Question

For any real number x, let $ \left[ x \right] $ denotes the largest integer less than or equal to x. Let f be a real valued function defined on the interval $ \left[ -10,10 \right] $ by:
 $ f\left( x \right)=\left\{ \begin{matrix}
   x-\left[ x \right]\text{ if }\left[ x \right]\text{ is odd} \\
   1+\left[ x \right]-x\text{ if }\left[ x \right]\text{ is even} \\
\end{matrix} \right\} $
Then the value of $ \dfrac{{{\pi }^{2}}}{10}\int\limits_{-10}^{10}{f\left( x \right)\cos \pi xdx} $ is:
(a) 0
(b) 1
(c) 2
(d) 3
(e) 4

Answer 1

Hint: First of all, write the integral, $ \int\limits_{-10}^{10}{f\left( x \right)\cos \pi xdx} $ by splitting the integral from -10 to -9, -9 to -8, -8 to -7 and so on till 9 to 10. Then as f(x) contains $ \left[ x \right] $ which will change its value in different limits. For e.g., from -10 to -9, the value of $ \left[ x \right] $ is even and f(x) is equal to $ 1+\left[ x \right]-x\text{ } $ . Similarly, for -9 to -8, the value of $ \left[ x \right] $ is odd and then f(x) is equal to $ x-\left[ x \right] $ . In this way, put different values of f(x) in different limits and hence, simplify.

