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If $\left [ \: \: \right ]$ denotes the greatest integer function, then the integral $\int_{0}^{\pi }\left [ \cos x \right ]dx$ is equal to :
Option: 1
$\frac{\pi }{2}$

Option: 2
$0$

Option: 3
$-1$

Option: 4
$-\frac{\pi }{2}$

Answers (1)

As we have learned

Introduction of area under the curve -

The area between the curve $y= f(x),x$ axis and two ordinates at the point $x=a\, and \,x= b\left ( b>a \right )$ is given by

$A= \int_{a}^{b}f(x)dx=\int_{a}^{b}ydx$

- wherein

$\\I=\int_{0}^{\pi}[\cos x]dx\\ \\I=\int_{0}^{\pi/2}[\cos x]dx + \int_{\pi/2}^{\pi}[\cos x]dx \; = I_{1} + I_{2}$

$Now, \; we \; know \; that \\ for\;\;\; 0< x\leq \pi/2\;\;\;\; cosx\in [0,1). \\Hence,\; [cosx] = 0. And. \;\;for \; \pi/2< x\leq \pi \;\; cosx \in [-1. 0 ). \\ therefore, [cosx] = -1.$

Hence, the integration can be calculate by area under the curve,

$where,\; f(x) = 0 \; when x \in (0,\pi/2] \\ f(x) = -1 \; when x\in (\pi/2, \pi]$

Hence, the value of integral will be given by area under the curve along with its sign.

$Hence, \\ I_{1} = 0 \\ I_{2} = -1\times(\pi - \frac{\pi}{2} ) = -\pi/2$

Hence, $\\ I = -\pi/2$

Hence, correct option is option (4)

Posted by

Sumit Saini

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