The mean square value of a function \(f(t)\) over the interval \(t=a\) to \(t=b\) is defined to be $$ \frac{1}{b-a} \int_{a}^{b}(f(t))^{2} \mathrm{~d} t $$ Find the mean square value of \(f(t)=\sin t\) over the interval \(t=0\) to \(t=2 \pi\).

Integral calculus is a branch of mathematics focused on the accumulation of quantities and the areas under and between curves. This mathematical toolset allows us to calculate the total size or value, such as finding the mean square value of a function over a given interval.
When we are dealing with functions such as sine or cosine, integral calculus becomes exceedingly useful in physics and engineering for determining waveforms and energy levels.
In the context of our exercise, integral calculus helps us find the mean square value of \( f(t) = \sin t \) over a period, which is essential in analyzing the performance of alternating currents in electrical engineering or the intensity of a sound wave in acoustics.

The double-angle formulas are a set of trigonometric identities that relate the sine and cosine functions of double angles to the square of those functions. One of the most frequently used double-angle formulas is \( \cos{2t} = 1 - 2\sin^2 t \).
These formulas are crucial in simplifying complex trigonometric expressions, making certain integrals more tractable. As we've seen in the solution, using the double-angle formula allowed the transformation of \( \sin^2 t \) into an expression involving \( \cos{2t} \), which was then integrated over the given interval with the aid of integral calculus to find the mean square value.

The sine function squared, denoted as \( \sin^2 t \), represents the square of the sine of an angle t. Calculating the integral of \( \sin^2 t \) is a common task in trigonometry and calculus.
Direct integration of \( \sin^2 t \) is not straightforward. However, by employing the double-angle formula, the squared sine function can be expressed in terms of cosine, which is easier to integrate. The exercise demonstrates how we convert \( \sin^2 t \) to \( \frac{1 - \cos{2t}}{2} \) to facilitate the evaluation of the definite integral. This technique streamlines calculations in various scientific fields, including signal processing and wave analysis.

Definite integral evaluation involves calculating the integral of a function between specified limits. This process results in a number that represents the net area between the function and the x-axis over that interval.
In our exercise, a definite integral is used to find the mean square value of \( f(t) = \sin t \) over one complete oscillation (from \( t = 0 \) to \( t = 2\pi \)). The evaluation of the definite integral, after applying the double-angle formula, gives us the mean value of the function squared over the given period, which in this case is \( \frac{1}{2} \).
Understanding how to evaluate a definite integral is essential for problems in physics, engineering, and economics, where it is necessary to calculate work done, area, or total profit over a specific interval.