The given problem asks us to find the antiderivative (or "stamfunktion" in Danish) of the function $f(x) = 2x^3 - 6x$, and it specifies that the graph passes through the point $P(1,10)$.

Step 1: Find the general antiderivative of $f(x)$

To find the antiderivative, we integrate the given function:

$F(x) = \int (2x^3 - 6x) \, dx$

For each term:

1. $\int 2x^3 \, dx = \frac{2}{4}x^4 = \frac{1}{2}x^4$
2. $\int -6x \, dx = -6 \cdot \frac{x^2}{2} = -3x^2$

Thus, the general form of the antiderivative is:

$F(x) = \frac{1}{2}x^4 - 3x^2 + C$

where $C$ is the constant of integration.

Step 2: Use the point $P(1,10)$ to find $C$

We know that $F(1) = 10$, so we substitute $x = 1$ and $F(x) = 10$ into the equation to find $C$:

$10 = \frac{1}{2}(1)^4 - 3(1)^2 + C$

Simplifying:

$10 = \frac{1}{2} - 3 + C$

$10 = -\frac{5}{2} + C$

Now, solve for $C$:

$C = 10 + \frac{5}{2} = \frac{20}{2} + \frac{5}{2} = \frac{25}{2}$

Thus, $C = \frac{25}{2}$.

Final answer:

The specific antiderivative is:

$F(x) = \frac{1}{2}x^4 - 3x^2 + \frac{25}{2}$

Let me know if you want more details on any steps, and here are some related questions:

1. How would you find the antiderivative of a more complex function?
2. What is the geometric meaning of the antiderivative in terms of the area under the curve?
3. How can you verify that $F(x)$ is the correct antiderivative?
4. How would you apply similar methods if the function involved trigonometric terms?
5. What if the point $P$ was different—how would that affect the result?

Tip: When integrating polynomials, always increase the power of the variable by 1 and divide by the new power!