Problem 25 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^{\pi}_{\pi/6} \sin \theta \,d\theta \)

Step-by-Step Solution

Answer

The integral evaluates to \( 1 + \frac{\sqrt{3}}{2} \).

Step 1: Identify the Integral

The given integral is \( \int_{\pi/6}^{\pi} \sin \theta \, d\theta \). We need to evaluate this definite integral.

Step 2: Find the Antiderivative

The antiderivative of \( \sin \theta \) is \( -\cos \theta \). So, the indefinite integral is \( -\cos \theta + C \), where \( C \) is the constant of integration.

Step 3: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). Here, \( F(\theta) = -\cos \theta \).

Step 4: Evaluate the Antiderivative at the Bounds

Evaluate \( F(\theta) \) at the upper limit \( \pi \) and the lower limit \( \pi/6 \). Compute:- \( F(\pi) = -\cos \pi = -(-1) = 1 \)- \( F(\pi/6) = -\cos (\pi/6) = -\left( \frac{\sqrt{3}}{2} \right) \).

Step 5: Calculate the Definite Integral

Subtract \( F(\pi/6) \) from \( F(\pi) \):\[\int_{\pi/6}^{\pi} \sin \theta \, d\theta = F(\pi) - F(\pi/6) = 1 - \left( -\frac{\sqrt{3}}{2} \right) = 1 + \frac{\sqrt{3}}{2}\]

Step 6: Conclusion

The evaluated integral is \( 1 + \frac{\sqrt{3}}{2} \).

Key Concepts

Antiderivative

In calculus, finding an antiderivative for a function is a crucial step towards solving integrals. An antiderivative is essentially the reverse of differentiation. It refers to a function whose derivative matches the function you started with. Knowing how to identify and use antiderivatives is important when evaluating integrals.
An antiderivative of a given function is not unique, as we can add any constant to it. This constant is symbolized as 'C' and represents the family of all possible antiderivatives. For example, the antiderivative of \( \sin \theta \) is \( -\cos \theta + C \).
Understanding antiderivatives allows us to transition from an indefinite integral to a definite one. In the given problem, we only care about the bounds, so the constant 'C' doesn't affect our result. By identifying \( -\cos \theta \) as the antiderivative, we can apply further calculus principles to solve the definite integral.

Fundamental Theorem of Calculus

This theorem bridges the world of differentiation and integration seamlessly, allowing us to compute definite integrals with precision. It tells us that if we have an antiderivative \( F \) of a function \( f \), then the definite integral of \( f \) from \( a \) to \( b \) is \( F(b) - F(a) \).
Here's the essential benefit: instead of summing an infinite number of areas under the curve, we simply plug the upper and lower bounds into the antiderivative. For our context, after finding the antiderivative \( F(\theta) = -\cos \theta \), we use this theorem to evaluate the integral: \( \int_{\pi/6}^{\pi} \sin \theta \, d\theta = F(\pi) - F(\pi/6) \).
By working through this formula, we're able to achieve the solution in a more efficient way than attempting to sum infinite small parts. Thus, understanding this theorem is key to successfully navigating calculus problems where both integration and differentiation are involved.

Trigonometric Integrals

When we encounter integrals, particularly trigonometric ones, it's vital to recognize their patterns and behaviors. Trigonometric functions like \( \sin \theta \) and \( \cos \theta \) are periodic, which plays a role in solving their integrals.
In this particular problem, \( \int \sin \theta \, d\theta \), trigonometric functions have standard antiderivatives: the antiderivative of \( \sin \theta \) is \( -\cos \theta \). Recognizing these can quickly simplify solving an integral.
Familiarity with trigonometric identities can also offer shortcuts or simplify problems further, particularly when functions are not alone in the integral. For beginners, being able to visualize these functions within their periodic curves—understanding where they start and stop within an interval—can provide deeper insight into the behavior of these integrals. Ultimately, grasping these fundamentals is essential for tackling more complex calculus problems successfully.