Problem 184 - Calculus Volume 1

Question

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{0}^{\pi / 4} \sec ^{2} \theta d \theta $$

Step-by-Step Solution

Answer

The value of the definite integral is 1.

Step 1: Identify the Integrand and Limits

The exercise presents the definite integral \( \int_{0}^{\pi / 4} \sec^{2} \theta \, d\theta \), where \( \sec^{2} \theta \) is the integrand, and the limits of integration are from \( 0 \) to \( \frac{\pi}{4} \).

Step 2: Determine the Antiderivative

The antiderivative of \( \sec^{2} \theta \) is \( \tan \theta \). This means that when we find the indefinite integral, we have \( \int \sec^{2} \theta \, d\theta = \tan \theta + C \), where \( C \) is the constant of integration.

Step 3: Apply the Fundamental Theorem of Calculus, Part 2

We apply the Fundamental Theorem of Calculus, Part 2, which states that if \( F \) is an antiderivative of \( f \) on an interval \( [a, b] \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). With \( F(\theta) = \tan \theta \), we evaluate \( F \) at the bounds: \( F(\frac{\pi}{4}) - F(0) \).

Step 4: Evaluate the Antiderivative at the Upper Limit

Calculate \( F(\frac{\pi}{4}) = \tan(\frac{\pi}{4}) = 1 \), since \( \tan(\frac{\pi}{4}) \) is \( 1 \).

Step 5: Evaluate the Antiderivative at the Lower Limit

Calculate \( F(0) = \tan(0) = 0 \), since \( \tan(0) \) is \( 0 \).

Step 6: Compute the Definite Integral

Now substitute back to obtain \( \int_{0}^{\pi/4} \sec^{2} \theta \, d\theta = F(\frac{\pi}{4}) - F(0) = 1 - 0 = 1 \). Therefore, the value of the definite integral is \( 1 \).

Key Concepts

Definite Integral

A definite integral is a way of finding the total area under a curve that is defined by a function. In our example, we are considering the definite integral \( \int_{0}^{\pi/4} \sec^{2} \theta \, d\theta \). The definite integral is described by two main components:

• The integrand, which is the function you're integrating, here being \( \sec^2 \theta \)
• The limits of integration, which indicate the interval over which the integration occurs, which in this case are from \(0\) to \(\frac{\pi}{4}\).

The process of evaluating a definite integral involves taking the antiderivative of the integrand and then applying the limits. Using the Fundamental Theorem of Calculus, Part 2, this gives us the net result of integrating over a specific interval.

Antiderivative

The antiderivative, also known as an indefinite integral, is a function whose derivative is the original function you started with. For example, the antiderivative of \( \sec^2 \theta \) is \( \tan \theta \). This means that \( \frac{d}{d\theta} \tan \theta = \sec^2 \theta \).When you find the antiderivative, it includes a constant of integration, denoted by \( C \). However, when calculating a definite integral, this constant cancels out, so you don't need to worry about it. In our problem, we identify that \( \int \sec^2 \theta \, d\theta = \tan \theta + C \). This step is crucial because, to evaluate the definite integral, we need the antiderivative to apply the Fundamental Theorem of Calculus.

Integrand

The integrand is the function you are integrating over a specified range. In our example, the integrand is \( \sec^2 \theta \). The choice of integrand determines the shape of the curve whose area you're calculating between the limits of integration.

• For example, if the integrand is a constant, you are essentially computing the area of a rectangle.
• If the integrand is a linear function, the area might be a trapezoid.

Knowing the integrand helps you decide what kind of mathematical process, such as differentiation or trigonometric substitution, might be required to solve the integral.

Limits of Integration

The limits of integration are the two values at the lower and upper ends of the interval over which you perform the integration. In \( \int_{0}^{\pi/4} \sec^{2} \theta \, d\theta \), we have:

• The lower limit is \( 0 \)
• The upper limit is \( \frac{\pi}{4} \)

These limits represent the starting and ending points on the \( \theta \)-axis for your integration. By calculating the antiderivative at these points, you can determine the overall change in the function value from the starting to the ending point. This computes the definite integral, allowing you to understand the accumulated value of the integrand across the interval. Importantly, limits of integration aid in capturing the exact segment of the graph and ensure that the definite integral has a specific, finite value.