## Integration of the "red line" in order to calculate the second high school kinematic equation

To be able to find the integral we must first find the formula for the read line (velocity graph).

The assumption under which the high school equation is valid, is: The velocity increases linear from V_{0} to V_{1} over the time period Δt.

Δt = t_{1} - t_{0}

The slope m of the red line is:
V_{1} - V_{0}
m = ---------
t_{1} - t_{0}

The general formula of a line is: y = m * x + b. In our case we have t on the x-axis and v(t) on the y-axis.

The formula for the velocity over time for this specific line is therefore:

V_{1} - V_{0}
v(t) = -------- * t + b
t_{1} - t_{0}

b is the y-intersect:

V_{1} - V_{0}
b = V_{1} - ---------- * t_{1}
t_{1} - t_{0}

Thus the general formula for this red line, v(t), is:
V_{1} - V_{0} V_{1} - V_{0}
v(t) = -------- * t + V_{1} - --------- * t_{1}
t_{1} - t_{0} t_{1} - t_{0}

Now we have to find the integral of v(t) for the section from t=t_{0} to t=t_{1}.

Let's abbreviate the formula a bit by introducing the constant m for the slope and b for the y-intersect:

v(t) = m * t + b

The integral of v(t) is:

∫v(t)∂t = ½ * m * t^{2} + b * t

The integral from t=t_{0} to t=t_{1} is therefore:

½ * m * t_{1}^{2} + b * t_{1} - ( ½ * m * t_{0}^{2} + b * t_{0})

We put the values for b and m back into the equation and we get:

1 V_{1} - V_{0} V_{1} - V_{0} ( 1 V_{1} - V_{0} V_{1} - V_{0} )
- ------- t_{1}^{2} + V_{1}t_{1} - ------- t_{1}^{2} - ( - -------- t_{0}^{2} + V_{1}t_{0} - ------- t_{1}t_{0})
2 t_{1} - t_{0} t_{1} - t_{0} ( 2 t_{1} - t_{0} t_{1} - t_{0} )

this is equal to:

1 V_{1} - V_{0} 1 V_{1} - V_{0} V_{1} - V_{0}
- -- ------- t_{1}^{2} + V_{1}t_{1} - -- ------- t_{0}^{2} - V_{1}t_{0} + -------- t_{1}t_{0}
2 t_{1} - t_{0} 2 t_{1} - t_{0} t_{1} - t_{0}

since our aim is to eventually replace t_{1} - t_{0} with Δt we rewrite the equation like this:
1 V_{1} - V_{0} V_{1} - V_{0}
- -- ------- (t_{1}^{2} + t_{0}^{2}) + V_{1} (t_{1} - t_{0}) + -------- t_{1}t_{0}
2 t_{1} - t_{0} t_{1} - t_{0}

With t_{1}^{2} + t_{0}^{2} = (t_{1} - t_{0})^{2} + 2t_{1}t_{0} we can re-write the equation as follows:
V_{1} - V_{0}
------- (- ½(t_{1} - t_{0})^{2} - ½ 2t_{1}t_{0} + t_{1}t_{0} ) + V_{1} (t_{1} - t_{0})
t_{1} - t_{0}

this is equal to (replace t_{1} - t_{0} with Δt):
V_{1} - V_{0}
- ------- Δt + V_{1} Δt
2

this is equal to:
½ (V_{0} + V_{1}) Δt

and that is the solution of the integral ∫v(t)∂t with v(t) being a line starting at V_{0} and ending at V_{1} over the time period Δt.