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A definite integral is a way to calculate the area under a curve described by a function over a specific interval. In simpler terms, it allows us to find out how much space lies beneath a curve between two points on the x-axis. When we perform a definite integral, we are essentially summing an infinite number of infinitesimally small quantities to get a total quantity over the specified interval.

For instance, when we calculate the integral of a function from 0 to 4, such as:

• \( \int_{0}^{4} f(x) \, dx \)

we are finding the total area under the curve from x = 0 to x = 4. In this exercise, to find the average value of the function \( f(x) = 3 \sqrt{x} \), we compute its definite integral over the interval [0, 4] as one step in the process. This area serves as a core part of finding what the function's average value is across the chosen interval.

The power rule for integration is a fundamental technique used to solve integrals involving powers of x. When dealing with functions like \( x^n \), this rule is a real lifesaver because it gives us a straightforward way to find antiderivatives.

The power rule formula says:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

where \( C \) represents the constant of integration. However, when computing definite integrals, \( C \) is not needed since we are evaluating the result over a specific interval.

In step 3 of our solution, the function is \( 3 x^{1/2} \). By using the power rule we get:

• \( \int 3 x^{1/2} \, dx = 3 \cdot \frac{x^{3/2}}{3/2}= 2x^{3/2} \)

The power rule simplifies the integration process and allows us to easily obtain a solution for complex functions based on power expressions.

When calculating the definite integral to find the average value of a function, specifying the interval is crucial. An interval is simply the range over which we want to evaluate the function. In this exercise, we are working on the interval [0,4]. This means our integration will be performed from x = 0 to x = 4. The endpoints in the interval determine the bounds for our definite integral, which we can denote as:

• \( a = 0 \)
• \( b = 4 \)

This is mathematically expressed in the average value formula as:

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

Here, \( b - a \) calculates the length of the interval, which is 4 in this case. Including the correct interval in our calculations ensures we are finding the area or average we are interested in, as it signifies the section of the function's graph we are evaluating.