91Ó°ÊÓ

Integral calculus is a crucial part of mathematics that focuses on the concept of integration. Integration is the process of finding the integral of a function. It is the reverse process of differentiation. One of the main purposes of integration is to calculate areas under curves. When we integrate a function, we sum up an infinite number of small quantities to find a whole.
Another application of integral calculus is finding the accumulation of quantities, such as distance traveled over time, given a speed function.
The concept of integrals divides into two major types: definite and indefinite integrals, each serving different purposes.

A definite integral has both an upper and lower limit and provides a number that represents the net area under the curve of a function between these limits. It’s represented as: \[ \text{Definite integral from } a \text{ to } b \text{ of } f(x) \text{ dx, written as } \int_a^b f(x) \text{ dx} \]
In the exercise, we find the average value of the function \( f(x) = \frac{1}{\text{ sqrt(x)}} \) over the interval [1, 9]. To solve this, we need to evaluate the definite integral of \( f(x) \) from 1 to 9.
By solving the definite integral, we calculate the area under the curve \( f(x) \) from x = 1 to x = 9, which is essential when using the average value formula.

Antiderivatives are functions that reverse the process of differentiation. When we differentiate a function and then integrate it, we should return to the original function. Finding an antiderivative is also known as indefinite integration. For a function \( f(x) \), its antiderivative is a function \( F(x) \, such that \ F'(x) = f(x) \).
For example, in the given problem, we need the antiderivative of \( x^{-\frac{1}{2}} \), which is \ 2x^{\frac{1}{2}} \.
This antiderivative is evaluated at the upper and lower limits, 9 and 1 respectively, to find the value of the definite integral. Calculating these values correctly is crucial to arriving at the right average value.

The average value of a function over a certain interval is given by the formula:
\[ \text{Average value} = \frac{1}{b-a} \times \text{ definite integral from } a \text{ to } b \text{ of } f(x)\text{ dx}\]
This formula combines the concept of an integral with averaging. Here, \( b \) and \( a \) are the upper and lower bounds of the interval, respectively. First, we find the definite integral of \( f(x) \) over [a, b] and then divide it by the length of the interval \( (b-a) \).
In our exercise, with \( f(x) = \frac{1}{\text{ sqrt(x)}} \) over the interval [1, 9], we calculated the definite integral to be 4. Thus, applying the average value formula results in:
\[ \text{Average value} = \frac{1}{9-1} \times 4 = \frac{4}{8} = 0.5 \]