Q122 / 200 · WBJEE 2014 · Mathematics
Let $f(x)=\max \{x+|x|, x-[x]\}, \quad$ where $\quad \mid x]$ denotes the greateat integer $\leq x$. Then, the value of $\int_{-3}^{3} f(x) d x$ is
- A0
- B$51 / 2$
- Correct answerC$21 / 2$
- D1
Solution
Given, $f(x)=\max \{x+|x|, x-[x]\}$
$$
\begin{aligned}
=\left\{\begin{array}{ll}
2 x, & x \geq 0 \\
x-[x], & x \leq 0
\end{array}\right.\\
\begin{aligned}
\int_{-3}^{3} f(x) d x &=\int_{-3}^{0} x-[x] d x+\int_{0}^{3} 2 x d x \\
&=3 \int_{-1}^{0}(1+x) d x+2 \int_{0}^{3} x d x
\end{aligned}
\end{aligned}
$$
[$\because$ $x-[x]$ is a periodic function at $x=1]$
$$
\begin{array}{l}
=3\left[x+\frac{x^{2}}{2}\right]_{-1}^{0}+2\left[\frac{x^{2}}{2}\right]_{0}^{3} \\
=3\left[0-0-\left(-1+\frac{1}{2}\right)\right]+3^{2}-0 \\
=3\left[\frac{1}{2}\right]+9=\frac{3}{2}+9 \\
=\frac{21}{2}
\end{array}
$$
Hence, the correct answer is option: 'C'.