What is the difference between net area and total area from the curve to the x-axis

What is the difference between net area and total area from the curve to the x-axis? What is the difference in using definite integrals to calculate the total area from the curve to the x-axis from x = a to x = b rather than the net area? What would the solutions to integral(s) calculating the net area as well as total area of a velocity function tell you about the situation? Include examples in your detailed explanations.

The Correct Answer and Explanation is:

The key difference between net area and total area from the curve to the x-axis lies in how areas above and below the x-axis are treated.

1. Net Area:
   The net area refers to the algebraic sum of the areas between the curve and the x-axis over a given interval [a, b]. It takes into account both positive and negative areas, depending on whether the curve is above or below the x-axis. Areas above the x-axis are positive, while areas below the x-axis are negative. Mathematically, this is calculated using a definite integral: Net Area=∫abf(x) dx\text{Net Area} = \int_a^b f(x) \, dxNet Area=∫ab​f(x)dx Example:
   For f(x)=xf(x) = xf(x)=x from a=−1a = -1a=−1 to b=1b = 1b=1: ∫−11x dx=0\int_{-1}^1 x \, dx = 0∫−11​xdx=0 This is because the areas above and below the x-axis cancel out.
2. Total Area:
   The total area refers to the sum of the absolute areas between the curve and the x-axis, regardless of whether the curve is above or below the x-axis. In this case, negative areas are treated as positive. To find the total area, you would compute the integral of the absolute value of the function: Total Area=∫ab∣f(x)∣ dx\text{Total Area} = \int_a^b |f(x)| \, dxTotal Area=∫ab​∣f(x)∣dx Example:
   For f(x)=xf(x) = xf(x)=x from a=−1a = -1a=−1 to b=1b = 1b=1, the total area is: ∫−11∣x∣ dx=2(since the area above and below the x-axis is treated as positive)\int_{-1}^1 |x| \, dx = 2 \quad (\text{since the area above and below the x-axis is treated as positive})∫−11​∣x∣dx=2(since the area above and below the x-axis is treated as positive)
3. Definite Integrals for Net and Total Areas of a Velocity Function:
   When calculating the net area under a velocity function (say, v(t)v(t)v(t) over a time interval), the net area would give the displacement, which accounts for both the motion in the positive and negative directions. If v(t)v(t)v(t) is positive, the object moves in one direction, and if v(t)v(t)v(t) is negative, the object moves in the opposite direction. The total area would give the total distance traveled, irrespective of direction, because all negative velocities are considered as positive. Example:
   For a velocity function v(t)=5−t2v(t) = 5 – t^2v(t)=5−t2 over the interval [0,2][0, 2][0,2]:
   • Net area (displacement):
   ∫02(5−t2) dt=[5t−t33]02=(10−83)−0=303−83=223≈7.33\int_0^2 (5 – t^2) \, dt = \left[ 5t – \frac{t^3}{3} \right]_0^2 = (10 – \frac{8}{3}) – 0 = \frac{30}{3} – \frac{8}{3} = \frac{22}{3} \approx 7.33∫02​(5−t2)dt=[5t−3t3​]02​=(10−38​)−0=330​−38​=322​≈7.33
   • Total area (total distance traveled):
   ∫02∣5−t2∣ dt\int_0^2 |5 – t^2| \, dt∫02​∣5−t2∣dt (This would require splitting the integral at points where 5−t25 – t^25−t2 changes sign, but essentially the result gives the total distance.)

In summary:

• Net area accounts for direction, leading to displacement.
• Total area treats all movement as positive, giving total distance traveled.