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Definite Integration question

2010 · Shift 1 · Q46
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Definite Integration question

2010 · Shift 1 · Q46

JEE AdvancedMathematicsDefinite IntegrationNumerical+3 / −1
For any real number x,x,x, let [x]\left[ x \right][x] denote the largest integer less than or equal to x.x.x. Let fff be a real valued function defined on the interval [−10,10]\left[ { - 10,10} \right][−10,10] by f(x)={x−[x]if[x]is odd,1+[x]−xif[x]is evenf\left( x \right) = \left\{ {\begin{matrix} {x - \left[ x \right]} & {if\left[ x \right]is\,odd,} \\ {1 + \left[ x \right] - x} & {if\left[ x \right]is\,even} \\ \end{matrix} } \right.f(x)={x−[x]1+[x]−x​if[x]isodd,if[x]iseven​ Then the value of π210∫−1010f(x)cos⁡ πx dx{{{\pi ^2}} \over {10}}\int\limits_{ - 10}^{10} {f\left( x \right)\cos \,\pi x\,dx}10π2​−10∫10​f(x)cosπxdx is
Numerical answer
View written solutionFree

Correct answer: 4

  1. Understand the function on each unit interval

For any integer nnn, on the interval [n,n+1)[n,n+1)[n,n+1) we have [x]=n[x]=n[x]=n. So

f(x)={x−n,n odd,1+n−x,n even.f(x)= \begin{cases} x-n,& n \text{ odd},\\ 1+n-x,& n \text{ even}. \end{cases}f(x)={x−n,1+n−x,​n odd,n even.​

Thus on each unit interval, f(x)f(x)f(x) is a triangular/sawtooth function.

  1. Break the integral into unit intervals

We need to compute

I=∫−1010f(x)cos⁡(πx) dx.I=\int_{-10}^{10} f(x)\cos(\pi x)\,dx.I=∫−1010​f(x)cos(πx)dx.

Write it as

I=∑n=−109∫nn+1f(x)cos⁡(πx) dx.I=\sum_{n=-10}^{9}\int_n^{n+1} f(x)\cos(\pi x)\,dx.I=n=−10∑9​∫nn+1​f(x)cos(πx)dx.

Now let x=n+tx=n+tx=n+t, where t∈[0,1)t\in[0,1)t∈[0,1). Then dx=dtdx=dtdx=dt and

cos⁡(πx)=cos⁡(π(n+t))=cos⁡(nπ+πt)=(−1)ncos⁡(πt).\cos(\pi x)=\cos(\pi(n+t))=\cos(n\pi+\pi t)=(-1)^n\cos(\pi t).cos(πx)=cos(π(n+t))=cos(nπ+πt)=(−1)ncos(πt).

Also,

  • if nnn is odd, then f(x)=tf(x)=tf(x)=t,
  • if nnn is even, then f(x)=1−tf(x)=1-tf(x)=1−t.

So the contribution from interval [n,n+1)[n,n+1)[n,n+1) is

In=∫01gn(t)(−1)ncos⁡(πt) dt,I_n=\int_0^1 g_n(t)(-1)^n\cos(\pi t)\,dt,In​=∫01​gn​(t)(−1)ncos(πt)dt,

where

gn(t)={t,n odd,1−t,n even.g_n(t)=\begin{cases}t,& n\text{ odd},\\1-t,& n\text{ even}. \end{cases}gn​(t)={t,1−t,​n odd,n even.​
  1. Show every interval gives the same integral

If nnn is even, then (−1)n=1(-1)^n=1(−1)n=1, so

In=∫01(1−t)cos⁡(πt) dt.I_n=\int_0^1 (1-t)\cos(\pi t)\,dt.In​=∫01​(1−t)cos(πt)dt.

If nnn is odd, then (−1)n=−1(-1)^n=-1(−1)n=−1, so

In=−∫01tcos⁡(πt) dt.I_n=-\int_0^1 t\cos(\pi t)\,dt.In​=−∫01​tcos(πt)dt.

