Been searching the net for awhile and everything just comes back about doing the definite integral. So just thought to ask here.

Title says it all. Is there a closed form solution for the indefinite integral \$int |x| dx\$ ? Using integration by parts

\$\$int |x|~dx=int hedailysplash.tvrmsgn(x)x~dx=|x|x-int |x| ~dx\$\$

since \$fracddx |x|= hedailysplash.tvrmsgn(x)\$ on non-zero sets. This yields

\$\$int |x| ~dx = fracx2~.\$\$ You are looking for a function \$f(x)\$ so that \$\$int_a^b |x|dx=f(b)-f(a).\$\$ This is what is meant by \$int |x|dx\$. I propose that \$f(x)=x|x|/2\$ is such a function. Let us test it. If both \$a\$ and \$b\$ are both positive, then \$\$int_a^b |x|dx=int_a^b x,dx=b^2/2-a^2/2=b|b|/2-a|a|/2=f(b)-f(a).\$\$If \$a\$ and \$b\$ are both negative, then \$\$int_a^b |x|dx=-int_a^b x,dx=-b^2/2-(-a^2/2)=b|b|/2-a|a|/2=f(b)-f(a).\$\$Finally, if \$a and \$b>0\$, we get\$\$int_a^b |x|dx=-int_a^0 x,dx+int_0^b x,dx=b^2/2+a^2/2=b|b|/2-a|a|/2=f(b)-f(a).\$\$Of course, we could have \$b and \$a>0\$, but then we could switch the limits, and this reduces to the third case.

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Thus, \$f(x)=x|x|/2\$ is an indefinite integral, or antiderivative of \$|x|\$. You can use \$frac)dx=fracx\$ and \$int|x|dx = int fracxxdx\$.

\$int|x|dx = int xd(|x|)\$, using integration by parts \$int|x|dx = x|x| - int|x|dx \$

\$2int|x|dx = x|x|\$

\$int|x|dx = fracx2\$

\$fracxx\$ is a better way to define the sign function. Thanks for contributing an answer to thedailysplash.tvematics Stack Exchange!

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Asking for help, clarification, or responding to other answers.Making statements based on opinion; back them up with references or personal experience.

Use thedailysplash.tvJax to format equations. thedailysplash.tvJax reference.

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