Complete step by step answer:
In the above problem, we are asked to solve the integral:
  $ \dfrac{{{\pi }^{2}}}{10}\int\limits_{-10}^{10}{f\left( x \right)\cos \pi xdx} $
The f(x) in the above integral varies as follows:
 $ f\left( x \right)=\left\{ \begin{matrix}
   x-\left[ x \right]\text{ if }\left[ x \right]\text{ is odd} \\
   1+\left[ x \right]-x\text{ if }\left[ x \right]\text{ is even} \\
\end{matrix} \right\} $
First of all, we are going to simplify the integral given above:
 $ \int\limits_{-10}^{10}{f\left( x \right)\cos \pi xdx} $
In the above integral, $ \cos \pi x $ is even function means $ g\left( -x \right)=g\left( x \right) $ and the function f(x) is also an even function which we are going to show below:
In the f(x), if we put x as –x then when $ \left[ x \right] $ is odd then f (-x) is equal to:
 $ f\left( -x \right)=-x-\left[ -x \right] $
In the above function, we can write $ \left[ -x \right]=-1-\left[ x \right] $ then above equation becomes:
 $ \begin{align}
  & f\left( -x \right)=-x-\left( -1-\left[ x \right] \right) \\
 & \Rightarrow f\left( -x \right)=-x+1+\left[ x \right] \\
\end{align} $
As you can see that the value of the function which we are getting is the function when $ \left[ x \right] $ is even.
Now, writing x as –x in f(x) when $ \left[ x \right] $ is even we get,
 $ \begin{align}
  & f\left( -x \right)=1+\left[ -x \right]-\left( -x \right) \\
 & \Rightarrow f\left( -x \right)=1-1-\left[ x \right]+x \\
 & \Rightarrow f\left( -x \right)=x-\left[ x \right] \\
\end{align} $
As you can see that the value of the function which we are getting is the function when $ \left[ x \right] $ is odd.
From the above, we have shown that the function f(x) is an even function. This means that the function written in the integral is completely even so we can write the above integral as:
 $ 2\int\limits_{0}^{10}{f\left( x \right)\cos \pi xdx} $
For now, we are integrating $ \int\limits_{0}^{10}{f\left( x \right)}\cos \pi xdx $ and we are going to rewrite this integral in the following way:
 $ \begin{align}
  & \int\limits_{0}^{1}{f\left( x \right)\cos \pi xdx}+\int\limits_{1}^{2}{f\left( x \right)\cos \pi xdx}+\int\limits_{2}^{3}{f\left( x \right)\cos \pi xdx+\int\limits_{3}^{4}{f\left( x \right)\cos \pi xdx+\int\limits_{4}^{5}{f\left( x \right)\cos \pi xdx+}}} \\
 & \int\limits_{5}^{6}{f\left( x \right)\cos \pi xdx+\int\limits_{6}^{7}{f\left( x \right)\cos \pi xdx}}+\int\limits_{7}^{8}{f\left( x \right)\cos \pi xdx}+\int\limits_{8}^{9}{f\left( x \right)\cos \pi xdx}+\int\limits_{9}^{10}{f\left( x \right)\cos \pi xdx} \\
\end{align} $
In the above integral, when the limit is from 0 to 1 then $ \left[ x \right] $ takes value 0 and as $ \left[ x \right] $ is even then f(x) is equal to $ x-\left[ x \right] $ and from 1 to 2 then $ \left[ x \right] $ takes value 1 which is the odd value so f(x) is equal to $ 1+\left[ x \right]-x $ . Similarly, we can write the above integral in different limits.
\[\begin{align}
  & \int\limits_{0}^{1}{\left( 1+0-x \right)\cos \pi xdx}+\int\limits_{1}^{2}{\left( x-1 \right)\cos \pi xdx}+\int\limits_{2}^{3}{\left( 1+2-x \right)\cos \pi xdx+\int\limits_{3}^{4}{\left( x-3 \right)\cos \pi xdx+\int\limits_{4}^{5}{\left( 1+4-x \right)\cos \pi xdx+}}} \\
 & \int\limits_{5}^{6}{\left( x-5 \right)\cos \pi xdx+\int\limits_{6}^{7}{\left( 1+6-x \right)\cos \pi xdx}}+\int\limits_{7}^{8}{\left( x-7 \right)\cos \pi xdx}+\int\limits_{8}^{9}{\left( 1+8-x \right)\cos \pi xdx}+\int\limits_{9}^{10}{\left( x-9 \right)\cos \pi xdx} \\
 & =\int\limits_{0}^{1}{\left( 1-x \right)\cos \pi xdx}+\int\limits_{1}^{2}{\left( x-1 \right)\cos \pi xdx}+\int\limits_{2}^{3}{\left( 3-x \right)\cos \pi xdx+\int\limits_{3}^{4}{\left( x-3 \right)\cos \pi xdx+\int\limits_{4}^{5}{\left( 5-x \right)\cos \pi xdx+}}} \\
 & +\int\limits_{5}^{6}{\left( x-5 \right)\cos \pi xdx+\int\limits_{6}^{7}{\left( 7-x \right)\cos \pi xdx}}+\int\limits_{7}^{8}{\left( x-7 \right)\cos \pi xdx}+\int\limits_{8}^{9}{\left( 9-x \right)\cos \pi xdx}+\int\limits_{9}^{10}{\left( x-9 \right)\cos \pi xdx} \\
