For any real number x, let [[x]] denote the largest integer less than equal to x.

Question

For any real number $x$, let $$[x]$$ denote the largest integer less than equal to $x$. Let $f$ be a real valued function defined on the interval $$[-10,10]$$ by $$f(x)=\left\{\[\begin{gathered} x-[x], \text { if }[x] \text { is odd } \ 1+[x]-x, \text { if }[x] \text { is even } . \end{gathered}$$\right.\]

Then the value of $$\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x \,d x$$ is :

Accepted Answer

Case 1 :

Let $$0 \le x < 1$$

then $$\left[ x \right] = 0$$, which is even

$$\therefore$$ $$f(x) = 1 + \left[ x \right] - x$$

$= 1 + 0 - x$

$= 1 - x$

Case 2 :

Let $$1 \le x < 2$$

then $$\left[ x \right] = 1$$, which is odd

$$\therefore$$ $$f(x) = x - \left[ x \right]$$

$= x - 1$

Case 3 :

Let $$2 \le x < 3$$

then $$\left[ x \right] = 2$$, which is even

$$\therefore$$ $$f(x) = 1 + \left[ x \right] - x$$

$= 1 + 2 - x$

$= 3 - x$

Case 4 :

Let $$3 \le x < 4$$

then $$\left[ x \right] = 3$$, which is odd

$$\therefore$$ $$f(x) = x - \left[ x \right]$$

$= x - 3$

$$\therefore$$ $$f(x) = \left\{ {\begin{pmatrix} {1 - x} & ; & {0 \le x < 1}  \
   {x - 1} & ; & {1 \le x < 2}  \
   {3 - x} & ; & {2 \le x < 3}  \
   {x - 3} & ; & {3 \le x < 4}  \

\end{pmatrix} } \right.$$

$$\therefore$$ $f(x)$ is periodic and period of $f(x) = 2$

And period of $$\cos \pi x = {{2\pi } \over \pi } = 2$$

$$\therefore$$ Period of $$f(x)\cos \pi x = 2$$

Now,

$$I = {{{\pi ^2}} \over {10}}\int_{ - 10}^{10} {f(x)\cos \pi x\,dx} $$

$$ = {{{\pi ^2}} \over {10}}\int_{ - 10}^{ - 10 + 10 \times 2} {f(x)\cos \pi x\,dx} $$

$$ = {{{\pi ^2}} \over {10}}\int_0^{10 \times 2} {f(x)\cos \pi x\,dx} $$

$$ = {{{\pi ^2}} \over {10}} \times 10\int_0^2 {f(x)\cos \pi x\,dx} $$

$$ = {\pi ^2}\int_0^2 {f(x)\cos \pi x\,dx} $$

$$\therefore$$ $$I = {\pi ^2}\left[ {\int_0^1 {f(x)\cos \pi x\,dx + \int_1^2 {f(x)\cos \pi x\,dx} } } \right]$$

$$ = {\pi ^2}\left[ {\int_0^1 {(1 - x)\cos \pi x\,dx + \int_1^2 {(x - 1)\cos \pi x\,dx} } } \right]$$

$$ = {\pi ^2}\left[ {\int_0^1 {\cos \pi x\,dx - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - \int_1^2 {\cos \pi x\,dx} } } } } \right]$$

$$ = {\pi ^2}\left[ {{1 \over \pi }\left[ {\sin \pi x} \right]_0^1 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - {1 \over \pi }\left[ {\sin \pi x} \right]_1^2} } } \right]$$

$$ = {\pi ^2}\left[ {0 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - 0} } } \right]$$

$$ = {\pi ^2}\left[ { - \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_0^1 + \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_1^2} \right]$$

$$\left[ {\mathrm{As}\,\int {x\cos \pi x\,dx = x\,.\,\int {\cos \pi x - \int {\left( {1\,.\,{{\sin \pi x} \over \pi }} \right)dx = x\,.\,{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x + c} } } } \right]$$

$$ = {\pi ^2}\left[ { - \left[ {\left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\,.\,\cos \pi } \right) - \left( {0 + {1 \over {{\pi ^2}}}\,.\,\cos 0} \right)} \right] + \left[ {\left( {2\,.\,{{\sin 2\pi } \over \pi } + {1 \over {{\pi ^2}}}\cos 2\pi } \right) - \left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\cos \pi } \right)} \right]} \right]$$

$$ = {\pi ^2}\left[ { - \left\{ {\left( { - {1 \over {{\pi ^2}}}} \right) - \left( {{1 \over {{\pi ^2}}}} \right)} \right\} + \left( {\left( { + {1 \over {{\pi ^2}}}} \right) - \left( { - {1 \over {{\pi ^2}}}} \right)} \right.} \right]$$

$$ = {\pi ^2}\left[ { - \left( { - {2 \over {{\pi ^2}}}} \right) + {2 \over {{\pi ^2}}}} \right]$$

$$ = {\pi ^2}\left[ {{2 \over {{\pi ^2}}} + {2 \over {{\pi ^2}}}} \right]$$

$$ = {\pi ^2} \times {4 \over {{\pi ^2}}}$$

$= 4$

Solutions · 1