Question

If [ ] denotes the greatest integer function then the integral \[\int\limits_{0}^{\pi }{\left[ \cos x \right]dx}\] is equal to
[a] $\dfrac{\pi }{2}$
[b] 0
[c] -1
[d] $-\dfrac{\pi }{2}$

Answer 1

Hint: Use the property that if x is not an integer then [x]+[-x] = -1. Use the property of definite integrals $\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(a+b-x)dx}}$.

Complete step-by-step solution -
We will solve the above question using the properties of G.I.F function and definite integrals.
Let $I=\int\limits_{0}^{\pi }{\left[ \cos x \right]dx}\text{ (i)}$
We know that $\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{f(a+b-x)dx}}$.
Using the above formula, we get
$I=\int\limits_{0}^{\pi }{\left[ \cos \left( \pi +0-x \right) \right]dx}\text{ }$
$\Rightarrow I=\int\limits_{0}^{\pi }{\left[ \cos \left( \pi -x \right) \right]dx}$
We know $\cos \left( \pi -x \right)=-\cos x$
Substituting in the value of I we get
$I=\int\limits_{0}^{\pi }{\left[ -\cos x \right]dx}\text{ (ii)}$
Adding equation (i) and (ii) we get
$2I=\int\limits_{0}^{\pi }{\left( \left[ \cos x \right]+\left[ -\cos x \right] \right)dx}$
We know $\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{c}{f(x)dx+\int\limits_{c}^{b}{f(x)dx}}}$
Taking $c=\dfrac{\pi }{2}$in the above formula and using in I we get
$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \left[ \cos x \right]+\left[ -\cos x \right] \right)dx}+\int\limits_{\dfrac{\pi }{2}}^{\pi }{\left( \left[ \cos x \right]+\left[ -\cos x \right] \right)dx}$
Since in the interval $\left( 0,\pi \right)$$\cos x$takes an integral value at $\dfrac{\pi }{2}$ only we get
$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{-1dx}+\int\limits_{\dfrac{\pi }{2}}^{\pi }{-1dx}$
Integrating we get
$\Rightarrow 2I=\left. -x \right|_{0}^{\dfrac{\pi }{2}}+\left. -x \right|_{\dfrac{\pi }{2}}^{\pi }$
Substituting proper limits
$\begin{align}
  & \Rightarrow 2I=-\dfrac{\pi }{2}-0+\left( -\pi -\left( \dfrac{-\pi }{2} \right) \right) \\
 & \Rightarrow 2I=\dfrac{-\pi }{2}-\pi +\dfrac{\pi }{2} \\
 & \Rightarrow 2I=-\pi \\
 & \Rightarrow I=\dfrac{-\pi }{2} \\
\end{align}$
Option (d) is correct

Note: [1] We can also solve this question by plotting the graph of $\cos x$

As is evident from the graph $0<\cos x<1$ in the interval $\left( 0,\dfrac{\pi }{2} \right)$ and $-1<\cos x<0$ in the interval $\left( \dfrac{\pi }{2},\pi \right)$.
Hence $\left[ \cos x \right]=0$ in the interval $\left( 0,\dfrac{\pi }{2} \right)$and $\left[ \cos x \right]=-1$ in the interval $\left( \dfrac{\pi }{2},\pi \right)$
Hence total area under $\left[ \cos x \right]$ in the interval \[\left( 0,\pi \right)=0\times \dfrac{\pi }{2}+\left( -1 \right)\dfrac{\pi }{2}=\dfrac{-\pi }{2}\].
Hence the value of the definite integral = $\dfrac{-\pi }{2}$.
[2] Sometimes solving questions using graphs makes the question very easy in definite integrals.
[3] Some identities of [x] to remember
     [a] $[x]+[-x]=\left[ \begin{matrix}
   0\text{ if }x\in I \\
   -1\text{ if }x\notin I \\
\end{matrix} \right.$
     [b] $[x+a]=[x]+a\ \text{if }a\in I$
     [c] power of prime in n! = $\sum\limits_{r=1}^{\infty }{\left[ \dfrac{n}{{{p}^{r}}} \right]}$
     [d] $\left[ nx \right]\ne n\left[ x \right]$ in general
     [e] $x-[x]$ is called fractional part of x and is denoted by $\left\{ x \right\}$
     [f] $[x]\le x$
[4] Some properties of definite integrals to remember
     [a] $\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}$
     [b] $\int\limits_{a}^{b}{f(x)dx}=-\int\limits_{b}^{a}{f(x)dx}$
     [c] $\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(a+b-x)dx}$
     [d] $\int\limits_{-a}^{a}{f(x)}=\int\limits_{0}^{a}{\left( f(x)+f(-x) \right)dx}$