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We are tasked with calculating the resultant force and its location for a distributed load, represented by the given function \( y = a \cdot f(x) \), where:

\[
f(x) = 1 + A\left(\frac{x}{L}\right) + B\left(\frac{x^2}{L^2}\right) + C\left(\frac{x^3}{L^3}\right) + D\left(\frac{x^4}{L^4}\right)
\]

The problem provides the following parameters:

- \( a = 2.0 \, \text{kN/m} \)
- \( L = 2 \, \text{m} \)
- \( A = -1 \)
- \( B = 3 \)
- \( C = 1 \)
- \( D = 1 \)

### 1. **Resultant Force \( R \)**

The resultant force \( R \) is calculated by integrating the load function over the domain from \( x = 0 \) to \( x = 2L \).

\[
R = \int_0^{2L} a \cdot f(x) \, dx
\]

First, substitute the expression for \( f(x) \) into the equation:

\[
R = \int_0^{2L} 2.0 \cdot \left[ 1 + \left(\frac{-1 \cdot x}{L}\right) + \left(\frac{3 \cdot x^2}{L^2}\right) + \left(\frac{x^3}{L^3}\right) + \left(\frac{x^4}{L^4}\right) \right] dx
\]

Now, let's integrate term by term. Each term will involve powers of \( x \), and we can compute this explicitly.

### 2. **Location of the Resultant Force \( x_R \)**

The location \( x_R \) of the resultant force is found using the moment about the origin:

\[
x_R = \frac{\int_0^{2L} x \cdot a \cdot f(x) \, dx}{R}
\]

Again, substitute the load function into the integral and calculate both the numerator and denominator.

I will now proceed with the integration and the final calculations.The calculations yield the following results:

1. **Resultant Force \( R \)**: \( 73.6 \, \text{kN} \)
2. **Location of Resultant Force \( x_R \)**: \( 3.09 \, \text{m} \)

This means the total load is 73.6 kN, and it acts at a point approximately 3.09 meters from the origin along the x-axis.

Would you like a detailed breakdown of the integration steps or have any other questions? Here are five related questions you might find useful:

1. How does varying the parameters \( A \), \( B \), \( C \), and \( D \) affect the resultant force?
2. Can you explain how to derive the moment about the origin for irregular loads?
3. What is the significance of the resultant location \( x_R \) in practical applications?
4. How does the length \( L \) influence the load distribution and resultant?
5. How would this result change if the load function was discontinuous?

**Tip**: When calculating the location of a resultant force, ensure the integration for both force and moment are over the same limits for consistent results.

Ohjeinen pinta-ala edustaa jakaantunutta kuormitusta. Laske integroimalla kuormituksen resultantti ja sen sijainti x-akselin suunnassa, kun kuormitusfunktion parametrit ovat a = 2.0 kN/m, L = 2 m, A = -1, B = 3, C = 1 ja D = 1.

Learn how to calculate the resultant force and its position for a distributed load using integration. Parameters provided: a = 2.0 kN/m, L = 2 m, A = -1, B = 3, C = 1, and D = 1. Suitable for university-level statics or physics problems.

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