Question

Let \(F(x)=\int_{0}^{x} \sqrt{a^{2}-t^{2}} d t .\) The figure shows that \(F(x)=\) area of sector \(O A B+\) area of triangle \(O B C\) a. Use the figure to prove that $$ F(x)=\frac{a^{2} \sin ^{-1}(x / a)}{2}+\frac{x \sqrt{a^{2}-x^{2}}}{2} $$ b. Conclude that $$ \int \sqrt{a^{2}-x^{2}} d x=\frac{a^{2} \sin ^{-1}(x / a)}{2}+\frac{x \sqrt{a^{2}-x^{2}}}{2}+C $$

Short answer

Answer: The indefinite integral of \(\sqrt{a^2 - x^2}\) is \(\frac{a^{2} \sin^{-1}(x / a)}{2}+\frac{x\sqrt{a^{2}-x^{2}}}{2}+C\), where C is the constant of integration.

Step-by-step solution

Find the Area of Sector OAB

Since the area of F(x) is given as the sum of the area of sector OAB and triangle OBC, we start by finding an expression for the area of sector OAB. The radius of the circle is a, and we use the angle \(\theta\) formed by the lines OA and OB. The area of sector OAB can be represented as: $$ Area_{OAB} = \frac{1}{2} a^{2} \theta $$

Express \(\theta\) in terms of x and a

In order to find a relationship between \(\theta\), x, and a, we can use the sine function. Observing from the figure, we see that \(\sin(\theta) = \frac{x}{a}\). Now, we can express \(\theta\) in terms of x and a as follows: $$\theta = \sin^{-1}(\frac{x}{a})$$

Substitute \(\theta\) into the Area of Sector OAB

Now, we will substitute the expression for \(\theta\) from step 2 into the formula for the area of sector OAB: $$ Area_{OAB} = \frac{1}{2} a^{2}\sin^{-1}(\frac{x}{a}) $$

Find the Area of Triangle OBC

Next, we need to find an expression for the area of triangle OBC. We observe that the base of triangle OBC is x, and the height is \(\sqrt{a^2 - x^2}\). Hence, the area of triangle OBC can be found using the formula: $$ Area_{OBC} = \frac{1}{2} x \sqrt{a^{2} - x^{2}} $$

Combine Area Formulas to Create F(x)

Now that we have expressions for the areas of the sector OAB and triangle OBC, we can add them together to arrive at the formula for F(x): $$ F(x) = Area_{OAB} + Area_{OBC} = \frac{1}{2} a^{2}\sin^{-1}(\frac{x}{a}) + \frac{1}{2} x \sqrt{a^{2} - x^{2}} $$

Conclude the Indefinite Integral

Given that we have found the appropriate formula for F(x), we can conclude that the indefinite integral of \(\sqrt{a^2 - x^2}\) is equal to: $$ \int \sqrt{a^{2}-x^{2}} dx = \frac{a^{2} \sin^{-1}(x / a)}{2}+\frac{x\sqrt{a^{2}-x^{2}}}{2}+C $$ where C is the constant of integration.

Key concepts

Indefinite Integral

An indefinite integral, often called an antiderivative, represents the collection of all functions whose derivative is the integrand. When we express the indefinite integral, we include a constant of integration, denoted as "C."

This constant exists because integration is the reverse process of differentiation, and any constant added to a function disappears when differentiated. For example, if \(F(x)\) is an antiderivative of \(f(x)\), then \(abla rac{d}{dx}F(x) = f(x)abla\). Hence, \(\int f(x) \, dx = F(x) + C\).

In the given problem, the indefinite integral \(\int \sqrt{a^{2} - x^{2}} \, dx\) is resolved to combine familiar functions, particularly involving inverse trigonometric expressions.

Geometric Interpretation

The geometric interpretation of an integral provides a visual understanding of what the integral represents. In this exercise, we have a geometric setup involving a sector of a circle and a triangle. This allows us to express the integral in terms of these geometric areas.

• Area of Sector OAB: Formulated using the angle \(\theta\) and radius \(a\), the area is derived as \( \frac{1}{2} a^2 \theta\).
• Area of Triangle OBC: Calculated using base \(x\) and height \(\sqrt{a^2 - x^2}\), the area is given by \(\frac{1}{2} x \sqrt{a^2 - x^2}\).

By adding these areas, we match the problem's formulation of \(F(x)\), visually aiding our understanding of the integral as summing these geometric components.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to find angles when specific trigonometric values are known. Here, \( \sin^{-1}(x/a) \) is employed to find \( \theta \), the angle subtended by the radius of the circle.

These functions can be tricky, but they open up solutions in integration and differentiation. For the given function, knowing that \sin(\theta) = x/a allows us to derive: \(\theta = \sin^{-1}(x/a)\).

These relations help transform complex integrals into manageable algebraic expressions that can be integrated or solved.

Integration Techniques

Integration techniques are strategies to simplify the integration of complex functions. This exercise showcases:

• Substitution: By recognizing \( \sin^{-1}(\frac{x}{a})\) as a necessary component, substitution aids in simplifying expressions to derive areas.
• Geometric methods: Utilized here by interpreting areas under curves and integrating to find the expressions for the integral.

These techniques make handling integrals of intricate functions, such as \(\sqrt{a^2 - x^2} \), more approachable. Learning and applying these tactics pave the way for solving more advanced calculus problems.