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2023 PEA Q21

f(x)=⌊x⌋f(x)=\lfloor x\rfloor. Evaluate ∫05f(x2) dx\displaystyle\int_0^{\sqrt 5}f(x^2)\,dx.

The available source text does not include a full option block for this item, so the question is marked needsReview.
Reveal answer and solution

Answer

C

Solution

  1. 1

    f(x2)=⌊x2⌋f(x^2)=\lfloor x^2\rfloor. On [0,5][0,\sqrt 5],

  2. 2
    ⌊x2⌋={0,0≤x<11,1≤x<22,2≤x<33,3≤x<24,2≤x<5 \lfloor x^2\rfloor=\begin{cases}0,& 0\le x<1\\ 1,& 1\le x<\sqrt 2\\ 2,& \sqrt 2\le x<\sqrt 3\\ 3,& \sqrt 3\le x<2\\ 4,& 2\le x<\sqrt 5\end{cases}
  3. 3

    So

  4. 4
    ∫05⌊x2⌋dx=0+(2−1)⋅1+(3−2)⋅2+(2−3)⋅3+(5−2)⋅4. \int_0^{\sqrt 5}\lfloor x^2\rfloor dx=0+(\sqrt 2-1)\cdot 1+(\sqrt 3-\sqrt 2)\cdot 2+(2-\sqrt 3)\cdot 3+(\sqrt 5-2)\cdot 4.
  5. 5

    Compute:

  6. 6
    =2−1+23−22+6−33+45−8=(2−22)+(23−33)+45+(−1+6−8)=−2−3+45−3=45−3−2−3.\begin{aligned} &=\sqrt 2-1+2\sqrt 3-2\sqrt 2+6-3\sqrt 3+4\sqrt 5-8\\ &=(\sqrt 2-2\sqrt 2)+(2\sqrt 3-3\sqrt 3)+4\sqrt 5+(-1+6-8)\\ &=-\sqrt 2-\sqrt 3+4\sqrt 5-3\\ &=4\sqrt 5-\sqrt 3-\sqrt 2-3. \end{aligned}

Answer structure / marking notes

Needs review: source TeX does not provide a full four-option MCQ block.

Content note

Imported from public/resources/isi/msqe/solutions/pea/2023/ISI_MSQE_PEA_2023_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.