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Definite Integration question

2021 · 17 Mar · Shift 1 · Q40
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Definite Integration question

2021 · 17 Mar · Shift 1 · Q40

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
If [ . ] represents the greatest integer function, then the value of ∣∫0π2[[x2]−cos⁡x]dx∣\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|​0∫2π​​​[[x2]−cosx]dx​ is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 1

  1. We need to evaluate
I=∣∫0π/2[ [x2]−cos⁡x ]dx∣,I=\left|\int_0^{\sqrt{\pi/2}} \left[\,[x^2]-\cos x\,\right]dx\right|,I=​∫0π/2​​[[x2]−cosx]dx​,

where [ ⋅ ][\,\cdot\,][⋅] denotes the greatest integer function.

  1. First simplify the expression inside the greatest integer.

Since 0≤x≤π/20\le x\le \sqrt{\pi/2}0≤x≤π/2​, we have

0≤x2≤π2<2.0\le x^2\le \frac{\pi}{2}<2.0≤x2≤2π​<2.

Hence

[x2]={0,0≤x<1,1,1≤x≤π/2.[x^2]= \begin{cases} 0, & 0\le x<1,\\ 1, & 1\le x\le \sqrt{\pi/2}. \end{cases}[x2]={0,1,​0≤x<1,1≤x≤π/2​.​

Also, for 0≤x≤π/2<π20\le x\le \sqrt{\pi/2}<\frac{\pi}{2}0≤x≤π/2​<2π​,

0<cos⁡x≤1.0<\cos x\le 1.0<cosx≤1.

So we split into two intervals.

  1. For 0≤x<10\le x<10≤x<1:
[x2]=0  ⟹  [x2]−cos⁡x=−cos⁡x.[x^2]=0 \implies [x^2]-\cos x = -\cos x.[x2]=0⟹[x2]−cosx=−cosx.

Since −1≤−cos⁡x<0-1\le -\cos x<0−1≤−cosx<0 and in fact −cos⁡x∈[−1,0)-\cos x\in[-1,0)−cosx∈[−1,0),

[[x2]−cos⁡x]=[−cos⁡x]=−1\big[[x^2]-\cos x\big]=[-\cos x]=-1[[x2]−cosx]=[−cosx]=−1

for 0≤x<10\le x<10≤x<1.

At x=0x=0x=0, the value is [−1]=−1[-1]=-1[−1]=−1, so this is valid on the whole interval [0,1)[0,1)[0,1).

  1. For 1≤x≤π/21\le x\le \sqrt{\pi/2}1≤x≤π/2​:
[x2]=1  ⟹  [x2]−cos⁡x=1−cos⁡x.[x^2]=1 \implies [x^2]-\cos x = 1-\cos x.[x2]=1⟹[x2]−cosx=1−cosx.

Now for x≥1x\ge 1x≥1, we have 0<cos⁡x<10<\cos x<10<cosx<1, so

0<1−cos⁡x<1.0<1-\cos x<1.0<1−cosx<1.

Therefore,

[[x2]−cos⁡x]=[1−cos⁡x]=0\big[[x^2]-\cos x\big]=[1-\cos x]=0[[x2]−cosx]=[1−cosx]=0

for 1≤x≤π/21\le x\le \sqrt{\pi/2}1≤x≤π/2​.

  1. Thus the integrand is
[[x2]−cos⁡x]={−1,0≤x<1,0,1≤x≤π/2.\big[[x^2]-\cos x\big]= \begin{cases} -1, & 0\le x<1,\\ 0, & 1\le x\le \sqrt{\pi/2}. \end{cases}[[x2]−cosx]={−1,0,​0≤x<1,1≤x≤π/2​.​

So

∫0π/2[[x2]−cos⁡x]dx=∫01(−1) dx+∫1π/20 dx.\int_0^{\sqrt{\pi/2}} \big[[x^2]-\cos x\big]dx =\int_0^1 (-1)\,dx + \int_1^{\sqrt{\pi/2}} 0\,dx.∫0π/2​​[[x2]−cosx]dx=∫01​(−1)dx+∫1π/2​​0dx.

Hence

=−1+0=−1.= -1 + 0 = -1.=−1+0=−1.

Therefore,

I=∣−1∣=1.I=|-1|=1.I=∣−1∣=1.
  1. Final answer:
1\boxed{1}1​
  1. Comparison with stored correct answer: Stored correct answer = 111, which matches our result.
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