Problem 407 - Calculus Volume 1

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 \(F(x)=\int_{a}^{x} f(t) d t .\) $$[\mathbf{T}] \int \frac{1}{\sqrt{9-x^{2}}} d x \text { over }[-3,3]$$

Step-by-Step Solution

Answer

Antiderivative is \( \arcsin\left(\frac{x}{3}\right) \). Add \(C = \frac{\pi}{2}\) to align with the definite integral.

Step 1: Recognize the Function

The integral to solve is \( \int \frac{1}{\sqrt{9-x^2}} \, dx \). This is a standard integral that represents the inverse trigonometric function, specifically \( \arcsin(x) \) with some adjustments.

Step 2: Solve the Indefinite Integral

The integral \( \int \frac{1}{\sqrt{9-x^2}} \, dx \) can be solved using the formula for \( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C \). In our case, \( a = 3 \). Thus, the antiderivative is \( \arcsin\left(\frac{x}{3}\right) \). Since we're given \( C = 0 \), the antiderivative is \( \arcsin\left(\frac{x}{3}\right) \).

Step 3: Graph Function and Antiderivative

Using a graphing calculator, plot the function \( f(x) = \frac{1}{\sqrt{9-x^2}} \) and its antiderivative \( F(x) = \arcsin\left(\frac{x}{3}\right) \) over the interval \([-3, 3]\). The function \( f(x) \) represents the derivative and exhibits radical behavior approaching vertical asymptotes near \( x = 3 \) and \( x = -3 \). The antiderivative \( F(x) \) is the inverse sine function, smoothly increasing from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) over the interval.

Step 4: Identify \(C\) for Definite Integral

To find a value of \(C\) such that adding \( C \) to the antiderivative recovers the definite integral, we need to evaluate the definite integral from \(-3\) to some point \(x\) for \(\int_{-3}^{x} \frac{1}{\sqrt{9-t^2}} \, dt\). At \(x = -3\), the definite integral should be 0, hence the adjusted antiderivative \( F(x) + C \) must equal the definite integral result. Evaluate \( F(-3) = \arcsin(-1) = -\frac{\pi}{2} \). Thus, set \( C = \frac{\pi}{2} \) to cancel out \( F(-3) \), ensuring \( F(x) + C \) equals the definite integral.

Key Concepts

definite integral

A definite integral is a fundamental concept in calculus that helps us find the exact area under a curve between two specified points, \(a\) and \(b\). In mathematical terms, it is expressed as \( \int_{a}^{b} f(x) \, dx \). This operation considers both the function and its antiderivative over a specific interval.
To solve definite integrals, we use the Fundamental Theorem of Calculus, which states that if \(F(x)\) is an antiderivative of \(f(x)\), then the definite integral of \(f(x)\) from \(a\) to \(b\) is \( F(b) - F(a) \).
In the exercise, we're dealing with \( \int_{-3}^{x} \frac{1}{\sqrt{9-t^2}} \, dt \), where we identify a suitable constant \(C\) so that when added to the antiderivative, it mimics this definite integral. By calculating \( F(-3) = \arcsin(-1) = -\frac{\pi}{2} \), we see that setting \(C = \frac{\pi}{2}\) adjusts our antiderivative \(F(x)\) to reflect the definite integral accurately. This ensures \(F(x) + C\) doesn't result in discrepancies at specific points.

inverse trigonometric function

Inverse trigonometric functions are used to determine angles when the values of the trigonometric functions are known. They are simply the inverses of the standard trigonometric functions like sine, cosine, and tangent.
In calculus, and specifically in solving integrals, recognizing inverse trigonometric forms is crucial. For instance, the integral \( \int \frac{1}{\sqrt{9-x^2}} \, dx \) utilizes the form of arcsine \( \arcsin(x) \). This is due to its resemblance to \( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C \).
Applying this formula to the given integral, with \( a = 3\), the solution becomes \( \arcsin\left(\frac{x}{3}\right) \). This shows how understanding inverse trigonometric identities helps transition from a complex integral to a solution involving familiar mathematical concepts. The arcsine function, specifically, translates the expression into an angle measure ranging from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), which aligns with the behavior of the corresponding trigonometric function.

graphing calculator

A graphing calculator is a highly versatile tool in mathematics, allowing students and professionals to visualize functions, calculate integrals, and much more.
In our exercise, a graphing calculator serves the purpose of graphing both the function \( f(x) = \frac{1}{\sqrt{9-x^2}} \) and its antiderivative \( F(x) = \arcsin\left(\frac{x}{3}\right) \) over the interval \([-3, 3]\). This visualization helps uncover their behavior over this range, especially recognizing the asymptotic nature of \(f(x)\) as it approaches \(+3\) and \(-3\).
Graphing helps in understanding key concepts such as:

• The drastic rise in the values of \(f(x)\) depicting vertical asymptotes.
• The smooth increase and range of \(F(x)\) denoting the arcsine function.
• The clear interval dependency of these functions.

Overall, using graphing calculators makes it easier to verify solutions, understand complex behaviors visually, and affirm the calculations in real-time for better comprehension.