Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, 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 value of the integral is \( \sqrt{3} \).

Step 1: Identify the Functions

Identify the function inside the integral. In this case, the function is \( 2 \sin t \). The limits of integration are from \( \frac{\pi}{6} \) to \( \frac{\pi}{2} \).

Step 2: Find the Antiderivative

According to the Second Fundamental Theorem of Calculus, we first need to find the antiderivative of \( 2 \sin t \). The antiderivative of \( \sin t \) is \(-\cos t \), so the antiderivative of \( 2 \sin t \) is \(-2 \cos t \).

Step 3: Evaluate the Antiderivative at Upper Limit

Substitute the upper limit \( t = \frac{\pi}{2} \) into the antiderivative: \(-2 \cos \frac{\pi}{2} = -2 \times 0 = 0 \).

Step 4: Evaluate the Antiderivative at Lower Limit

Substitute the lower limit \( t = \frac{\pi}{6} \) into the antiderivative: \(-2 \cos \frac{\pi}{6} = -2 \times \frac{\sqrt{3}}{2} = -\sqrt{3} \).

Step 5: Subtract Lower Limit from Upper Limit

Apply the Second Fundamental Theorem of Calculus, which tells us to subtract the antiderivative evaluated at the lower limit from the antiderivative evaluated at the upper limit: \(0 - (-\sqrt{3}) = \sqrt{3} \). Thus, the value of the integral is \( \sqrt{3} \).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that help us find the accumulated sum of a function over a specific interval. They are represented with integral signs that include lower and upper limits. In essence, a definite integral gives you the net area between the function and the horizontal axis, from the lower to the upper limit.
When dealing with definite integrals, you evaluate the antiderivative at these two limits and take the difference. This subtraction process, guided by the Second Fundamental Theorem of Calculus, is crucial as it simplifies finding areas under curves effectively. In our example, it converts the problem of finding the exact area or accumulation from \( \frac{\pi}{6} \) to \( \frac{\pi}{2} \) into a straightforward calculation using antiderivatives.

Antiderivative

An antiderivative is a function that "reverses" differentiation, allowing us to retrieve the original function from its derivative. If differentiation is about finding slopes, then antiderivation helps recover the actual function from those slopes. It can be thought of as the "opposite" of taking a derivative.
In practical terms, to solve an integral problem, we often need the antiderivative. For example, in our integral \( \int 2 \sin t \), the antiderivative is \(-2 \cos t\). Here, \(-2 \cos t\) is the function whose derivative gives \(2 \sin t\). This reversal is vital in solving integrals and finding exact areas or accumulated values over a specified interval.

Calculus Integration

Calculus integration involves combining the methods of finding sums (integration) and differences (differentiation) to solve problems involving curves and areas. This process is at the heart of analyzing and understanding the behavior of functions over specific intervals.
Integration can be indefinite or definite, with indefinite providing a general formula without specific limits and definite integrating over a specific range, such as \( \frac{\pi}{6} \) to \( \frac{\pi}{2} \) in our exercise. In calculus, integration allows us to handle various real-world applications, from calculating areas and volumes to solving differential equations. It is essentially about accumulation and understanding how changes build up over a given period.

Trigonometric Integrals

Integrals involving trigonometric functions, known as trigonometric integrals, are a special class within calculus. These functions include sine, cosine, tangent, and their inverses, among others. They often appear in physics and engineering due to their periodic nature which models waves and oscillations.
When integrating trigonometric functions, it is crucial to determine their antiderivatives correctly, as these form the basis of solving the integrals. For example, knowing that the antiderivative of \( \sin t \) is \(- \cos t \) lets us solve integrals like \( \int 2 \sin t \) efficiently. Mastering trigonometric integrals expands our toolkit for tackling a wide range of mathematical problems in various scientific fields.