Question

Here, $[x]$ denotes the greatest integer less than or equal to $x$ . Given that $f(x) = [x] + x$ . The value obtained when this function is integrated with respect to $x$ with lower limit as $\frac{3}{2}$ and upper limit as $\frac{9}{2}$ , is

• A. $12$
• B. $10.5$
• C. $8$
• D. $16.5$

Answer 1

Answer: (D)

$16.5$

Answer 2

The correct option is(D): 16.5

We have, $f \left(x\right)=\left[x\right]+x$
Now, $\int\limits_{3/ 2}^{9/ 2}f\left(x\right)dx$
$=\int\limits^{9/ 2}_{ 3/ 2}\left[x\right]dx +\int\limits_{3 /2}^{9 /2} x\, dx$
$=\int\limits_{3/ 2}^{2}1dx+\int\limits_{2}^{3}2dx+\int\limits_{3}^{4}3dx+\int\limits^{9 /2}_{4}4dx+\int\limits_{3/ 2}^{9 2} x\,dx$
$=\left[x\right]_{3 2}^{2}+2\left[x\right]_{2}^{3}+3\left[x\right]_{3}^{4}+4\left[x\right]_{4}^{9 2}+ \left[\frac{x^{2}}{2}\right]_{3/ 2}^{9/ 2}$
$=\left(2-\frac{3}{2}\right)+2\left(3-2\right)+3\left(4-3\right)+4\left(\frac{9}{2}-4\right)$
$+\frac{1}{2}\left[\left(\frac{9}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}\right]$
$=\frac{1}{2}+2+3+\frac{4}{2}+\frac{72}{8}=\frac{1}{2}+16$
$=16.5$