JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 2)
The value of the integral, $$\int\limits_1^3 {[{x^2} - 2x - 2]dx} $$, where [x] denotes the greatest integer less than or equal to x, is :
$$-$$ 5
$$ - \sqrt 2 - \sqrt 3 + 1$$
$$-$$ 4
$$ - \sqrt 2 - \sqrt 3 - 1$$
Explanation
$$I = \int\limits_1^3 { - 3dx + \int\limits_1^3 {\left[ {{{(x - 1)}^2}} \right]dx} } $$
Put x $$-$$ 1 = t ; dx = dt
$$I = ( - 6) + \int\limits_0^2 {\left[ {{t^2}} \right]} dt$$
$$I = - 6 + \int\limits_0^1 {0dt} + \int\limits_1^{\sqrt 2 } {1dt} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dt} + \int\limits_{\sqrt 3 }^2 {3dt} $$
$$I = - 6 + \left( {\sqrt 2 - 1} \right) + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 $$
$$I = - 1 - \sqrt 2 - \sqrt 3 $$
Put x $$-$$ 1 = t ; dx = dt
$$I = ( - 6) + \int\limits_0^2 {\left[ {{t^2}} \right]} dt$$
$$I = - 6 + \int\limits_0^1 {0dt} + \int\limits_1^{\sqrt 2 } {1dt} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dt} + \int\limits_{\sqrt 3 }^2 {3dt} $$
$$I = - 6 + \left( {\sqrt 2 - 1} \right) + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 $$
$$I = - 1 - \sqrt 2 - \sqrt 3 $$
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