JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 20)
The value of $$\int\limits_{ - 2}^2 {|3{x^2} - 3x - 6|dx} $$ is ___________.
Answer
19
Explanation
x2 – x – 2 = (x - 2)(x + 1)
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$$\int\limits_{ - 2}^2 {3|{x^2} - x - 2|dx} $$
$$ = 3\int\limits_{ - 2}^2 {|{x^2} - x - 2|dx} $$
$$ = 3\left[ {\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - x - 2} \right)dx} + \int\limits_{ - 1}^2 { - \left( {{x^2} - x - 2} \right)dx} } \right]$$
$$ = 3\left[ {\left. {\left( {{{{x^3}} \over 3} - {{{x^2}} \over 2} - 2x} \right)} \right|_{ - 2}^{ - 1} - \left( {{{{x^3}} \over 3} - {{{x^2}} \over 2} - 2x} \right)_{ - 1}^2} \right]$$
$$ = 3\left[ {7 - {2 \over 3}} \right]$$
= 19
$$\int\limits_{ - 2}^2 {3|{x^2} - x - 2|dx} $$
$$ = 3\int\limits_{ - 2}^2 {|{x^2} - x - 2|dx} $$
$$ = 3\left[ {\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - x - 2} \right)dx} + \int\limits_{ - 1}^2 { - \left( {{x^2} - x - 2} \right)dx} } \right]$$
$$ = 3\left[ {\left. {\left( {{{{x^3}} \over 3} - {{{x^2}} \over 2} - 2x} \right)} \right|_{ - 2}^{ - 1} - \left( {{{{x^3}} \over 3} - {{{x^2}} \over 2} - 2x} \right)_{ - 1}^2} \right]$$
$$ = 3\left[ {7 - {2 \over 3}} \right]$$
= 19
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