Verifying $W = \int \vec{F} \cdot d\vec{x}$

Question

Verifying formula for Work;

$$W = \int \vec{F} \cdot d\vec{x} \quad(i)$$

Let us consider a very simple scenario; I will solve first by simple Maths and then by calculus.

Scenario 1: (Force vector is always $5\vec{i} N$) [Solving by simple maths,]

A block of mass 2kg is initially at origin($\vec{x}=0\vec{i}$) is displaced to $\vec{x}=8m\vec{i}$

$W_{1}=\vec{F} \cdot \vec{s} \quad$ ; where $\quad \vec{s}=displacement \ vector$

$W_{1}=5*8* \cos(0^{\circ})=40 J$

$-------------------------------------------------$

Scenario 2: (Force vector is always $5\vec{i} N$) [Solving by simple maths,]

A block of mass 2kg is initially at $x=8m\vec{i}$ is displaced to origin ($\vec{x}=0\vec{i}$)

$W_{2}=\vec{F} \cdot \vec{s} $

$W_{2}=5*8* \cos(180^{\circ})=-40 J$

Work came negative as displacement and force are in opposite direction, here force is opposing in nature like frictional force.

$------------------------------------------------$

Now solving same above two scenarios using calculus $$W = \int \vec{F} \cdot d\vec{x} \quad(i)$$

Scenario 1C: (Force vector is always $5\vec{i} N$)

A block of mass 2kg is initially at origin($\vec{x}=0\vec{i}$) is displaced to $\vec{x}=8m\vec{i}$

$$W = \int \vec{F} \cdot d\vec{x} $$ $$W = \int^8_0 F dx\cos(0^{\circ}) $$ $$W_{1c}=F(8-0)=40$$

$-------------------------------------------------$

Scenario 2C: (Force vector is always $5\vec{i} N$)

A block of mass 2kg is initially at $\vec{x}=8m\vec{i}$ is displaced to origin($\vec{x}=0\vec{i}$).

$$W = \int \vec{F} \cdot d\vec{x} \quad(i)$$ $$W = \int^0_8 F dx\cos(180^{\circ}) $$ $$W_{2c}=-F(0-8)=40J$$

But, previously we obtained $W_2=-40J$ $-------------------------------------------------$

Why are answer for $W_{2}$ and $W_{2C}$ different though they represent same scenario?

This question is simplified form for confusion in Potential energies for Gravity and Electrostatics.

For Sign convention I read this also Potential energy sign conventions

Answer 1

$$\mathbf{F}\cdot d\mathbf{s}=|\mathbf{F}||d\mathbf{s}|\cos\theta$$ if $ds<0$, then $|d\mathbf{s}|=-ds$ ($d\mathbf{s}$ is a vector, notice the bolded $s$).
We suppose that $x$ will vary from $8$ to $0$, obviously $dx<0$ (because $x$ is decreasing). $$\begin{align*} \int_{\mathbf{s}_i}^{\mathbf{s}_f}\mathbf{F}\cdot d\mathbf{s}&=\int_{\mathbf{s}_i}^{\mathbf{s}_f}|\mathbf{F}||d\mathbf{s}|\cos(\pi)\\ &=\int_8^0-|\mathbf{F}|(-dx)\\ &=\int_8^0|\mathbf{F}|dx\\ &=-8|\mathbf{F}| \end{align*}$$ You can also use $(a,b,c)\cdot(d,e,f)=ad+be+cf$.
Here, $d\mathbf{s}=(dx,0,0)$, and its direction depends on the sign of $dx$ (to the left, since $dx<0$).

Answer 2

Your error is because your limits of integration don't agree with the sign convention for $\mathrm{d}x$ in 2C. When you do the work line integral from point '8' to point '0', the direction of $\mathrm{d}\mathbf{s}$ is in the direction $8\to0$, and is in the $-\mathbf{x}$ direction, which is the opposite direction as $\mathbf{F}$; hence, your equation should read:

$$W=\int_8^0F\mathrm{d}x\cos(0)=-40\text{J}$$