x^2+2-[x^2+2])+(x^3+3-[x^3-3])(x^4+4-[x^4-4]))dx{I}_1={\int }_{-1}^0((x+1)\left(x^2\right)+\left(x^3\right)\left(x^4\right))dx+{\int }_0^1(\left(x\right)\left(x^2\right)+\left(x^3\right)\left(x^4\right))dx{I}_1={\int }_{-1}^0(x^3+x^2+x^7)dx+{\int }_0^1(x^3+x^7)dx$
$={[\frac{x^4}{4}+\frac{x^3}{3}+\frac{x^8}{8}]}_{-1}^0+{[\frac{x^4}{4}+\frac{x^8}{8}]}_0^1$
$=\frac{1}{4}+\frac{1}{3}-\frac{1}{8}+\frac{1}{4}+\frac{1}{8}=\frac{1}{3}$