Download App

7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
$\int_0^\pi {|\cos x|\,dx = } $
  • A
    $\pi $
  • B
    $0$
  • ✓
    $2$
  • D
    $1$
✓

Answer

Correct option: C.
$2$
c
(c) $\int_0^\pi {\,\,|\cos x|} \,dx = 2\int_0^{\pi /2} {|\cos x|dx  = 2[\sin x]_0^{\pi /2} = 2} $.

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 7.2 definite integral→7.2 definite integral — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $y=y(x)$ is the solution of the differential equation $x \frac{d y}{d x}+2 y=x e^{x}, y(1)=0$ then the local maximum value of the function $z(x)=x^{2} y(x)-e^{x}$, $x \in R$ is
→
If $x=f(t)$ and $y=g(t)$ are differentiable functions of $t$, then $\frac{d^2 y}{d x^2}$ is
→
If $A=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$, then
→

If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is:

    →
    If the domain of the function $f(x)=\log _e$ $\left(\frac{2 x+3}{4 x^2+x-3}\right)+\cos ^{-1}\left(\frac{2 x-1}{x+2}\right)$ is $(\alpha, \beta]$, then the value of $5 \beta-4 \alpha$ is equal to
    →
    A value of ${\tan ^{ - 1}}\,\,\left( {\sin \,\left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{2}{3}} } \right)} \right)} \right)$ is
    →
    If $ABCD $ is a parallelogram, $\overrightarrow {AB} = 2\,i + 4\,j - 5\,k$ and $\overrightarrow {AD} = \,i + 2\,j + 3\,k,$ then the unit vector in the direction of $BD $ is
    →
    If $P = \left[ {\begin{array}{*{20}{c}}1&2&3\\2&3&4\\3&4&5\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\{ - 2}&0\\0&{ - 4}\end{array}} \right]$$\left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 5}&{ - 6}\\0&0&1\end{array}} \right]$ then ${P_{22}} = $
    →
    If $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix},$ then:
    1. f(a) = 0
    2. f(b) = 0
    3. f(0) = 0
    4. f(1) = 0
    →
    The set of points where $f(x) = \frac{x}{4+|x|}$ is differentiable is
    →