The definite integral is a powerful tool in calculus that allows us to find the average value of a function over a specified interval. To understand this concept, we can start by recalling the process of Riemann sums, which approximate the area under a curve by dividing it into rectangles. As the number of rectangles increases (approaching infinity), the Riemann sum converges to the definite integral, which represents the exact area under the curve.
To find the average value of a function \( f(x) \) over the interval \([a, b]\), we can use the formula:
\[\text{Average Value} = \frac{1}{b - a} \int_a^b f(x) \, dx\]
In this formula, \( b - a \) represents the width of the interval, and the integral \( \int_a^b f(x) \, dx \) calculates the total area under the curve from \( a \) to \( b \). The average value thus gives us a single output that represents the mean of all function values over that interval.
To illustrate this, consider the function \( f(x) = x + 2 \) over the interval \([0, 4]\). We can apply the average value formula as follows:
\[\text{Average Value} = \frac{1}{4 - 0} \int_0^4 (x + 2) \, dx\]
First, we compute the integral:
\[\int (x + 2) \, dx = \frac{x^2}{2} + 2x\]
Next, we evaluate this from \( 0 \) to \( 4 \):
\[\left[ \frac{4^2}{2} + 2(4) \right] - \left[ \frac{0^2}{2} + 2(0) \right] = \left[ 8 + 8 \right] - 0 = 16\]
Now, substituting back into the average value formula gives us:
\[\text{Average Value} = \frac{1}{4} \cdot 16 = 4\]
This result indicates that the average output of the function \( f(x) = x + 2 \) over the interval from \( 0 \) to \( 4 \) is \( 4 \). Understanding how to derive and apply this formula is essential for solving problems related to the average value of functions, making it a straightforward process once the foundational concepts are grasped.
