Problem 12 - Calculus

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 Integral

The given integral is \( \int_{\pi/6}^{\pi/2} 2 \sin t \, dt \). This is a definite integral which needs to be evaluated using the Second Fundamental Theorem of Calculus.

Step 2: Understand the Second Fundamental Theorem of Calculus

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

Step 3: Find the Antiderivative

Identify an antiderivative of \( 2 \sin t \). Since the antiderivative of \( \sin t \) is \( -\cos t \), an antiderivative of \( 2 \sin t \) is \( -2 \cos t \).

Step 4: Evaluate the Antiderivative at the Bounds

Using the antiderivative \( F(t) = -2 \cos t \), evaluate it at both bounds of the integral: \[ F\left(\frac{\pi}{2}\right) = -2 \cos\left(\frac{\pi}{2}\right) = -2 \times 0 = 0 \]\[ F\left(\frac{\pi}{6}\right) = -2 \cos\left(\frac{\pi}{6}\right) = -2 \times \frac{\sqrt{3}}{2} = -\sqrt{3} \]

Step 5: Apply the Second Fundamental Theorem

Subtract the value of the antiderivative at \( \pi/6 \) from the value at \( \pi/2 \): \[ \int_{\pi/6}^{\pi/2} 2 \sin t \, dt = F\left(\frac{\pi}{2}\right) - F\left(\frac{\pi}{6}\right) = 0 - (-\sqrt{3}) = \sqrt{3} \]

Step 6: Conclusion

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

Key Concepts

Definite Integral

A definite integral is a type of integral that provides the accumulated area under a curve, within a specific range on the x-axis. In other words, it is the integral of a function with upper and lower limits. For the exercise provided, we evaluated the integral from \( \pi/6 \) to \( \pi/2 \). This process includes finding the area under the curve of the function from one point to the other.

When calculating a definite integral, it is important to understand the:

• Limits of integration: These are the bounds, \( a \) and \( b \), that define the interval \( [a, b] \) over which the integration is performed.
• Integrand: The function to be integrated, which in this case is \( 2 \sin t \).

Using the Second Fundamental Theorem of Calculus, we can simplify this process by evaluating the antiderivative at these limits and taking the difference: \( F(b) - F(a) \). This gives you the exact value of the integral within the given bounds, thus providing the "net area" of the curve above the t-axis.

Antiderivative

An antiderivative, often referred to as a primitive function, is a function whose derivative is the original function you started with. Finding the antiderivative is a key part of solving definite integrals, as elucidated by the Second Fundamental Theorem of Calculus.

For instance, when considering the function \( 2 \sin t \), we need to find a function \( F(t) \) such that when differentiated, returns \( 2 \sin t \). By knowing that the derivative of \( \cos t \) is \( -\sin t \), we realize:

• The antiderivative of \( \sin t \) is \( -\cos t \).
• Therefore, the antiderivative of \( 2 \sin t \) is \( -2 \cos t \).

The antiderivative allows us to compute the definite integral by evaluating it at the bounds, producing values that help us find the total area under the curve between these bounds. This is central to solving problems involving integrals as seen in the exercise.

Trigonometric Integration

Trigonometric integration involves integrating functions that are composed of trigonometric functions like sine, cosine, etc. Such integrals often appear in calculus problems and rely on known antiderivatives of trigonometric functions.

In the given problem, integrating \( 2 \sin t \) involves recognizing the basic antiderivative of the sine function:

• The antiderivative of \( \sin t \) is \( -\cos t \).
• By extension, \( 2 \sin t \) has the antiderivative \( -2 \cos t \).

Understanding these basic relationships is essential in trigonometric integration, as it allows us to manipulate the integrals of more complex expressions effectively. Moreover, using trigonometric identities can also aid in evaluating these integrals by simplifying the expressions. In scenarios requiring application, it's these foundational antiderivatives that enable efficient solving of integrals involving sine and cosine, as demonstrated in this exercise.