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Question

Question: What is the integral of \[x\left( {\bmod x} \right)\]? Here mod is modulus of x....

What is the integral of x(modx)x\left( {\bmod x} \right)? Here mod is modulus of x.

Explanation

Solution

Here in this question, we have to find the integral value of x(modx)x\left( {\bmod x} \right) i.e., xxx\left| x \right|. To solve this first we have to define xx and x\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 solution:
The \bmod x$$$$\left( {\left| x \right|} \right)depends on the domain in which it is being integrated.
For x>0x > 0, x=x\left| x \right| = x
For x<0x < 0, x=x\left| x \right| = - x
Consider the given question: integral of x(modx)x\left( {\bmod x} \right)
i.e., taking integration to x(modx)x\left( {\bmod x} \right) with respect to xx, we have
x(modx)dx\Rightarrow \,\,\int {x\left( {\bmod x} \right)} \,dx
Or
xxdx\Rightarrow \,\,\int {x\left| x \right|} \,dx--------(1)
This can be integrated by two cases.
Case 1:
Let x=αx = - \alpha where α<0 - \alpha < 0--------(2)
Then, dx=dαdx = - d\alpha
On substituting in equation (1), we have
ααd(α)\Rightarrow \int { - \alpha \left| { - \alpha } \right|} \,d( - \alpha )
αα(dα)\Rightarrow \int { - \alpha \cdot \alpha } \,\left( { - d\alpha } \right)
On by sign convention, we get
α2dα\Rightarrow \int {{\alpha ^2}d\alpha }
By using integral formula xndx=xn+1n+1+C\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C we get
α33+C\Rightarrow \,\,\dfrac{{{\alpha ^3}}}{3} + C ---------(3)
where C is integral constant.
From equation (2), x=αα=xx = - \alpha \Rightarrow \alpha = - x then, equation (3) becomes
(x)33+C\Rightarrow \,\,\dfrac{{{{\left( { - x} \right)}^3}}}{3} + C
x33+C\Rightarrow \,\, - \dfrac{{{x^3}}}{3} + C
Therefore, xxdx=x33+C \Rightarrow \,\,\int {x\left| x \right|} \,dx = - \dfrac{{{x^3}}}{3} + C-----------(4)
Case 2:
Now let consider x=βx = \beta where β>0\beta > 0 ……………… (5)
Then, dx=dβdx = d\beta
On substituting in equation (1), we have
ββd(β)\Rightarrow \,\,\int {\beta \left| \beta \right|} \,d(\beta )
β2dβ\Rightarrow \,\,\int {{\beta ^2}\,d\beta }
Again, by using integral formula xndx=xn+1n+1+C\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C we get
β33+C\Rightarrow \,\,\dfrac{{{\beta ^3}}}{3} + C -----------(6)
where C is integral constant.
From equation (5), x=βx = \beta then, equation (6) becomes
x33+C\Rightarrow \,\,\dfrac{{{x^3}}}{3} + C
Therefore, xxdx=x33+C \Rightarrow \,\,\int {x\left| x \right|} \,dx = \dfrac{{{x^3}}}{3} + 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 sgn\operatorname{sgn} .
By using sign function, we can write the integral as
xxdx=sgn(x)x33+C \Rightarrow \,\,\int {x\left| x \right|} \,dx = \operatorname{sgn} (x)\dfrac{{{x^3}}}{3} + C. Hence the solution.

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.