Remember that \(\displaystyle{F}\cdot{d}{s}={F}\cdot\hat{{n}}{d}{S}{w}{h}{e}{r}{e}\hat{{n}}\) is the unit outward normal vector. Luckily that vector will be constant across each of the 6 faces of the cube. So we just need to break this integral into 6 parts:

\(\displaystyle\int_{{S}}{F}\cdot{d}{S}=\int_{{\top}}{F}\cdot\hat{{n}}_{{\top}}{d}{S}+\int_{{\le{f}{t}}}{F}\cdot\hat{{n}}_{{\le{f}{t}}}{d}{S}+\int_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{F}\cdot\hat{{n}}_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{d}{S}+\ldots\)

Let's start with the top side of the cube. Along that face, the outward normal points up. I.e.\(\displaystyle\hat{{n}}={k}\).

Also along the top face we have \(\displaystyle-{1}\le{x}\le{1},-{1}\le{y}\le{1},{\quad\text{and}\quad}{z}={1}\).Hence we have

\(\displaystyle\int_{\top}{F}\cdot\hat{{n}}_{{\top}}={\int_{{{y}=-{1}}}^{{{y}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{\left[{x}{i}+{2}{y}{j}+{3}{\left({1}\right)}{k}\right]}\cdot{k}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{{y}=-{1}}}^{{{y}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{3}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={3}{\int_{{{y}=-{1}}}^{{{y}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle{3}{\left({2}\right)}{\int_{{{y}=-{1}}}^{{{y}={1}}}}{\left.{d}{y}\right.}\)

=3(2)(2)=12 Along the left face (pretend you're sitting somewhat out on the positive\(\displaystyle\times{a}\xi{s}{l}\infty{k}\in{g}{a}{t}{t}{h}{e}{c}{u}{b}{e}{)},{t}{h}{e}{o}{u}{t}{w}{a}{r}{d}{\left\|{a}\right\|}{l}{i}{s}-{j}{\quad\text{and}\quad}{t}{h}{e}{v}{a}{l}{u}{e}{s}{o}{f}{x},{y},{\quad\text{and}\quad}{z}{a}{r}{e}-{1}\le{x}\le{1},{y}=-{1},{\quad\text{and}\quad}-{1}\le{z}\le{1}\).Thus

\(\displaystyle\int_{{\le{f}{t}}}{F}\cdot\hat{{n}}_{{\le{f}{t}}}{d}{S}={\int_{{{z}=-{1}}}^{{{z}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{\left[{x}{i}+{2}{\left(-{1}\right)}{j}+{3}{z}{k}\right]}\cdot-{j}{\left.{d}{x}\right.}{\left.{d}{z}\right.}={8}\)

At this point hopefully you get the idea so I'm just going to quickly list the rest of the integrals and their values:

\(\displaystyle\int_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{F}\cdot\hat{{n}}_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{d}{S}={4}\)

\(\displaystyle\int_{{()}}{F}\cdot\hat{{n}}_{{()}}{d}{S}={8}\)

\(\displaystyle\int{\left({b}{a}{c}{k}\right)}{F}\cdot\hat{{n}}_{{{b}{a}{c}{k}}}{d}{S}={4}\)

\(\displaystyle\int{\left(\bot\to{m}\right)}{F}\cdot\hat{{n}}_{{\bot\to{m}}}{d}{S}={12}\)

So if the flux through each face is 12, 8, 4, 8, 4, and 12, respectively, then the total flux from the cube is

12+8+4+8+4+12=48

\(\displaystyle\int_{{S}}{F}\cdot{d}{S}=\int_{{\top}}{F}\cdot\hat{{n}}_{{\top}}{d}{S}+\int_{{\le{f}{t}}}{F}\cdot\hat{{n}}_{{\le{f}{t}}}{d}{S}+\int_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{F}\cdot\hat{{n}}_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{d}{S}+\ldots\)

Let's start with the top side of the cube. Along that face, the outward normal points up. I.e.\(\displaystyle\hat{{n}}={k}\).

Also along the top face we have \(\displaystyle-{1}\le{x}\le{1},-{1}\le{y}\le{1},{\quad\text{and}\quad}{z}={1}\).Hence we have

\(\displaystyle\int_{\top}{F}\cdot\hat{{n}}_{{\top}}={\int_{{{y}=-{1}}}^{{{y}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{\left[{x}{i}+{2}{y}{j}+{3}{\left({1}\right)}{k}\right]}\cdot{k}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{{y}=-{1}}}^{{{y}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{3}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={3}{\int_{{{y}=-{1}}}^{{{y}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle{3}{\left({2}\right)}{\int_{{{y}=-{1}}}^{{{y}={1}}}}{\left.{d}{y}\right.}\)

=3(2)(2)=12 Along the left face (pretend you're sitting somewhat out on the positive\(\displaystyle\times{a}\xi{s}{l}\infty{k}\in{g}{a}{t}{t}{h}{e}{c}{u}{b}{e}{)},{t}{h}{e}{o}{u}{t}{w}{a}{r}{d}{\left\|{a}\right\|}{l}{i}{s}-{j}{\quad\text{and}\quad}{t}{h}{e}{v}{a}{l}{u}{e}{s}{o}{f}{x},{y},{\quad\text{and}\quad}{z}{a}{r}{e}-{1}\le{x}\le{1},{y}=-{1},{\quad\text{and}\quad}-{1}\le{z}\le{1}\).Thus

\(\displaystyle\int_{{\le{f}{t}}}{F}\cdot\hat{{n}}_{{\le{f}{t}}}{d}{S}={\int_{{{z}=-{1}}}^{{{z}={1}}}}{\int_{{{x}=-{1}}}^{{{x}={1}}}}{\left[{x}{i}+{2}{\left(-{1}\right)}{j}+{3}{z}{k}\right]}\cdot-{j}{\left.{d}{x}\right.}{\left.{d}{z}\right.}={8}\)

At this point hopefully you get the idea so I'm just going to quickly list the rest of the integrals and their values:

\(\displaystyle\int_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{F}\cdot\hat{{n}}_{{{f}{\quad\text{or}\quad}{w}{a}{r}{d}}}{d}{S}={4}\)

\(\displaystyle\int_{{()}}{F}\cdot\hat{{n}}_{{()}}{d}{S}={8}\)

\(\displaystyle\int{\left({b}{a}{c}{k}\right)}{F}\cdot\hat{{n}}_{{{b}{a}{c}{k}}}{d}{S}={4}\)

\(\displaystyle\int{\left(\bot\to{m}\right)}{F}\cdot\hat{{n}}_{{\bot\to{m}}}{d}{S}={12}\)

So if the flux through each face is 12, 8, 4, 8, 4, and 12, respectively, then the total flux from the cube is

12+8+4+8+4+12=48