Let $\phi(t) = \phi_1(t)+i\phi_2(t)$ be a complex function in a real interval $t$  In order to prove: $$\left|\int_a^b \phi(t) \ dt \right|\le \int_a^b |\phi(t)| \ dt$$  I tried: $$\int_a^b \phi(t) \ dt  = \int_a^b \phi_1(t) + i\phi_2(t) \ dt  = \int_a^b \phi_1(t) \ dt + \int_a^b \phi_2(t) \ dt \rightarrow$$  $$\left|\int_a^b \phi(t) \ dt \right| = \sqrt{\left(\int_a^b \phi_1(t) \ dt\right)^2+\left(\int_a^b \phi_2(t) \ dt\right)^2}\le \sqrt{\left(\int_a^b \phi_1(t) \ dt\right)^2}+\sqrt{\left(\int_a^b \phi_2(t) \ dt\right)^2} = $$  $$\left|\int_a^b \phi_1(t) \ dt\right|+\left|\int_a^b \phi_2(t) \ dt\right|\le \int_ a^b |\phi_1(t)| \ dt + \int_a^b |\phi_2(t)| \ dt = \int_a^b |\phi_1(t)|+|\phi_2(t)| \ dt$$  But how to arrive at $\int_a^b |\phi(t)| \ dt$?
I've been given a proof that shows the following:  If $f:[a,b]\to \mathbb C$ is a continuous function and $f(t)=u(t)+iv(t)$ then $$\left| \int_a^b f(t) dt \right| \leq \int_a^b \left| f(t) \right| dt$$ The proof begins by letting $\theta$ be the principle argument of the complex number $\int_a^b f(t)dt$ and there is one step in the proof I don't understand; can anyone explain to me why $$\int_a^b e^{-i\theta}f(t) dt=\Re\bigg(\int_a^be^{-i\theta}f(t)dt\bigg)$$  Would this not imply that the imaginary part of the left hand side is equal to $0$? I'm not sure why the LHS and RHS would be equal here.
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