
What is the value of \[\int\limits_{ - 2}^2 {\left| x \right|dx} \]?
A. 0
B. 1
C. 2
D. 4
Answer
162.9k+ views
Hint: First we will find where the \[\left| x \right|\] is positive and negative in the interval \[ - 2 < x < 2\]. Then we will break the interval of the integration according to the positive and negative of \[\left| x \right|\]. Then solve it to get solution.
Formula Used:Absolute value function:
\[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x}&{{\rm{if }}\,x < 0}\\x&{{\rm{if }}x > 0}\end{array}} \right.\]
Integration formula:
\[\int {xdx} = \dfrac{{{x^2}}}{2} + c\]
Complete step by step solution:Given integration is
\[\int\limits_{ - 2}^2 {\left| x \right|dx} \]
We know that, \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x}&{{\rm{if }}\,x < 0}\\x&{{\rm{if }}x > 0}\end{array}} \right.\]
Rewrite the equation when \[ - 2 < x < 2\]
\[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x}&{ - 2 < \,x < 0}\\x&{0 < x < 0}\end{array}} \right.\]
Now we will break the interval of the integration into two intervals those are -2 to 0 and 0 to 2.
Therefore, \[\int\limits_{ - 2}^2 {\left| x \right|dx} \]
\[ = \int\limits_{ - 2}^0 {\left( { - x} \right)dx} + \int\limits_0^2 {xdx} \]
Now applying integration formula:
\[ = \left[ { - \dfrac{{{x^2}}}{2}} \right]_{ - 2}^0 + \left[ {\dfrac{{{x^2}}}{2}} \right]_0^2\]
Now putting lower limit and upper limit:
\[ = \left[ { - \dfrac{{{0^2}}}{2} + \dfrac{{{2^2}}}{2}} \right] + \left[ {\dfrac{{{2^2}}}{2} - \dfrac{{{0^2}}}{2}} \right]\]
Simplify the above expression:
\[ = \left[ {0 + 2} \right] + \left[ {2 - 0} \right]\]
= 4
Option ‘D’ is correct
Additional Information:An absolute value function is a function that contains an expression in absolute value symbol that is ||.
Note: Students often make mistakes when they integrate an absolute function. They don’t consider the positive value and negative value of \[\left| x \right|\]. They take only positive value and solve it. To integrate an absolute function we have to follow the condition for absolute function.
Formula Used:Absolute value function:
\[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x}&{{\rm{if }}\,x < 0}\\x&{{\rm{if }}x > 0}\end{array}} \right.\]
Integration formula:
\[\int {xdx} = \dfrac{{{x^2}}}{2} + c\]
Complete step by step solution:Given integration is
\[\int\limits_{ - 2}^2 {\left| x \right|dx} \]
We know that, \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x}&{{\rm{if }}\,x < 0}\\x&{{\rm{if }}x > 0}\end{array}} \right.\]
Rewrite the equation when \[ - 2 < x < 2\]
\[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{ - x}&{ - 2 < \,x < 0}\\x&{0 < x < 0}\end{array}} \right.\]
Now we will break the interval of the integration into two intervals those are -2 to 0 and 0 to 2.
Therefore, \[\int\limits_{ - 2}^2 {\left| x \right|dx} \]
\[ = \int\limits_{ - 2}^0 {\left( { - x} \right)dx} + \int\limits_0^2 {xdx} \]
Now applying integration formula:
\[ = \left[ { - \dfrac{{{x^2}}}{2}} \right]_{ - 2}^0 + \left[ {\dfrac{{{x^2}}}{2}} \right]_0^2\]
Now putting lower limit and upper limit:
\[ = \left[ { - \dfrac{{{0^2}}}{2} + \dfrac{{{2^2}}}{2}} \right] + \left[ {\dfrac{{{2^2}}}{2} - \dfrac{{{0^2}}}{2}} \right]\]
Simplify the above expression:
\[ = \left[ {0 + 2} \right] + \left[ {2 - 0} \right]\]
= 4
Option ‘D’ is correct
Additional Information:An absolute value function is a function that contains an expression in absolute value symbol that is ||.
Note: Students often make mistakes when they integrate an absolute function. They don’t consider the positive value and negative value of \[\left| x \right|\]. They take only positive value and solve it. To integrate an absolute function we have to follow the condition for absolute function.
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