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

The Correct Option is D

The correct option is(D): 16.5

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