Question

The value of $\int\limits_{ - 2}^2 {\left| {3{x^2} - 3x - 6} \right|dx} $ is

Answer 1

Hint: Find the roots of quadratic equations to know in which limit the value of modulus will be positive and negative. Then, write the modulus in expanded form with their appropriate limits and integrate the function.

Formula Used:
Integration formula –
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $

Complete step by step solution:
Given that,
$\int\limits_{ - 2}^2 {\left| {3{x^2} - 3x - 6} \right|dx} - - - - - (1)$
Here, $3{x^2} - 3x - 6 = 0$
${x^2} - x - 2 = 0$
${x^2} - 2x + x - 2 = 0$
$x(x - 2) + 1(x - 2) = 0$
$(x + 1)(x - 2) = 0$
$x = - 1,x = 2$
$ \Rightarrow $ the value of $\left| {3{x^2} - 3x - 6} \right|$ will be negative from $ - 1 \leqslant x \leqslant 2$
Now, Equation (1) will be
$ = 3\left[ {\int\limits_{ - 2}^2 {\left| {{x^2} - x - 2} \right|dx} } \right]$
$ = 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[ {\dfrac{{{x^3}}}{3} - \dfrac{{{x^2}}}{2} - 2x} \right]_{ - 2}^{ - 1} + \left[ { - \dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2} + 2x} \right]_{ - 1}^2} \right]$
$ = 3\left[ {\left[ {\left( {\dfrac{{{{\left( { - 1} \right)}^3}}}{3} - \dfrac{{{{\left( { - 1} \right)}^2}}}{2} - 2\left( { - 1} \right)} \right) - \left( {\dfrac{{{{\left( { - 2} \right)}^3}}}{3} - \dfrac{{{{\left( { - 2} \right)}^2}}}{2} - 2\left( { - 2} \right)} \right)} \right] + \left[ {\left( { - \dfrac{{{{\left( 2 \right)}^3}}}{3} + \dfrac{{{{\left( 2 \right)}^2}}}{2} + 2\left( 2 \right)} \right) - \left( { - \dfrac{{{{\left( { - 1} \right)}^3}}}{3} + \dfrac{{{{\left( { - 1} \right)}^2}}}{2} + 2\left( { - 1} \right)} \right)} \right]} \right]$
$ = 3\left[ {\left( {\dfrac{{ - 1}}{3} - \dfrac{1}{2} + 2} \right) - \left( {\dfrac{{ - 8}}{3} - \dfrac{4}{2} + 4} \right) + \left( { - \dfrac{8}{3} + \dfrac{4}{2} + 4} \right) - \left( {\dfrac{1}{3} + \dfrac{1}{2} - 2} \right)} \right]$
$ = 3\left[ {\dfrac{7}{3} + \dfrac{3}{2} - 2 + \dfrac{{\left( { - 9} \right)}}{3} + \dfrac{3}{2} + 6} \right]$
$ = 3\left[ {\dfrac{{ - 2}}{3} + 7} \right]$
$ = 3\left[ {\dfrac{{ - 2 + 21}}{3}} \right]$
$ = 19$
Hence, the value of $\int\limits_{ - 2}^2 {\left| {3{x^2} - 3x - 6} \right|dx} $ is $19$.

Note: In such questions, students must remember that after finding the roots always check whether those values or that range are giving positive value or negative value then find the limit accordingly. Put the limit only till the given range. While taking any number common during integration, take that constant outside and don't remove that directly.