Question

How do you evaluate the definite integral $\int {\left| {{x^2} - 4} \right|} dx$from $\left[ {0,3} \right]$?

Answer 1

We will open the modulus and break the integration of the function into a sum of two integration functions with different limits. Use the general formula of integration and solve the two integrals from given limits.
Modulus of a function ‘x’ opens up as \left| x \right| = \left\\{ {\begin{array}{*{20}{c}} x&{;x \geqslant 0} \\\ { - x}&{;x < 0} \end{array}} \right.
$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$

Complete step-by-step solution:
We have to solve the definite integral $\int {\left| {{x^2} - 4} \right|} dx$ from $\left[ {0,3} \right]$
We will break the function of modulus i.e. $\left| {{x^2} - 4} \right|$ between the limits from 0 to 3
We equate the function inside the modulus to 0
$\Rightarrow {x^2} - 4 = 0$
Shift constant value to right hand side of the equation
$\Rightarrow {x^2} = 4$
Take square root on both sides of the equation
$\Rightarrow \sqrt {{x^2}} = \sqrt 4$
$\Rightarrow \sqrt {{x^2}} = \sqrt {{2^2}}$
Cancel square root by square power on both sides of the equation
$\Rightarrow x = \pm 2$
Since we have to separate limits between 0 and 3, we will ignore the value -2
So, we can break down the limit $\left[ {0,3} \right]$into two parts i.e. $\left[ {0,2} \right]$and $\left[ {2,3} \right]$.
From $\left[ {0,2} \right]$:
$\Rightarrow \left| {{x^2} - 4} \right| = - ({x^2} - 4)$.............… (1)
From $\left[ {2,3} \right]$:
$\Rightarrow \left| {{x^2} - 4} \right| = ({x^2} - 4)$...............… (2)
From equations (1) and (2) we can write the function \left| {{x^2} - 4} \right| = \left\\{ {\begin{array}{*{20}{c}} { - ({x^2} - 4)}&{;x \in \left[ {0,2} \right]} \\\ {({x^2} - 4)}&{;x \in \left[ {2,3} \right]} \end{array}} \right.
So, we can write the definite integral as
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = \int\limits_0^2 { - ({x^2} - 4)dx} + \int\limits_2^3 {({x^2} - 4)dx}$
We can bring out the negative sign from the integration on right hand side of the equation
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = - \int\limits_0^2 {({x^2} - 4)dx} + \int\limits_2^3 {({x^2} - 4)dx}$
Now we use the general integration formula i.e.$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = - \left[ {\dfrac{{{x^{2 + 1}}}}{{2 + 1}} - 4x} \right]_0^2 + \left[ {\dfrac{{{x^{2 + 1}}}}{{2 + 1}} - 4x} \right]_2^3$
Add the terms in power on right hand side of the equation
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = - \left[ {\dfrac{{{x^3}}}{3} - 4x} \right]_0^2 + \left[ {\dfrac{{{x^3}}}{3} - 4x} \right]_2^3$
Apply limits on both brackets i.e. substitute the value of x as limits.
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = - \left[ {\left( {\dfrac{{{2^3}}}{3} - 4 \times 2} \right) - \left( {\dfrac{{{0^3}}}{3} - 4 \times 0} \right)} \right] + \left[ {\left( {\dfrac{{{3^3}}}{3} - 4 \times 3} \right) - \left( {\dfrac{{{2^3}}}{3} - 4 \times 2} \right)} \right]$
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = - \left( {\dfrac{{{2^3}}}{3} - 8} \right) + \left( {\dfrac{{{3^3}}}{3} - 12} \right) - \left( {\dfrac{{{2^3}}}{3} - 8} \right)$
Calculate cube values on right hand side of the equation
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = - \left( {\dfrac{8}{3} - 8} \right) + \left( {\dfrac{{27}}{3} - 12} \right) - \left( {\dfrac{8}{3} - 8} \right)$
Cancel possible terms from numerator and denominator on right hand side of the equation
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = - \left( {\dfrac{8}{3} - 8} \right) + \left( {9 - 12} \right) - \left( {\dfrac{8}{3} - 8} \right)$
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = - \left( {\dfrac{8}{3} - 8} \right) + ( - 3) - \left( {\dfrac{8}{3} - 8} \right)$
Open all the brackets on right hand side of the equation
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = - \dfrac{8}{3} + 8 - 3 - \dfrac{8}{3} + 8$
Add the positive values
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = 16 - \left( {\dfrac{8}{3} + 3 + \dfrac{8}{3}} \right)$
Take LCM of negative terms on right hand side of the equation
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = 16 - \left( {\dfrac{{8 + 8 + 9}}{3}} \right)$
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = 16 - \left( {\dfrac{{8 + 8 + 9}}{3}} \right)$
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = 16 - \dfrac{{25}}{3}$
Take LCM of terms on right hand side of the equation
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = \dfrac{{48 - 25}}{3}$
$\Rightarrow \int\limits_0^3 {\left| {{x^2} - 4} \right|dx} = \dfrac{{23}}{3}$

$\therefore$The value of definite integral $\int {\left| {{x^2} - 4} \right|} dx$ from $\left[ {0,3} \right]$ is $\dfrac{{23}}{3}$.

Note: Many students make the mistake of ignoring the modulus sign and directly solve the integral which ends up giving the value -3 which is wrong. Keep in mind we break the interval according to the modulus function and then apply integration to it. Also, when adding or subtracting the terms in the end, always apply the BODMAS rule and collect addition and subtraction separately.