Now compute

∫01(1−t)cos⁡(πt) dt=∫01cos⁡(πt) dt−∫01tcos⁡(πt) dt.\int_0^1 (1-t)\cos(\pi t)\,dt =\int_0^1 \cos(\pi t)\,dt-\int_0^1 t\cos(\pi t)\,dt.∫01​(1−t)cos(πt)dt=∫01​cos(πt)dt−∫01​tcos(πt)dt.

But

∫01cos⁡(πt) dt=[sin⁡(πt)π]01=0.\int_0^1 \cos(\pi t)\,dt=\left[\frac{\sin(\pi t)}{\pi}\right]_0^1=0.∫01​cos(πt)dt=[πsin(πt)​]01​=0.

Hence

∫01(1−t)cos⁡(πt) dt=−∫01tcos⁡(πt) dt.\int_0^1 (1-t)\cos(\pi t)\,dt=-\int_0^1 t\cos(\pi t)\,dt.∫01​(1−t)cos(πt)dt=−∫01​tcos(πt)dt.

So both even and odd cases are equal. Therefore every interval contributes the same value

J=∫01(1−t)cos⁡(πt) dt.J=\int_0^1 (1-t)\cos(\pi t)\,dt.J=∫01​(1−t)cos(πt)dt.

There are 202020 unit intervals from −10-10−10 to 101010, so

I=20J.I=20J.I=20J.
  1. Compute JJJ

We evaluate

J=∫01(1−t)cos⁡(πt) dt.J=\int_0^1 (1-t)\cos(\pi t)\,dt.J=∫01​(1−t)cos(πt)dt.

Integrate by parts for

∫01tcos⁡(πt) dt.\int_0^1 t\cos(\pi t)\,dt.∫01​tcos(πt)dt.

Let

u=t,dv=cos⁡(πt) dt.u=t,\quad dv=\cos(\pi t)\,dt.u=t,dv=cos(πt)dt.

Then

du=dt,v=sin⁡(πt)π.du=dt,\quad v=\frac{\sin(\pi t)}{\pi}.du=dt,v=πsin(πt)​.

Thus

∫01tcos⁡(πt) dt=[tsin⁡(πt)π]01−1π∫01sin⁡(πt) dt.\int_0^1 t\cos(\pi t)\,dt =\left[\frac{t\sin(\pi t)}{\pi}\right]_0^1-\frac{1}{\pi}\int_0^1 \sin(\pi t)\,dt.∫01​tcos(πt)dt=[πtsin(πt)​]01​−π1​∫01​sin(πt)dt.

The boundary term is 000, so

∫01tcos⁡(πt) dt=−1π[−cos⁡(πt)π]01=−1π2(−cos⁡π+cos⁡0).\int_0^1 t\cos(\pi t)\,dt =-\frac{1}{\pi}\left[-\frac{\cos(\pi t)}{\pi}\right]_0^1 =-\frac{1}{\pi^2}\left(-\cos\pi+\cos0\right).∫01​tcos(πt)dt=−π1​[−πcos(πt)​]01​=−π21​(−cosπ+cos0).

Since cos⁡π=−1\cos\pi=-1cosπ=−1 and cos⁡0=1\cos0=1cos0=1,

∫01tcos⁡(πt) dt=−1π2(1+1)=−2π2.\int_0^1 t\cos(\pi t)\,dt =-\frac{1}{\pi^2}(1+1)=-\frac{2}{\pi^2}.∫01​tcos(πt)dt=−π21​(1+1)=−π22​.

Therefore

J=−∫01tcos⁡(πt) dt=2π2.J=-\int_0^1 t\cos(\pi t)\,dt=\frac{2}{\pi^2}.J=−∫01​tcos(πt)dt=π22​.

Hence

I=20⋅2π2=40π2.I=20\cdot \frac{2}{\pi^2}=\frac{40}{\pi^2}.I=20⋅π22​=π240​.
  1. Compute the required value

We need

π210∫−1010f(x)cos⁡(πx) dx=π210⋅40π2=4.\frac{\pi^2}{10}\int_{-10}^{10} f(x)\cos(\pi x)\,dx =\frac{\pi^2}{10}\cdot \frac{40}{\pi^2}=4.10π2​∫−1010​f(x)cos(πx)dx=10π2​⋅π240​=4.
  1. Final answer

The required integer is

4.\boxed{4}.4​.

This matches the stored correct answer.

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