QJEEM04Mad103: JEE Main 2004 Mathematics - Class 12 JEEM04Mad103 Solution

Question

Verified Solution • 2025–26 Session moderate

The value of

\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx}

is

Accepted Answer

\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|} dx = \int\limits_{ - 2}^3 {\left| {{x^2} - 1} \right|} dx

Now

| x^{2} - 1 | = {\begin{matrix} x^{2} - 1 & i f & x \leq - 1 \\ 1 - x^{2} & i f & - 1 \leq x \leq 1 \\ x^{2} - 1 & i f & x \geq 1 \end{matrix}

\therefore

Integral is

\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - 1} \right)dx + \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)} } dx + \int\limits_1^3 {\left( {{x^2} - 1} \right)dx}

\left[ {{{{x^3}} \over 3} - x} \right]_{ - 2}^{ - 1} + \left[ {x - {{{x^3}} \over 3}} \right]_{ - 1}^1 + \left[ {{{{x^3}} \over 3} - x} \right]_1^3

= \left( { - {1 \over 3} + 1} \right) - \left( { - {8 \over 3} + 2} \right) + \left( {2 - {2 \over 3}} \right) + \left( {{{27} \over 3} - 3} \right) - \left( {{1 \over 3} - 1} \right)

= {2 \over 3} + {2 \over 3} + {4 \over 3} + 6 + {2 \over 3} = {{28} \over 3}

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