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Definite Integration question

2022 · 27 Jul · Shift 2 · Q38
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Definite Integration question

2022 · 27 Jul · Shift 2 · Q38

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
Let f(x)=min⁡{[x−1],[x−2],…,[x−10]}f(x)=\min \{[x-1],[x-2], \ldots,[x-10]\}f(x)=min{[x−1],[x−2],…,[x−10]} where [t] denotes the greatest integer ≤t\leq \mathrm{t}≤t. Then ∫010f(x)dx+∫010(f(x))2 dx+∫010∣f(x)∣dx\int\limits_{0}^{10} f(x) \mathrm{d} x+\int\limits_{0}^{10}(f(x))^{2} \mathrm{~d} x+\int\limits_{0}^{10}|f(x)| \mathrm{d} x0∫10​f(x)dx+0∫10​(f(x))2 dx+0∫10​∣f(x)∣dx is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 385

  1. Simplify the function f(x)f(x)f(x)

We are given

f(x)=min⁡{[x−1],[x−2],…,[x−10]}. f(x)=\min\{[x-1],[x-2],\ldots,[x-10]\}.f(x)=min{[x−1],[x−2],…,[x−10]}.

Now, for fixed xxx, the numbers

x−1, x−2, …, x−10x-1,\ x-2,\ \ldots,\ x-10x−1, x−2, …, x−10

are decreasing as the subtracted integer increases. Hence

x−10≤x−9≤⋯≤x−1.x-10 \le x-9 \le \cdots \le x-1.x−10≤x−9≤⋯≤x−1.

Since the greatest integer function is non-decreasing,

[x−10]≤[x−9]≤⋯≤[x−1].[x-10] \le [x-9] \le \cdots \le [x-1].[x−10]≤[x−9]≤⋯≤[x−1].

Therefore the minimum is simply

f(x)=[x−10].f(x)=[x-10].f(x)=[x−10].

So the expression becomes

I=∫010[x−10]dx+∫010([x−10])2dx+∫010∣[x−10]∣dx.I=\int_0^{10}[x-10]dx+\int_0^{10}([x-10])^2dx+\int_0^{10}|[x-10]|dx.I=∫010​[x−10]dx+∫010​([x−10])2dx+∫010​∣[x−10]∣dx.
  1. Find the values of [x−10][x-10][x−10] on [0,10][0,10][0,10]

For x∈[0,10)x\in[0,10)x∈[0,10), we have x−10∈[−10,0)x-10\in[-10,0)x−10∈[−10,0).

So:

  • on [0,1)[0,1)[0,1), [x−10]=−10[x-10]=-10[x−10]=−10
  • on [1,2)[1,2)[1,2), [x−10]=−9[x-10]=-9[x−10]=−9
  • on [2,3)[2,3)[2,3), [x−10]=−8[x-10]=-8[x−10]=−8
  • ⋯\cdots⋯
  • on [9,10)[9,10)[9,10), [x−10]=−1[x-10]=-1[x−10]=−1

At x=10x=10x=10, [x−10]=0[x-10]=0[x−10]=0, but a single point does not affect the integral.

Thus, on each interval [k,k+1)[k,k+1)[k,k+1) for k=0,1,…,9k=0,1,\dots,9k=0,1,…,9,

[x−10]=k−10.[x-10]=k-10.[x−10]=k−10.

Equivalently, the values are

−10,−9,−8,−7,−6,−5,−4,−3,−2,−1.-10,-9,-8,-7,-6,-5,-4,-3,-2,-1.−10,−9,−8,−7,−6,−5,−4,−3,−2,−1.

Each persists on an interval of length 111.

  1. Compute ∫010f(x) dx\int_0^{10} f(x)\,dx∫010​f(x)dx
∫010f(x) dx=∑j=110(−j)⋅1=−(1+2+⋯+10).\int_0^{10} f(x)\,dx = \sum_{j=1}^{10} (-j)\cdot 1 =-(1+2+\cdots+10).∫010​f(x)dx=j=1∑10​(−j)⋅1=−(1+2+⋯+10).

Now,

1+2+⋯+10=10⋅112=55.1+2+\cdots+10=\frac{10\cdot 11}{2}=55.1+2+⋯+10=210⋅11​=55.

Hence,

∫010f(x) dx=−55.\int_0^{10} f(x)\,dx=-55.∫010​f(x)dx=−55.
  1. Compute ∫010(f(x))2 dx\int_0^{10} (f(x))^2\,dx∫010​(f(x))2dx
∫010(f(x))2 dx=∑j=110j2⋅1=12+22+⋯+102.\int_0^{10} (f(x))^2\,dx = \sum_{j=1}^{10} j^2\cdot 1 =1^2+2^2+\cdots+10^2.∫010​(f(x))2dx=j=1∑10​j2⋅1=12+22+⋯+102.

Using

12+22+⋯+n2=n(n+1)(2n+1)6,1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6},12+22+⋯+n2=6n(n+1)(2n+1)​,

for n=10n=10n=10,

12+22+⋯+102=10⋅11⋅216=385.1^2+2^2+\cdots+10^2=\frac{10\cdot 11\cdot 21}{6}=385.12+22+⋯+102=610⋅11⋅21​=385.

Therefore,

∫010(f(x))2 dx=385.\int_0^{10} (f(x))^2\,dx=385.∫010​(f(x))2dx=385.
  1. Compute ∫010∣f(x)∣ dx\int_0^{10} |f(x)|\,dx∫010​∣f(x)∣dx

Since f(x)f(x)f(x) takes values −10,−9,…,−1-10,-9,\ldots,-1−10,−9,…,−1 on the intervals above,

∣f(x)∣=10,9,…,1.|f(x)|=10,9,\ldots,1.∣f(x)∣=10,9,…,1.

Thus,

∫010∣f(x)∣ dx=1+2+⋯+10=55.\int_0^{10} |f(x)|\,dx = 1+2+\cdots+10=55.∫010​∣f(x)∣dx=1+2+⋯+10=55.
  1. Add all three integrals

So,

I=(−55)+385+55=385.I=(-55)+385+55=385.I=(−55)+385+55=385.

Hence the required integer is

385.\boxed{385}.385​.
  1. Comparison with stored answer

Stored correct answer: 385385385.

Our derived answer is also 385385385, so it agrees.

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