The integral you've provided is:

$\int_0^{x} 1 \, \mathrm{d}x$

To solve this, we integrate the constant $1$ over the interval from $0$ to $x$.

Solution:

1. Since the integrand is $1$, integrating $1$ with respect to $x$ yields $x$: $\int 1 \, \mathrm{d}x = x$

2. Now, we evaluate this from $0$ to $x$: $\int_0^x 1 \, \mathrm{d}x = \left[ x \right]_0^x$

3. Substitute the limits: $= x - 0 = x$

Final Answer:

$\int_0^{x} 1 \, \mathrm{d}x = x$

Would you like further details or have any questions?

Here are some related questions that expand on this concept:

1. What happens when we integrate a constant other than 1 over a similar interval?
2. How would this integral change if the limits were from $a$ to $b$ instead of $0$ to $x$?
3. How do definite and indefinite integrals differ conceptually?
4. What are some applications of constant integrals in real-world contexts?
5. How does this integral represent the area under a constant function?

Tip: Integrating a constant over an interval gives the length of the interval times the value of the constant, which represents the area of a rectangle.