\end{align}\]
To solve the above integral, we need to solve the following integral:
 $ \int{x\cos \pi xdx} $
We are integrating the above integral by using integrating by parts:
 $ \begin{align}
  & x\int{\cos \pi x}dx-\int{\left( x \right)'\int{\left( \cos \pi x \right)}}dx \\
 & =x\left( \dfrac{\sin \pi x}{\pi } \right)-\int{1}\left( \dfrac{\sin \pi x}{\pi } \right)dx \\
 & =\dfrac{x\sin \pi x}{\pi }-\int{\left( \dfrac{\sin \pi x}{\pi } \right)dx} \\
 & =\dfrac{x\sin \pi x}{\pi }-\dfrac{1}{\pi }\left( -\dfrac{\cos \pi x}{\pi } \right) \\
 & =\dfrac{x\sin \pi x}{\pi }+\dfrac{1}{{{\pi }^{2}}}\left( \cos \pi x \right) \\
\end{align} $
Now, simplifying the above integral by using $ \int{x\cos \pi xdx} $ we get,
\[\begin{align}
  & =\int\limits_{0}^{1}{\cos \pi xdx}-\int\limits_{0}^{1}{x\cos \pi xdx}+\int\limits_{1}^{2}{x\cos \pi xdx}-\int\limits_{1}^{2}{\cos \pi xdx}+3\int\limits_{2}^{3}{\cos \pi xdx}-\int\limits_{2}^{3}{x\cos \pi xdx}+\int\limits_{3}^{4}{x\cos \pi xdx} \\
 & -3\int\limits_{2}^{3}{\cos \pi xdx}+9\int\limits_{8}^{9}{\cos \pi xdx}-\int\limits_{8}^{9}{x\cos \pi xdx}+\int\limits_{9}^{10}{x\cos \pi xdx}-9\int\limits_{9}^{10}{\cos \pi xdx} \\
 & =\left( \dfrac{\sin \pi x}{\pi } \right)_{0}^{1}-\left( \dfrac{x\sin \pi x}{\pi }+\dfrac{1}{{{\pi }^{2}}}\left( \cos \pi x \right) \right)_{0}^{1}+\left( \dfrac{x\sin \pi x}{\pi }+\dfrac{1}{{{\pi }^{2}}}\left( \cos \pi x \right) \right)_{1}^{2}-\left( \dfrac{\sin \pi x}{\pi } \right)_{1}^{2}+....... \\
 & -\left( \dfrac{x\sin \pi x}{\pi }+\dfrac{1}{{{\pi }^{2}}}\left( \cos \pi x \right) \right)_{8}^{9}+\left( \dfrac{x\sin \pi x}{\pi }+\dfrac{1}{{{\pi }^{2}}}\left( \cos \pi x \right) \right)_{9}^{10}-9\left( \dfrac{\sin \pi x}{\pi } \right)_{9}^{10} \\
\end{align}\]
In the above expression, sine terms will be 0 so all the sine terms become zero because we will get sine multiples of $ \pi $ is 0 so we are left with cosine terms.
 $ \begin{align}
  & -\dfrac{1}{{{\pi }^{2}}}\left( \cos \pi -\cos 0 \right)+\dfrac{1}{{{\pi }^{2}}}\left( \cos 2\pi -\cos \pi \right)-\dfrac{1}{{{\pi }^{2}}}\left( \cos 3\pi -\cos 2\pi \right)+\dfrac{1}{{{\pi }^{2}}}\left( \cos 4\pi -\cos 3\pi \right)-\dfrac{1}{{{\pi }^{2}}}\left( \cos 5\pi -\cos 4\pi \right) \\
 & +\dfrac{1}{{{\pi }^{2}}}\left( \cos 6\pi -\cos 5\pi \right)-\dfrac{1}{{{\pi }^{2}}}\left( \cos 7\pi -\cos 6\pi \right)+\dfrac{1}{{{\pi }^{2}}}\left( \cos 8\pi -\cos 7\pi \right)-\dfrac{1}{{{\pi }^{2}}}\left( \cos 9\pi -\cos 8\pi \right)+\dfrac{1}{{{\pi }^{2}}}\left( \cos 10\pi -\cos 9\pi \right) \\
 & =-\dfrac{1}{{{\pi }^{2}}}\left( -1-1 \right)+\dfrac{1}{{{\pi }^{2}}}\left( 1+1 \right)+-\dfrac{1}{{{\pi }^{2}}}\left( -1-1 \right)+\dfrac{1}{{{\pi }^{2}}}\left( 1+1 \right)-\dfrac{1}{{{\pi }^{2}}}\left( -1-1 \right)+\dfrac{1}{{{\pi }^{2}}}\left( 1+1 \right)-\dfrac{1}{{{\pi }^{2}}}\left( -1-1 \right)+\dfrac{1}{{{\pi }^{2}}}\left( 1+1 \right)-\dfrac{1}{{{\pi }^{2}}}\left( -1-1 \right)+\dfrac{1}{{{\pi }^{2}}}\left( 1+1 \right) \\
 & =5\left( \dfrac{2}{{{\pi }^{2}}}+\dfrac{2}{{{\pi }^{2}}} \right) \\
 & =\dfrac{20}{{{\pi }^{2}}} \\
\end{align} $
Now, multiplying the above equation by 2 we get,
 $ \dfrac{40}{{{\pi }^{2}}} $
As we have to solve $ \dfrac{{{\pi }^{2}}}{10}\int\limits_{-10}^{10}{f\left( x \right)\cos \pi xdx} $ so we have to multiply $ \dfrac{{{\pi }^{2}}}{10} $ by $ \dfrac{40}{{{\pi }^{2}}} $ .
 $ \begin{align}
  & \dfrac{{{\pi }^{2}}}{10}\left( \dfrac{40}{{{\pi }^{2}}} \right) \\
 & =4 \\
\end{align} $
Hence, the correct option is (e).

Note:
 In the above problem, you can track whether the solution you are doing is correct or not. In the above solution, after solving the integral $ \int\limits_{-10}^{10}{f\left( x \right)\cos \pi xdx} $ if you are not getting $ {{\pi }^{2}} $ in the denominator. This means that after multiplying the solution of the integral with $ \dfrac{{{\pi }^{2}}}{10} $ we won’t get the integer which you can see in the option.