Question

In the following exercises, solve for the antiderivative \(\int f\) of \(f\) with \(C=0\), then use a calculator to graph \(f\) and the antiderivative over the given interval \([a, b] .\) Identify a value of \(C\) such that adding \(C\) to the antiderivative recovers the definite integral \(F(x)=\int_{a}^{x} f(t) d t\) [T] \(\int \frac{1}{\sqrt{9-x^{2}}} d x\) over \([-3,3]\)

Short answer

Antiderivative: \(\arcsin\left(\frac{x}{3}\right)\), C = \(\frac{\pi}{2}\).

Step-by-step solution

Recognize the Function

The given function is the integral \[\int \frac{1}{\sqrt{9-x^2}} \, dx\]. This is a standard form that resembles the derivative of the arcsine function.

Identify the Corresponding Antiderivative

Recall that the antiderivative of \(\frac{1}{\sqrt{a^2-x^2}}\)is \(\arcsin\left(\frac{x}{a}\right)\). Here, \(a = 3\), so the antiderivative is\[\arcsin\left(\frac{x}{3}\right)\].

Assign C and Evaluate

According to the problem, we initially set \(C = 0\). Thus, the current antiderivative solution is \[\arcsin\left(\frac{x}{3}\right)\]with \(C = 0\).

Graph f(x) and the Antiderivative

Using a calculator or graphing tool, plot \(f(x) = \frac{1}{\sqrt{9-x^2}}\) and its antiderivative \(\arcsin\left(\frac{x}{3}\right)\) over the interval \([-3, 3]\). Both graphs should be apparent, one as a semi-circle and the other as the inverse sine curve.

Determine the Constant C

To identify the constant \(C\) that makes \(F(x) = \int_{a}^{x} f(t) \, dt\), use the definite integral from \(-3\) to \(x\). Evaluate a specific point, e.g., \(x = 0\): \[F(0) = \int_{-3}^{0} \frac{1}{\sqrt{9-t^2}} \, dt = \arcsin\left(0\right) - \arcsin\left(-1\right) = 0 - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2}\]Thus, \(C = \frac{\pi}{2}\) to make the expression have the integral property.

Key concepts

Antiderivative

An antiderivative, also known as an indefinite integral, represents the reverse process of differentiation. When you find an antiderivative of a function, you are essentially determining a function whose derivative would give you the original function back. For the given exercise, we encounter the function \( \int \frac{1}{\sqrt{9-x^2}} \, dx \). This resembles the form commonly associated with the derivative of the arcsine function. By identifying this standard pattern, we know that the antiderivative is \( \arcsin\left(\frac{x}{3}\right) \).
To solidify your understanding, always aim to identify the format and pattern of given functions, relating them back to known derivatives or antiderivatives. Practicing different variations will enhance your ability to recognize these patterns quickly. It is crucial to remember that the antiderivative is a family of functions, generalizable by adding a constant \( C \). However, in this exercise, we initially set \( C = 0 \). This particular antiderivative lacks any adjustment from constants, but we later adjust \( C \) based on conditions like definite integrals.

Definite Integral

A definite integral is used to compute the net area under a curve over a specific interval. It's written in the form \( \int_{a}^{b} f(t) \, dt \) and results in a specific numerical value, quantifying the accumulated area between a function and the x-axis.
In the exercise, the definite integral helps us find a specific constant \( C \) when adding to our antiderivative such that it equates \( F(x) = \int_{a}^{x} f(t) \, dt \). Using the interval \([-3,3]\), our task is to calculate the net area over a section of \( \int \frac{1}{\sqrt{9-t^2}} \, dt \) and correlate it with the arcsine integral. We determined \( C \) by calculating at the point \( x = 0 \), where \( F(0) \) equates \( \frac{\pi}{2} \). Thus, it shows that the adjustment \( C = \frac{\pi}{2} \) aligns with accumulating those integral properties.

Arcsine Function

The arcsine function is the inverse of the sine function, denoted as \( \arcsin(x) \). This function returns the angle whose sine value is \( x \). It’s typically restricted to the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\) due to the nature of inverse functions, ensuring each input has a unique output.
In calculus, recognizing when the function \( \int \frac{1}{\sqrt{a^2-x^2}} \, dx \) appears is vital because of its direct association with arcsine derivatives. In our exercise, the function \( \int \frac{1}{\sqrt{9-x^2}} \, dx \) leads us to \( \arcsin\left(\frac{x}{3}\right) \) because of this association. It serves as a classic example of how trigonometric ideas contribute heavily to calculus, especially in generating standard functions to determine derivatives and antiderivatives. Understanding the inference and calculating such functions adds precision when tackling more complex integrals and equations.

Graphing Functions

Graphing functions provides a visual insight into their behavior and interaction. It's a valuable skill in understanding and interpreting mathematical information. When working with \( f(x) = \frac{1}{\sqrt{9-x^2}} \) and its antiderivative \( \arcsin\left(\frac{x}{3}\right) \), graphing becomes crucial. It illustrates how the original function forms a semicircle, a well-known geometrical shape hinting its mathematical properties tied to circles.
Utilizing graphing calculators or software, you can plot these functions to see the difference in values and shapes over the interval \([-3, 3]\). The function \( f(x) \) will appear as a semicircle on the interval, while the antiderivative, resembling the inverse sine, provides a smooth curve hitting its end points at approximately \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \). Observing these graphs alongside calculations offers a greater sense of the equations' impacts and derivations, providing a bridge between numerical results and real-world visualization.