91Ó°ÊÓ

Understanding integral calculus is essential for finding the average value of a function over a specific interval. Integral calculus involves finding the area under the curve of a function, which is represented by the definite integral. To calculate the average value of a function, we use the formula:

• \( \text{average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \)

Here, we need to calculate the definite integral \( \int_{a}^{b} f(x) \, dx \), which gives us the total accumulated value over the interval \([a, b]\). The factor \( \frac{1}{b-a} \) normalizes this accumulated value over the length of the interval. By arranging the integral correctly, we can evaluate the average value, which is particularly useful in understanding how the function behaves across different intervals.

Inverse trigonometric functions, such as the inverse sine function used in this exercise, play a crucial role in solving integrals involving trigonometric expressions. The function \( \frac{1}{\sqrt{1-x^2}} \) can be recognized as the derivative of the inverse sine function, \( \sin^{-1}(x) \). This recognition allows us to evaluate the integral more easily. The general formula for finding the inverse sine integral is:

• \( \int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1}(x) + C \)

In our case, evaluating over the interval \([-\frac{1}{2}, 0]\), we calculate the specific values:

• \( \sin^{-1}(0) = 0 \)
• \( \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \)

This information helps us to perform calculations accurately. Knowing how to apply these inverse functions allows for a more straightforward process when evaluating certain types of integrals.

Definite integral evaluation refers to computing the integral of a function over a specified interval, which gives us a specific numerical result rather than a general expression involving a constant of integration. This process is essential for real-world applications, including finding areas, accumulated values, and average values.To evaluate the definite integral in this exercise, \( \int_{-\frac{1}{2}}^{0} \frac{1}{\sqrt{1-x^2}} \, dx \), you need to:

• Identify the antiderivative, here \( \sin^{-1}(x) \).
• Substitute the limits of integration into the antiderivative: \( \sin^{-1}(0) - \sin^{-1}\left(-\frac{1}{2}\right) \).
• Calculate the result of the subtraction, ensuring each trigonometric value is correctly evaluated.

Once evaluated, the definite integral produces a result, which, when multiplied by the factor \( \frac{1}{b-a} \), gives the average value over the interval. This technique pairs calculus with trigonometry effectively, offering precise mathematical insights into the behavior of functions.