What is value of integral |x|/x?

[vkash]

What is value of integral |x|/x. (x is not equal to zero)
For x>0 this integral became integral of 1 that is equal to x.
for x<0 this integral became integral of -1 that is equal to -x.
for x>0 it is x & for x<0 it is-x so it should mode |x|.

is it correct or incorrect?

NOTE: here |x| means modulus of x.

 

[Dickfore]

Use:
$$\frac{\vert x \vert}{x} = \left\lbrace \begin{array}{ccl} 1&,&x > 0 \\ -1&,&x < 0 \end{array}\right.$$

Then, the primitive function of 1 is x, and of -1 is -x. If you look at the conditions where each case holds, and compare it to the definition of $\vert x \vert$, what should you get?

 

[HallsofIvy]

Or draw a graph- horizontal lines. The integral can be interpreted as "area under the graph" and the graph is either one rectangle or two rectangles depending upon where you start and where you end. And the area of a rectangle of height 1 and width x is just x.

If x<0 then
$$\int_0^x\frac{|x|}{x}dx= -x$$
(which is positive, of course, because x<0)
and if x>0 then
$$\int_0^x\frac{|x|}{x}dx= x$$

If a< 0 and 0< b, then
$$\int_a^b\frac{|x|}{x}dx= \int_a^0\frac{|x|}{x}dx+\int_0^b\frac{|x|}{x}dx$$
$$= \int_0^a \frac{|x|}{x}dx+ \int_0^b\frac{|x|}{x}dx= - (-a)+ b= a+ b.$$

(Note that is legal to integrate across x= 0 because the itegral is a "smoothing" operation.)