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If [ ] denotes the greatest integer function, then the integral $\int_0^\pi[\cos x d x$ is equal to:
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Verified Answer
The correct answer is:
$-\frac{\pi}{2}$
$-\frac{\pi}{2}$
Let $\mathrm{I}=\int_0^\pi[\cos x] d x$
$$
\mathrm{I}=\int_0^\pi[\cos (\pi-x)] d x=\int_0^\pi[-\cos x] d x
$$
On adding (1) and (2), we get
$$
\begin{aligned}
&2 \mathrm{I}=\int_0^\pi[\cos x] d x+\int_0^\pi[-\cos x] d x \\
&2 \mathrm{I}=\int_0^\pi[\cos x]+[-\cos x] d x \\
&2 \mathrm{I} \\
&=\int_0^\pi-1 d x \quad(\because[x]+[-x]=-1 \text { if } x \notin Z) \\
&2 \mathrm{I}=-\left.x\right|_0 ^\pi=-\pi \\
&\Rightarrow \mathrm{I}=\frac{-\pi}{2}
\end{aligned}
$$
$$
\mathrm{I}=\int_0^\pi[\cos (\pi-x)] d x=\int_0^\pi[-\cos x] d x
$$
On adding (1) and (2), we get
$$
\begin{aligned}
&2 \mathrm{I}=\int_0^\pi[\cos x] d x+\int_0^\pi[-\cos x] d x \\
&2 \mathrm{I}=\int_0^\pi[\cos x]+[-\cos x] d x \\
&2 \mathrm{I} \\
&=\int_0^\pi-1 d x \quad(\because[x]+[-x]=-1 \text { if } x \notin Z) \\
&2 \mathrm{I}=-\left.x\right|_0 ^\pi=-\pi \\
&\Rightarrow \mathrm{I}=\frac{-\pi}{2}
\end{aligned}
$$
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