Problem 40 - Calculus Early Transcendentals

Question

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{1 / 2} \frac{d x}{\sqrt{1-x^{2}}}$$

Step-by-Step Solution

Answer

Question: Evaluate the integral $$\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}} dx $$ Answer: $$\frac{\pi}{6}$$

Step 1: Identify the given function and interval

The given function is $$\frac{1}{\sqrt{1-x^2}}$$ and the interval of integration is $$[0, \frac{1}{2}]$$.

Step 2: Find the antiderivative of the given function

We will find the antiderivative of the given function, which is the function whose derivative is the given function. The antiderivative of our given function is well-known and can be found in a table of standard integrals or derived using a trigonometric substitution. The antiderivative of $$\frac{1}{\sqrt{1-x^2}}$$ is $$\arcsin(x) + C$$, where C is the constant of integration.

Step 3: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that, if a function F(x) is an antiderivative of the function f(x) on the interval [a, b], then: $$\int_{a}^{b} f(x) dx = F(b) - F(a)$$ In our case, F(x) is $$\arcsin(x)+C$$, f(x) is $$\frac{1}{\sqrt{1-x^2}}$$, a is 0, and b is 1/2. Using these values and applying the Fundamental Theorem of Calculus: $$\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}} dx = (\arcsin(\frac{1}{2}) + C) - (\arcsin(0) + C)$$ Since C is a constant and independent of x, we have: $$\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}} dx = \arcsin(\frac{1}{2}) - \arcsin(0) $$

Step 4: Calculate the final result

Now, we just need to plug in the values and find the difference: $$\arcsin(\frac{1}{2}) - \arcsin(0) = \frac{\pi}{6} - 0 = \frac{\pi}{6} $$ So, the result of the integral is: $$\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1-x^2}} dx = \frac{\pi}{6}$$

Key Concepts

Antiderivative

An antiderivative of a function is another function whose derivative gives back the original function. It is the reverse process of differentiation. When finding the indefinite integral of a function, you are essentially looking for its antiderivative. For example, for the function \( \frac{1}{\sqrt{1-x^2}} \), its antiderivative is \( \arcsin(x) + C \). The constant \( C \) represents an entire family of functions that differ only by a constant.​

To find the antiderivative, you might use various methods, such as basic integration formulas, substitution, or more complex techniques like trigonometric substitution. Understanding antiderivatives is crucial for evaluating integrals over a specific interval—especially when applying the Fundamental Theorem of Calculus. This theorem links the concept of the antiderivative with the process of evaluating definite integrals.

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to simplify integrals involving square roots, especially those like \( \frac{1}{\sqrt{1-x^2}} \). By transforming the variable \( x \) into a trigonometric function, you can often simplify the integral into a more manageable form. For instance:

• Use \( x = \sin(\theta) \) when faced with \( \sqrt{1-x^2} \).
• The derivative \( dx = \cos(\theta) d\theta \) follows from this substitution.

After substitution, the integral changes in terms of \( \theta \):
\[ \int \frac{1}{\sqrt{1-(\sin(\theta))^2}} \cos(\theta) d\theta \]
This simplifies using the identity \( 1-(\sin(\theta))^2 = (\cos(\theta))^2 \). Using trigonometric identities can help transform the integral into a simpler trigonometric form. Once simplified, you perform the integral in terms of \( \theta \) and then substitute back to get the result in terms of the original variable \( x \). Using trigonometric substitution can also make finding antiderivatives more intuitive and straightforward.

Definite Integral Evaluation

Evaluating a definite integral involves calculating the exact area under the curve of a function between two points. This is where the Fundamental Theorem of Calculus is particularly useful. This theorem states that if \( F(x) \) is an antiderivative of \( f(x) \), then:
\[ \int_{a}^{b} f(x)\, dx = F(b) - F(a) \]
For the integral \( \int_{0}^{1/2} \frac{1}{\sqrt{1-x^2}} \, dx \), we use \( F(x) = \arcsin(x) \). Then evaluate \( F(1/2) - F(0) \):

• At \( x = \frac{1}{2} \), \( \arcsin(\frac{1}{2}) = \frac{\pi}{6} \).
• At \( x = 0 \), \( \arcsin(0) = 0 \).

Thus, the definite integral from 0 to \( \frac{1}{2} \) simplifies to \( \frac{\pi}{6} - 0 = \frac{\pi}{6} \). This means that the area under the curve \( \frac{1}{\sqrt{1-x^2}} \) from \( x = 0 \) to \( x = \frac{1}{2} \) is \( \frac{\pi}{6} \). This application illustrates how definite integral evaluation not only provides a numerical result but also describes the mathematical importance of functions over intervals.