Proving two integral inequalities Can anyone help me to prove that these integral inequalities hold?
Here $x$ is a real value:
$$
\left| \int_a^b\ f(x) dx \right| \leq \int_a^b\ |f(x)| dx
$$
Here $z$ is a complex value:
$$
\left| \int_C^ \ f(z) dz \right| \leq \int_C^\ |f(z)| |dz|
$$
 A: Notice by the triangle inequality
$$ \Big|\sum_{i=1}^n f(\xi_i)\Big|\Delta x_i\ \leq \sum_{i=1}^n |f(\xi_i)|\Delta x_i $$ Now when applying limits as $\Delta x_i$ go to $0$, then we obtain the first part of your question.
A: Let's denote
$$
f_+(x) = \max\{0, f(x)\}\\
f_-(x) = \max\{0, -f(x)\}
$$
We have
$$
f(x) = f_+(x) - f_-(x)\\
\lvert f(x) \rvert = f_+(x) + f_-(x)\\
\int_a^b f_{\pm} (x) dx \geq 0
$$
and so
$$
\left\lvert \int_a^b f(x)dx \right\rvert = \left\lvert \int_a^b f_+(x)dx - \int_a^b f_-(x)dx\right\rvert \leq\\
\left\lvert \int_a^b f_+(x)dx \right\rvert + \left\lvert\int_a^b f_-(x)dx\right\rvert =\\
\int_a^b f_+(x)dx + \int_a^b f_-(x)dx = \\
 \int_a^b (f_+(x) + f_-(x))dx = \int_a^b\lvert f(x)\rvert dx\\  
$$
A: Hint for the 2nd integration inequality:
Set $\overline{w}:=\int f(z) dz$ and have a look at $\int \overline{w} f(z) dz$. What happens to the lhs? Write the rhs as a double integral.
Hope this helps.

Complete proof of the 2nd inequality:
We know
$\displaystyle \int f(z) dz = \int \text{Re } f(x) dx + i \int \text{Im } f(x) dx$
Set $\overline{w}:=\int f(z) dz$. Then we get $|w|=|\overline{w}|=\left| \int f(x) dx \right|$ and further more
$$\overline{w} \int f(x) dx= \left( \int Re f(x) dx + i \int Im f(x) dx \right)\left( \int Re f(x) dx + i \int Im f(x) dx \right)=\left| \int f(x) \right|^2$$ and so $\displaystyle \overline{w} \int f(x) dx \in \mathbb{R}^+$. We finally have
$$\left| \int f(x) \right|^2 =
 \overline{w} \int f(x) dx =
 Re \left( \int \overline{w} f(x) dx \right) =
 \int Re(\overline{w}f(x))dx \leq\\
\int |\overline{w}f(x)|dx=
\int |\overline{w}||f(x)|dx=
|\overline{w}| \int |f(x)| dx
=\left| \int f(x) dx \right| \int |f(x)| dx \iff\\ \left| \int f(x) dx \right|  \leq
 \int |f(x)| dx$$
I hope I didn't mixed up some absolute values.
A: *

*Proving the first inequality


Suppose, WLOG, that 
$$\int_{a}^{b}f(x)dx \neq 0$$
Define
$$ \lambda = \frac{\left| \int_{a}^{b}f(x)dx \right|}{\int_{a}^{b}f(x)dx} $$

Remark 1: $\lambda \in \{\pm 1,\pm i\}$.
Remark 2: Integrals are homogeneous, i.e.,
$$\forall a\in \mathbb{C} :\int_{a}^{b}a f(x)dx = a \int_{a}^{b}f(x)dx$$
Remark 3: $\forall a\in \mathbb{C}: |Re(a)|\leq |a|$

So, we have
$$\mathbb{R^+} \ni \left|\int_{a}^{b}f(x)dx\right| = \lambda \int_{a}^{b}f(x)dx = \int_{a}^{b} \lambda f(x)dx = \int_{a}^{b}Re\{\lambda f(x)\}dx \leq \int_{a}^{b}|Re\{\lambda f(x)\}|dx \leq \int_{a}^{b}|\lambda f(x)|dx \leq |\lambda|\int_{a}^{b} |f(x)|dx$$
Finally,
$$ \left|\int_{a}^{b}f(x)dx\right| \leq \int_{a}^{b}\left|f(x)\right| dx $$


*Proving the second inequality:


$$\left| \int_{\gamma}f(z)dz \right| = \left| \int_{a}^{b}f(\gamma(t)) \gamma ' (t) dt \right| \leq \int_{a}^{b} \left| f(\gamma(t)) \gamma ' (t)\right| dt =  \int_{a}^{b} \left| f(\gamma(t))\right| \left| \gamma ' (t)\right| dt = \int_{\gamma}\left|f(z)\right||dz|$$
