Question

In Exercises \(31 - 36 ,\) find the average value of the function on the interval, using antiderivatives to compute the integral. $$y = \sin x , \quad [ 0 , \pi ]$$

Short answer

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

Step-by-step solution

Set up the integral

According to the average value formula, we need to evaluate the integral of \(y = \sin x\) from 0 to \(\pi\). The integral is represented as \(\int_{0}^{\pi} \sin x dx\).

Evaluate the integral

The antiderivative of \(\sin x\) is \(-\cos x\). Apply the Fundamental Theorem of Calculus to evaluate the definite integral. It states that if F(x) is an antiderivative of f(x) on the interval \([0, \pi]\), then \(\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 2.\)

Calculate the average value

The average value of \(y = \sin x\) over the interval [0, \pi] is given by the formula \(\frac{1}{b-a} \int_{a}^{b} f(x) dx\). Substitute \(a = 0\), \(b = \pi\), and the integral calculation into the formula, we get \(\frac{1}{\pi-0} \int_{0}^{\pi} \sin x dx = \frac{1}{\pi} * 2 = \frac{2}{\pi}.\)

Key concepts

Antiderivatives

An antiderivative, simply put, is the reverse of a derivative. It's a function that, when differentiated, yields the original function. For example, if we have a function such as \( f(x) = x^2 \), the derivative of \( f(x) \) would be \( 2x \). If we take the antiderivative of \( 2x \), we arrive back at our original function \( f(x) = x^2 \) (up to a constant).

In the context of the exercise, to find the antiderivative of \( y = \text{sin} x \), we consider what function, when differentiated, will give us \( \text{sin} x \). The function that meets this condition is \( -\text{cos} x \), since the derivative of \( -\text{cos} x \) is \( \text{sin} x \). Understanding antiderivatives is crucial because it allows us to evaluate definite integrals, which are essential for calculating areas under curves, volumes, and in this case, the average value of a function over an interval.

Definite Integral

The definite integral of a function provides a way to calculate the accumulation of quantities, such as area under a curve, between two points. It is usually represented as \( \text{from } a \text{ to } b \text{ of } f(x) dx \), or \( \textstyle\int_{a}^{b} f(x) dx \), with \( a \) and \( b \) being the limits of integration.

In our exercise, we are interested in the area under the function \( y = \text{sin} x \) from \( x = 0 \) to \( x = \pi \). To find this, we need to perform the integration of \( \text{sin} x \) over the interval \([0, \pi]\). The result of this integration reflects the total area swept by the curve, which is essential when determining the average value over the interval. Think of the definite integral as a means to accumulate all the tiny bits of area under the curve to find a total.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is what links differentiation and integration, two core concepts in calculus. It consists of two parts: the first part provides an antiderivative for a continuous function, while the second part allows us to evaluate the definite integral between two points.

According to this theorem, if we want to evaluate the integral \( \textstyle\int_{a}^{b} f(x) dx \), we can find an antiderivative of \( f(x) \), call it \( F(x) \), and then calculate the difference \( F(b) - F(a) \). The Fundamental Theorem of Calculus tells us that this difference will be equal to the definite integral of \( f(x) \) from \( a \) to \( b \).

This theorem simplifies the process of finding the area under a curve and plays a crucial role in our exercise. We used it to evaluate the integral \( \textstyle\int_{0}^{\pi} \text{sin} x dx \), which then allowed us to determine the average value of the function over the given interval by considering the total accumulated area normalized by the interval's length.