Question

In Exercises \(31 - 36 ,\) find the average value of the function on the interval, using antiderivatives to compute the integral. $$y = \frac { 1 } { 1 + x ^ { 2 } } , \quad [ 0,1 ]$$

Short answer

The average value of the function \( y = \frac { 1 } { 1 + x ^ { 2 } }\) on the interval \([ 0,1 ]\) is \( \frac{\pi}{4}. \)

Step-by-step solution

Write down the Function

The function provided is \( y = \frac { 1 } { 1 + x ^ { 2 } } \) and the interval is \([0,1]\).

Compute the Antiderivative

The antiderivative of the function \( y = \frac { 1 } { 1 + x ^ { 2 } } \) is \( F(x) = \arctan(x) \).

Use the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that the definite integral of a function can be computed using its antiderivative at the upper and lower bounds of the interval. Compute the integral over the interval by substituting these values into the antiderivative function. In this case, we calculate \( F(1) - F(0) = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \)

Compute the Average Value of the Function

The formula for the average value of a function over an interval \([a,b]\) is given by \( \frac {1} {b - a } \int _ { a } ^ { b } f(x) dx \). Substituting in the values, the average value is found to be \( \frac {1} {1 - 0 } \frac{\pi}{4} = \frac{\pi}{4}. \)

Key concepts

Antiderivatives

Understanding antiderivatives is key to mastering calculus, especially when dealing with the average value of functions. An antiderivative of a function is another function, whose derivative gives back the original function. Think of it as a process of reversing differentiation. If the original function describes the rate of change, the antiderivative represents the accumulation of those changes.

For example, the antiderivative of the function \( y = \frac { 1 } { 1 + x ^ { 2 } } \) is \( F(x) = \text{arctan}(x) \). In this context, for each \(x\), \( F(x) \) represents all the area accumulated under the curve of \(y\) from some starting point up to \(x\). It's important to remember that antiderivatives are determined up to an added constant, since differentiation of a constant is zero, so any constant could have been there before differentiating.

An easy way to remember this is to think of antiderivatives as the 'inverse' of derivatives, taking us from rates back to total values over intervals. In school, these are sometimes introduced as indefinite integrals because they are integrals without specified limits.

Definite Integral

The definite integral of a function gives us the total accumulation between two points, often representing area under the curve. Unlike the antiderivative, which is a function, the result of a definite integral is a number. It's a specific value that quantifies the total sum of changes between the starting point \(a\) and ending point \(b\) on the function's graph.

Let's consider the exercise function \( y = \frac { 1 } { 1 + x ^ { 2 } } \). The definite integral over the interval \[0, 1\] gives us the total area under the curve from \(x=0\) to \(x=1\). We compute this by evaluating the antiderivative \( F(x) \) that we found at the endpoints of our interval, and then subtract: \[ \text{Definite Integral} = F(1) - F(0) = \frac{\rm{\text{\textpi}}}{4} \].

Understanding this concept is essential in calculus and will help you interpret various physical phenomena, like distance travelled given a speed function, or the accumulated work done by a varying force.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a beautiful connection between differentiation and integration. In simple terms, it states that the definite integral of a function over an interval can be found using an antiderivative of that function. It's the bridge that allows us to move from the general antiderivative to the specific value of a definite integral.

In the context of the provided exercise, the Fundamental Theorem helps us calculate the area under the curve of \( y = \frac { 1 } { 1 + x ^ { 2 } } \) between \(x=0\) and \(x=1\). We first find the antiderivative, then apply it at the bounds of our interval to get the area: \( \text{Definite Integral} = F(b) - F(a) = \text{arctan}(1) - \text{arctan}(0) = \frac{\rm{\text{\textpi}}}{4} \).

This theorem is not only a foundational concept in calculus but also an incredibly practical tool because it allows us to compute areas, totals, and averages - in this case, the average value of the function on the interval.