When I looked up about absolute value on Wikipedia, I found that the antiderivative of $|x|$ is $\frac12 x|x|+C$. I am able to find the derivative of $|x|$ by treating the function as $\sqrt{x^2}$, but I am not able to integrate it.

When I put $\int_{-4}^{-1}|x|\,dx$ into Symbolab, the online calculator did not break the integral into piecewise function but calculate the indefinite integral first before using $F(b) - F(a)$. When I view the steps it used, it said:

If $\int f(x)\,dx = F(x)$ then $$\int \sqrt{(f(x))^2)}\,dx = \frac{\sqrt{f(x)^2}}{f(x)}$$ multiplied to $F(x)$ which becomes $\frac{\sqrt{x^2}}{x}$ multiplied to $\int x\,dx$

  • 4
    $\begingroup$ Forget online calculators and think. The function has basically two different expressions on two intervals, treat them separately. $\endgroup$ – Did Dec 4 '14 at 6:52
  • $\begingroup$ But unlike $\int (1/x)\,dx$, the constant in this case should be the same for the two parts. $\endgroup$ – GEdgar Dec 4 '14 at 15:08

The function ${\rm abs}$ is continuous on all of ${\mathbb R}$, hence should have primitives $F$ defined on all of ${\mathbb R}$. Given that ${\rm abs}$ is "special" at $x=0$ we should look for the primitive obtained by integrating from $0$ to $x$. In this way we obtain $$F(x)=\int_0^x |t|\>dt=\int_0^x t\>dt={x^2\over2}\qquad(x\geq0)$$ and $$F(x)=\int_0^x |t|\>dt=\int_0^x (-t)\>dt=-{x^2\over2}\qquad(x\leq0)\ .$$ The two partial results can be condensed into the single formula $$F(x)={x\>|x|\over 2}\qquad(-\infty<x<\infty)\ ,$$ and adding an arbitrary constant $C$ gives the general primitive of ${\rm abs}$.

  • $\begingroup$ Why can't we just write $|x|^2/2$ as the integral and integrate directly with the given limits, instead of writing two expressions and breaking up the limits? $\endgroup$ – HeWhoMustBeNamed Sep 26 '18 at 18:31
  • 2
    $\begingroup$ @MrReality: Because it is wrong. One has $|x|^2/2=x^2/2$, and the derivative of this is $x$, not $|x|$. $\endgroup$ – Christian Blatter Sep 26 '18 at 18:47

Here is my derivation for this,

$$\int |x|dx= \int \sqrt {x^2} dx = I$$

By integration by parts we know that,

$$ \int u(x)v(x)\,dx=u(x)\int v(x) \ dx-\int u'(x) \left [ \int v(x) \ dx \right ]dx$$ This is just a extension of the product rule in diffrentiation. Check out the wiki page.

So taking $u(x) = |x|$ and $v(x) =1$,

$$I = \int |x| \times 1 \ \ dx = |x| \int 1\ dx - \int \left [ \frac{d(|x|) }{dx}\int 1 \ dx \right ]dx $$

Now we can differentiate the absolute value of $x$ using chain rule,

$$\frac{d(|x|) }{dx} = \frac{d(\sqrt {x^2}) }{dx} = \frac{1}{2\sqrt{x^2}} (2x)= \frac{x}{\sqrt{x^2}} $$

Trivially we can say $ \int 1 \ dx = x$.

Substituting this in $I$,

$$I = |x|x \ - \ \int\frac{x}{\sqrt{x^2}} \ x \ dx$$

$$I = |x|x \ - \ \int\frac{x^2}{\sqrt{x^2}} \ dx$$

Because both $x^2$ and $\sqrt{x^2}$ are positive, we can rewrite this as,

$$I = |x|x \ - \ \int{\sqrt{x^2}} \ dx =|x|x \ - I $$

So as we have the same integral in the RHS, we take it to the LHS.

$$2I = |x|x$$

So we can conclude,

$$I(x) = \frac{x|x|}{2}$$

Try this out for yourself the area under the $|x|$ curve from $x=a$ to $x=b$ can be expressed as $I(b) - I(a)$.

The antiderivative of $|x|$ is a function $g(x)$ such that $g'(x) = |x|$. Note that for any value of $C$, $g(x) + C$ can also be such an antiderivative. So we add a Constant of integration.


In case the integration by parts formula I gave above is confusing, consider

$$\frac{d(a(x) \ b(x)) }{dx} = a'(x) \ b(x) + a(x) \ b'(x)$$

Integrating both sides,

$$\int \frac{d(a(x) \ b(x)) }{dx} = a(x) \ b(x) = \int a'(x) \ b(x) + \int a(x) \ b'(x)$$

If we substitute $a(x) = u(x)$ and $b'(x) = v(x)$, so $b(x) = \int v(x)$ and $a'(x) = u'(x)$

$$u(x)\int v(x) = \int u'(x) \left [ \int v(x) \ dx \right ]dx + \int u(x) \ v(x)$$

Thus we get the formula for integration by parts shown above.


enter image description here

∫|x|dx=∫〖√(x^2 ) dx〗=

=x√(x^2 )-∫〖x/√(x^2 ) dx〗


∫〖√(x^2 ) dx〗=x√(x^2 )-∫〖√(x^2 ) dx〗

∫〖√(x^2 ) dx〗=x√(x^2 )-∫〖√(x^2 ) dx〗


2∫〖√(x^2 ) dx〗=x√(x^2 )=x|x|

Finally we will have:

  ∫〖√(x^2 ) dx〗=x|x|/2

I can not put the formula correctly, If you guide how to do the job I will provide the answer in a better view

  • $\begingroup$ This seems like it could be a good contribution. If you want to learn how to format your answer correctly, you could go here for a tutorial and tips. Otherwise, perhaps some other user will come along and format it for you, but this is an old question and already answered. $\endgroup$ – 6005 Aug 11 '16 at 5:18
  • $\begingroup$ Please see how to use latex commands for a better format. $\endgroup$ – Babai Aug 11 '16 at 5:44







  • 1
    $\begingroup$ How do you justify the first equality? It's not immediately clear that the expression in the second line even makes sense. $\endgroup$ – Michael Albanese Dec 8 '14 at 16:59
  • $\begingroup$ Looks like integration by parts, but i'm not sure this is justified here. For this integral specifically, replace $abs(x)$ with $-x$ since $x$ lies in negative numbers only. $\endgroup$ – JacksonFitzsimmons Jan 13 '16 at 12:25
  • $\begingroup$ The calculation works (even if you replace bounds with $-4$ to $1$, so there must be some way to justify the $d(|x|)$. While I have seen this sort of notation before, I'm not familiar with the justification. And mathSE questions (1, 2) haven't really done justice to it. $\endgroup$ – 6005 Aug 11 '16 at 5:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.