
What is the integration of \[\bmod x\] ?
Answer
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Hint: Here in this question, we have to find the integral value of \[\bmod x\] i.e., \[\left| x \right|\] . To solve this first we have to define \[\left| x \right|\] , next integrate by using the substitution method and further simplify by using some standard modulo properties. The final solution of this question will be written by using signum function.
Complete step-by-step answer:
The \[\bmod x\] \[\left( {\left| x \right|} \right)\] depends on the domain in which it is being integrated.
For \[x > 0\] , \[\left| x \right| = x\]
For \[x < 0\] , \[\left| x \right| = - x\]
Consider the given question: integral of \[\bmod x\]
i.e., taking integration to \[\bmod x\] with respect to \[x\] , we have
\[ \Rightarrow \,\,\int {\bmod x} \,dx\]
Or
\[ \Rightarrow \,\,\int {\left| x \right|} \,dx\] --(1)
This can be integrated by two cases.
Complete step-by-step answer:
The \[\bmod x\] \[\left( {\left| x \right|} \right)\] depends on the domain in which it is being integrated.
For \[x > 0\] , \[\left| x \right| = x\]
For \[x < 0\] , \[\left| x \right| = - x\]
Consider the given question: integral of \[\bmod x\]
i.e., taking integration to \[\bmod x\] with respect to \[x\] , we have
\[ \Rightarrow \,\,\int {\bmod x} \,dx\]
Or
\[ \Rightarrow \,\,\int {\left| x \right|} \,dx\] --(1)
This can be integrated by two cases.
Case 1:
Let \[x = - \alpha \] where \[ - \alpha < 0\] --(2)
Then, \[dx = - d\alpha \]
On substituting in equation (1), we have
\[ \Rightarrow \int {\left| { - \alpha } \right|} \,d( - \alpha )\]
\[ \Rightarrow \int \alpha \,\left( { - d\alpha } \right)\]
On by sign convention, we get
\[ \Rightarrow \int { - \alpha \,d\alpha } \]
\[ \Rightarrow - \int {\alpha \,d\alpha } \]
By using integral formula \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C\] we get
\[ \Rightarrow \,\, - \dfrac{{{\alpha ^2}}}{2} + C\] ---(3)
where C is integral constant.
From equation (2), \[x = - \alpha \Rightarrow \alpha = - x\] then, equation (3) becomes
\[ \Rightarrow \,\, - \dfrac{{{{\left( { - x} \right)}^2}}}{2} + C\]
\[ \Rightarrow \,\, - \dfrac{{{x^2}}}{2} + C\]
Therefore, \[ \int {\left| x \right|} \,dx = - \dfrac{{{x^2}}}{2} + C\] ---(4)
Let \[x = - \alpha \] where \[ - \alpha < 0\] --(2)
Then, \[dx = - d\alpha \]
On substituting in equation (1), we have
\[ \Rightarrow \int {\left| { - \alpha } \right|} \,d( - \alpha )\]
\[ \Rightarrow \int \alpha \,\left( { - d\alpha } \right)\]
On by sign convention, we get
\[ \Rightarrow \int { - \alpha \,d\alpha } \]
\[ \Rightarrow - \int {\alpha \,d\alpha } \]
By using integral formula \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C\] we get
\[ \Rightarrow \,\, - \dfrac{{{\alpha ^2}}}{2} + C\] ---(3)
where C is integral constant.
From equation (2), \[x = - \alpha \Rightarrow \alpha = - x\] then, equation (3) becomes
\[ \Rightarrow \,\, - \dfrac{{{{\left( { - x} \right)}^2}}}{2} + C\]
\[ \Rightarrow \,\, - \dfrac{{{x^2}}}{2} + C\]
Therefore, \[ \int {\left| x \right|} \,dx = - \dfrac{{{x^2}}}{2} + C\] ---(4)
Case 2:
Now let consider \[x = \beta \] where \[\beta > 0\] ……………… (5)
Then, \[dx = d\beta \]
On substituting in equation (1), we have
\[ \Rightarrow \,\,\int {\left| \beta \right|} \,d(\beta )\]
\[ \Rightarrow \,\,\int {\beta \,d\beta } \]
Again, by using integral formula \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C\] we get
\[ \Rightarrow \,\,\dfrac{{{\beta ^2}}}{2} + C\] ---(6)
where C is integral constant.
From equation (5), \[x = \beta \] then, equation (6) becomes
\[ \Rightarrow \,\,\dfrac{{{x^2}}}{2} + C\]
Therefore, \[ \int {\left| x \right|} \,dx = \dfrac{{{x^2}}}{2} + C\] --(7)
We know that the sign function or signum function is an odd mathematical function that extracts the sign of a real number. In mathematical expressions the sign function is represented as \[\operatorname{sgn} \] .
By using sign function, we can write the integral as
\[ \Rightarrow \,\,\int {\left| x \right|} \,dx = \operatorname{sgn} (x)\dfrac{{{x^2}}}{2} + C\]
Therefore, the correct answer is “ \[ \int {\left| x \right|} \,dx = \operatorname{sgn} (x)\dfrac{{{x^2}}}{2} + C\] ”.
Note: By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So, simplification is needed. We must know the differentiation and integration formulas. And should know the definition of absolute value or mod function.
Now let consider \[x = \beta \] where \[\beta > 0\] ……………… (5)
Then, \[dx = d\beta \]
On substituting in equation (1), we have
\[ \Rightarrow \,\,\int {\left| \beta \right|} \,d(\beta )\]
\[ \Rightarrow \,\,\int {\beta \,d\beta } \]
Again, by using integral formula \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C\] we get
\[ \Rightarrow \,\,\dfrac{{{\beta ^2}}}{2} + C\] ---(6)
where C is integral constant.
From equation (5), \[x = \beta \] then, equation (6) becomes
\[ \Rightarrow \,\,\dfrac{{{x^2}}}{2} + C\]
Therefore, \[ \int {\left| x \right|} \,dx = \dfrac{{{x^2}}}{2} + C\] --(7)
We know that the sign function or signum function is an odd mathematical function that extracts the sign of a real number. In mathematical expressions the sign function is represented as \[\operatorname{sgn} \] .
By using sign function, we can write the integral as
\[ \Rightarrow \,\,\int {\left| x \right|} \,dx = \operatorname{sgn} (x)\dfrac{{{x^2}}}{2} + C\]
Therefore, the correct answer is “ \[ \int {\left| x \right|} \,dx = \operatorname{sgn} (x)\dfrac{{{x^2}}}{2} + C\] ”.
Note: By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So, simplification is needed. We must know the differentiation and integration formulas. And should know the definition of absolute value or mod function.
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