Title states it all. Is there a closed type solution because that the indefinite integral $\\int |x| dx$ ?
Using integration by parts
$$\\int |x|~dx=\\int \\inter-base.netrmsgn(x)x~dx=|x|x-\\int |x| ~dx$$
since $\\fracddx |x|=\\inter-base.netrmsgn(x)$ top top non-zero sets. This yields
$$\\int |x| ~dx = \\frac2~.$$
You are trying to find a role $f(x)$ so the $$\\int_a^b |x|dx=f(b)-f(a).$$ This is what is supposed by $\\int |x|dx$. Ns propose that $f(x)=x|x|/2$ is together a function. Let us test it. If both $a$ and $b$ space 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 $b$ space 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 might have $b and $a>0$, but then we might switch the limits, and also this reduces come the third case.
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Thus, $f(x)=x|x|/2$ is an indefinite integral, or antiderivative that $|x|$.
You deserve to use $\\frac)dx=\\fracx$ and also $\\int|x|dx = \\int \\fracxxdx$.
$\\int|x|dx = \\int xd(|x|)$, utilizing integration by components $\\int|x|dx = x|x| - \\int|x|dx $
$2\\int|x|dx = x|x|$
$\\int|x|dx = \\fracx2$
$\\fracx$ is a far better way to define the authorize function.
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