MCQ
$\int_{\, - \,2}^{\,2} {\,\left| {\,[x]\,} \right|\,dx = } $
  • A
    $1$
  • B
    $2$
  • C
    $3$
  • $4$

Answer

Correct option: D.
$4$
d
(d) $\int_{ - 2}^2 {|[x]|} \,dx = \int_{\, - 2}^{\, - 1} {\,|[x]|dx + \int_{ - 1}^0 {|[x]|dx + \int_0^1 {|[x]|dx| + \int_1^2 {|[x]|dx} } } } $

$ = \int_{ - 2}^{ - 1} {2dx\,\,} + \int_{ - 1}^0 {1dx + \int_0^1 {0\,dx + } } \int_1^2 {1dx} $

$ = 2[x]_{ - 2}^{ - 1} + [x]_{ - 1}^0 + 0 + [x]_1^2$

$ = 2( - 1 + 2) + (0 + 1) + (2 - 1) = 2 + 1 + 1 = 4.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Which of the following is true?
The parameter on which the value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\{\cos (p - d)x}&{\cos px}&{\cos (p + d)x}\\{\sin (p - d)x}&{\sin px}&{\sin (p + d)x}\end{array}\,} \right|$ does not depend upon
For any square matrix $A, A A^T$ is a
Let * be a binary operation on N defined by a * b = a + b + 10 for all a, b ∈ N. The identity element for * in N is:
The system of equations ${x_1} - {x_2} + {x_3} = 2,$ $\,3{x_1} - {x_2} + 2{x_3} = - 6$ and $3{x_1} + {x_2} + {x_3} = - 18$ has
If $\mathrm{p}(\mathrm{x})$ be a polynomial of degree three that has a local maximum value $8$ at $x=1$ and a local minimum value $4$ at $x=2 ;$ then $p(0)$ is equal to
If $y = {\log _{\cos x}}\sin x$, then ${{dy} \over {dx}}$ is equal to
If $f(x) = |3 − x| + (3 + x),$ where $(x) $ denotes the least integer greater than or equal to $x,$ then $f(x)$ is :
Let $A$ be a non-singular matrix of order $3$ . If $\operatorname{det}(\operatorname{adj}(2 \operatorname{adj}((\operatorname{det} A) A)))=3^{-13} \cdot 2^{-10}$ and det $(\operatorname{adj}(2 \mathrm{~A}))=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}$, then $|3 \mathrm{~m}+2 \mathrm{n}|$ is equal to...........
Let $f(x)=7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x$ for all $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the correct expression$(s)$ is(are)

$(A)$ $\int^{\pi / 4} x f(x) d x=\frac{1}{12}$

$(B)$ $\int_0^{\pi / 4} f(x) d x=0$

$(C)$ $\int_0^{\pi / 4} x f(x) d x=\frac{1}{6}$

$(D)$ $\int_0^{\pi / 4} f(x) d x=1$