QJEEM21Ma6404: JEE Main 2021 Mathematics - Class 12 JEEM21Ma6404 Solution: The value of , where [ t ] denotes the greatest integer t, is :

Question

Verified Solution • 2025–26 Session moderate

The value of

\int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}} dx

, where [ t ] denotes the greatest integer

\le

t, is :

Accepted Answer

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}}

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