_ -2 ^ 2 |x|dx= - Vidyadip

MCQ

$\int_{-2}^{2}|\text{x}|\text{dx}=$

A
$0$
B
$2$
C
$1$
✓
$4$

✓

Answer

Correct option: D.

$4$

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

Explore more

More "M.C.Q (1 Marks)" from Integrals→Integrals — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If the vector $i + j + k$ makes angles $\alpha ,\,\beta ,\,\gamma $ with vectors $i,\,j,k$ respectively, then

If A and B are two events such that $\text{P}(\text{A}|\text{B})=\text{p},\text{P(A)}=\text{p},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{9},$ then p =

The maximum value of $Z = 4x + 2y$ Subjected to the constraints $2\text{x}+3\text{y}\leq18,\text{x}+\text{y}\geq10,\text{x},\text{y}\geq0$ is:

$\int_{}^{} {\frac{{5({x^6} + 1)}}{{{x^2} + 1}}dx = } $

The function $\text{f}(\text{x})=\sum_{\text{r}=1}^5(\text{x}-\text{r})^{2}$ assume minimum value at $x =$

For a suitably chosen real constant $a$, let a function, $f: R-\{-a\} \rightarrow R$ be defined by $f(x)=\frac{a-x}{a+x} .$ Further suppose that for any real number $x \neq- a$ and $f( x ) \neq- a ,( fof )( x )= x .$ Then $f\left(-\frac{1}{2}\right)$ is equal to

If $\displaystyle \begin{vmatrix}\text{x} &\text{amp; } 1 \\ \text{y} &\text{amp; } 2 \end{vmatrix}-\displaystyle \begin{vmatrix}\text{y} &\text{amp; } 1 \\ 8&\text{amp; } 0\end{vmatrix}=\displaystyle \begin{vmatrix}2 &\text{amp; } 0 \\ \text{-x}&\text{amp; } 2\end{vmatrix}$ then the values of $x$ and $y$ respectively are :

The order and degree of the differential equation ${\left( {1 + 3\frac{{dy}}{{dx}}} \right)^{\frac{2}{3}}} = 4\frac{{{d^3}y}}{{d{x^3}}}$ are

The differential coefficient of $\text{f}(\log\text{x}) \ w.r.t. x,$ where $\text{f(x)}=\log\text{x}$ is:

If $f(x) = \left\{ \begin{array}{l}\frac{{{x^2} - 1}}{{x + 1}},\,{\rm{when \,\,}}x \ne - 1\\\,\,\,\,\,\,\,\, - 2,\,{\rm{when\,\, }}x = - 1\end{array} \right.$, then