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$$\int\limits_0^2 {\left[ {{x^2}} \right]dx} $$ is

JEE · Math · previous-year question

  1. A.$$2 - \sqrt 2 $$
  2. B.$$2 + \sqrt 2 $$
  3. C.$$\,\sqrt 2 - 1$$
  4. D.$$ - \sqrt 2 - \sqrt 3 + 5$$correct

Answer

D. $$ - \sqrt 2 - \sqrt 3 + 5$$

Explanation

$$\int\limits_0^2 {\left[ {{x^2}} \right]} dx = \int\limits_0^1 {\left[ {{x^2}} \right]dx} + \int\limits_1^{\sqrt 2 } {\left[ {{x^2}} \right]} dx + $$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$\int\limits_{\sqrt 2 }^{\sqrt 3 } {\left[ {{x^2}} \right]} + \int\limits_{\sqrt 3 }^2 {\left[ {{x^2}} \right]} dx$$ $$ = \int\limits_0^1 {0dx} + \int\limits_1^{\sqrt 2 } {1dx} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dx} + \int\limits_{\sqrt 3 }^2 {3dx} $$ $$ = \left[ x \right]_1^{\sqrt n } + \left[ {2x} \right]_{\sqrt 2 }^{\sqrt 3 } + \left[ {3x} \right]_{\sqrt 3 }^2$$ $$ = \sqrt 2 - 1 + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 $$ $$ = 5 - \sqrt 3 - \sqrt 2 $$

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