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In I proof I saw this inequality

$$ \int_0^x u'(y) \, dy \leq \int_0^x |u'(y)| \, dy $$

Why is this true? Is it also true without the derivative? Shouldn't it be a absolute value sign on the left hand side?

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    Yes, this is true without the derivative. This is because $|u(y)| - u(y) \ge 0$, and so $\int_{0}^{x} |u(y)| - u(y) \ge 0$. – 2017-02-16
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    Note that if $x$ is real, then $x \le |x|$. – 2017-02-16

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Hint:

If $f,g$ are Riemann Integrable on $[a,b]$ and $f \leq g$ on $[a,b]$

then $$\int_{a}^{b} f(x) \ dx \leq \int_{a}^{b} g(x) \ dx.$$

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    So $f(x) \le |f(x)|$ – 2017-02-16
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    Thanks! I saw this inequality, but is it true in my case? $f$ and $g$ are two different functions but in my inequality it is the same function, $u$, on both side. – 2017-02-16
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    Well let us pick $y \in [0,x].$ $u'(y)$ could perhaps be positive, 0, or even negative. But $|u'(y)|$ is always nonnegative. In the case $u'(y)<0,$ then $|u'(y)|>0,$ so $u'(y)<|u'(y)|.$ In the case $u'(y)=0,$ then $u'(y)=|u'(y)|=0,$ and in the case $u'(y)>0,$ we have $u'(y)=|u'(y)|>0.$ Thus, in any case $u'(y) \leq |u'(y)|$ and assuming Riemann integrability of both, what you say is true. – 2017-02-16
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Geometrically speaking, you can see the integral as the positive/negative area between the curve of a function and the x-axis.

So, given $f$, the integral of $|f|$ will be greater (in the same interval).