Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{\pi / 6}^{\pi / 2} 2 \sin t d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( \sqrt{3} \).

Step 1: Identify the Integrand and Limits of Integration

The exercise provides the definite integral \( \int_{\pi / 6}^{\pi / 2} 2 \sin t \, dt \). Here, the integrand is \( 2 \sin t \) and the limits of integration are \( t = \pi / 6 \) and \( t = \pi / 2 \).

Step 2: Find the Indefinite Integral

Determine the antiderivative of the integrand \( 2 \sin t \). The antiderivative of \( \sin t \) is \( -\cos t \). Thus, the antiderivative of \( 2 \sin t \) is \( -2 \cos t \).

Step 3: Apply the Second Fundamental Theorem of Calculus

According to the Second Fundamental Theorem of Calculus, the definite integral \( \int_{a}^{b} f(x) \, dx \) is equal to the difference of the antiderivative evaluated at the upper and lower limits.Here, evaluate \( -2 \cos t \) at the upper limit \( t = \pi / 2 \) and lower limit \( t = \pi / 6 \).

Step 4: Evaluate at the Upper Limit

Substitute \( t = \pi / 2 \) into the antiderivative:\[ -2 \cos(\pi / 2) = -2 \times 0 = 0 \](Since \( \cos(\pi / 2) = 0 \)).

Step 5: Evaluate at the Lower Limit

Substitute \( t = \pi / 6 \) into the antiderivative:\[ -2 \cos(\pi / 6) = -2 \times \frac{\sqrt{3}}{2} = -\sqrt{3} \](Since \( \cos(\pi / 6) = \frac{\sqrt{3}}{2} \)).

Step 6: Compute the Definite Integral

Subtract the evaluation at the lower limit from the evaluation at the upper limit:\[ 0 - (-\sqrt{3}) = \sqrt{3} \]

Step 7: Conclusion

Thus, the value of the definite integral \( \int_{\pi / 6}^{\pi / 2} 2 \sin t \, dt \) is \( \sqrt{3} \).

Key Concepts

Definite Integrals

A definite integral is a concept that calculates the total accumulation of quantities, such as area under a curve, between two specific points on the x-axis. In simpler terms, it can be thought of as the sum of infinitely small parts of an area or value, bounded by specified start and end points. The expression \[ \int_{a}^{b} f(x) \, dx \] denotes a definite integral where:

• \( a \) is the lower limit of integration, also the starting point.
• \( b \) is the upper limit of integration, also the endpoint.
• \( f(x) \) is the function being integrated, known as the integrand.

When solving a definite integral, such as the one presented in the original exercise, you find the total accumulation between these two points by first determining an antiderivative. Then, using the Second Fundamental Theorem of Calculus, you evaluate the antiderivative at both limits and find their difference.

Antiderivatives

An antiderivative, sometimes called a primitive function, is essentially the "reverse" process of differentiation. While differentiation helps find the rate of change, finding an antiderivative helps one recover the original function from its derivative. Consider a function \( F(x) \), if \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).
When tackling a definite integral, finding the correct antiderivative is crucial. For example, if the integrand is \( 2 \sin t \), the antiderivative involves identifying the inverse of the derivative of \( \sin t \). Knowing that the derivative of \( \cos t \) is \(-\sin t \), it follows that the antiderivative of \( \sin t \) is \(-\cos t \). Therefore, the antiderivative of \( 2 \sin t \) becomes \(-2 \cos t \).
The key is understanding that for every function, there can be multiple antiderivatives, differing by a constant, but for definite integrals, this constant cancels out, focusing solely on net changes between bounds.

Trigonometric Functions

Trigonometric functions are fundamental in calculus and are essential when dealing with problems involving angles, oscillations, and waves. The main trigonometric functions are sine, cosine, and tangent, each having certain key properties and derivatives.
The sine function, denoted \( \sin t \), represents the y-coordinate of a point on the unit circle as the angle \( t \) varies. Its derivative is \( \cos t \), indicating how the slope of \( \sin t \) changes as \( t \) increases. Conversely, the antiderivative of \( \sin t \) is \(-\cos t \).
In the given exercise, the integral involves the sine function. To solve it, understanding the trigonometric identities is vital:

• \( \cos(\pi/2) = 0 \)
• \( \cos(\pi/6) = \frac{\sqrt{3}}{2} \)

Recognizing these values allows for accurate evaluation of the antiderivative at given limits, which is a pivotal step in calculating the definite integral. This understanding reinforces the interplay between calculus and trigonometry, crucial for mastering integral calculus concepts.