Calculate for me shear force calculations and bending moment calculations, then draw for me shear force diagram and bending moment diagram

Let the fixed end be point A (left side) and the free end be point B (right side).
The total length of the beam is $L = 4 \, \text{m}$.
The uniformly distributed load (UDL) is $w = 3 \, \text{kN/m}$.
The point load at the free end is $P = 2 \, \text{kN}$.

The vertical reaction force $R_A$ at the fixed end is the sum of all downward loads:

The reaction moment $M_A$ at the fixed end is the sum of moments due to all loads about point A:

Consider a section at a distance $x$ from the fixed end (left side). Using the standard sign convention (upward forces positive for shear force when looking at the left part of the section):

At the fixed end ($x=0$):

At the free end ($x=4 \, \text{m}$):

Consider a section at a distance $x$ from the fixed end (left side). Using the standard sign convention (hogging moments are negative):

At the fixed end ($x=0$):

At the free end ($x=4 \, \text{m}$):

The shear force diagram starts at a positive value of $14 \, \text{kN}$ at the fixed end ($x=0$). It decreases linearly along the beam due to the uniformly distributed load. At the free end ($x=4 \, \text{m}$), the shear force is $2 \, \text{kN}$. The entire diagram is above the zero line, indicating positive shear force throughout the beam.
The equation for the SFD is $V(x) = \boxed{14 - 3x \, \text{kN}}$.

The bending moment diagram starts at a negative value of $-32 \, \text{kN} \cdot \text{m}$ at the fixed end ($x=0$), which is the maximum (in magnitude) bending moment. It then increases parabolically (becomes less negative) along the beam. At the free end ($x=4 \, \text{m}$), the bending moment is $0 \, \text{kN} \cdot \text{m}$. The entire diagram is below the zero line, indicating negative bending moment (hogging) throughout the beam.
The equation for the BMD is $M(x) = \boxed{-32 + 14x - 1.5x^2 \, \text{kN} \cdot \text{m}}$.