JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 3)
The value of $$\int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}} dx$$, where [ t ] denotes the greatest integer $$ \le $$ t, is :
$${{e + 1} \over 3}$$
$${{e - 1} \over {3e}}$$
$${1 \over {3e}}$$
$${{e + 1} \over {3e}}$$
Explanation
$$I = \int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}dx} $$
Here -1 $$ \le $$ x $$ \le $$ 1 then -1 $$ \le $$ x3 $$ \le $$ 1
Integer between -1 and 1 is 0. So integration will be divided into two parts, -1 to 0 and 0 to 1.
$$ = \int\limits_{ - 1}^0 {{x^2}{e^{[{x^3}]}}dx} + \int\limits_0^1 {{x^2}{e^{[{x^3}]}}dx} $$
$$ = \int\limits_{ - 1}^0 {{x^2}{e^{ - 1}}dx} + \int\limits_0^1 {{x^2}{e^0}dx} $$
= $${1 \over e} \times \left[ {{{{x^3}} \over 3}} \right]_{ - 1}^0 + \left[ {{{{x^3}} \over 3}} \right]_0^1$$
$$ = {1 \over e} \times \left( {0 - \left( {{{ - 1} \over 3}} \right)} \right) + {1 \over 3}$$
$$ = {1 \over {3e}} + {1 \over 3} = {{1 + e} \over {3e}}$$
Here -1 $$ \le $$ x $$ \le $$ 1 then -1 $$ \le $$ x3 $$ \le $$ 1
Integer between -1 and 1 is 0. So integration will be divided into two parts, -1 to 0 and 0 to 1.
$$ = \int\limits_{ - 1}^0 {{x^2}{e^{[{x^3}]}}dx} + \int\limits_0^1 {{x^2}{e^{[{x^3}]}}dx} $$
$$ = \int\limits_{ - 1}^0 {{x^2}{e^{ - 1}}dx} + \int\limits_0^1 {{x^2}{e^0}dx} $$
= $${1 \over e} \times \left[ {{{{x^3}} \over 3}} \right]_{ - 1}^0 + \left[ {{{{x^3}} \over 3}} \right]_0^1$$
$$ = {1 \over e} \times \left( {0 - \left( {{{ - 1} \over 3}} \right)} \right) + {1 \over 3}$$
$$ = {1 \over {3e}} + {1 \over 3} = {{1 + e} \over {3e}}$$
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