Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, use the Second Fundamental Theorem of Calculus to evaluate each definite integral. \(\int_{0}^{4} \sqrt{t} d t\)

Step-by-Step Solution

Answer

The definite integral is \( \frac{16}{3} \).

Step 1: Identify the Function and Limits

The definite integral given is \( \int_{0}^{4} \sqrt{t} \, dt \). Here, the integrand function is \( \sqrt{t} \), and the limits of integration are from 0 to 4.

Step 2: Find the Antiderivative

The antiderivative of \( \sqrt{t} \) is needed. Rewrite \( \sqrt{t} \) as \( t^{1/2} \). The antiderivative of \( t^{n} \) is \( \frac{t^{n+1}}{n+1} \). Thus, the antiderivative of \( t^{1/2} \) is \( \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2} \).

Step 3: Apply the Second Fundamental Theorem of Calculus

According to the Second Fundamental Theorem of Calculus, evaluate the antiderivative from the lower limit to the upper limit. Thus, calculate \( F(b) - F(a) \) where \( F(t) = \frac{2}{3} t^{3/2} \), \( b = 4 \), and \( a = 0 \).

Step 4: Evaluate at the Upper Limit

Substitute \( t = 4 \) into the antiderivative: \( F(4) = \frac{2}{3}(4)^{3/2} \).Calculate: \( 4^{3/2} = (\sqrt{4})^{3} = 2^{3} = 8 \).Thus, \( F(4) = \frac{2}{3} \cdot 8 = \frac{16}{3} \).

Step 5: Evaluate at the Lower Limit

Substitute \( t = 0 \) into the antiderivative: \( F(0) = \frac{2}{3}(0)^{3/2} = 0 \).

Step 6: Compute the Result

Subtract the result from Step 5 from the result in Step 4: \( F(4) - F(0) = \frac{16}{3} - 0 = \frac{16}{3} \).

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that represents the area under a curve between two points, known as the limits of integration. This type of integral is represented as \[ \int_{a}^{b} f(x) \, dx \]where \( a \) and \( b \) are the lower and upper limits, respectively. Here, \( f(x) \) is the function being integrated, also known as the integrand.

When evaluating a definite integral, the area is calculated precisely under the curve from \( x = a \) to \( x = b \). This is accomplished by using the antiderivative of the function, and applying the Second Fundamental Theorem of Calculus.

• The definite integral provides the net area, considering both above and below the x-axis.
• It is different from an indefinite integral, which does not have limits and results in a general form equation.
• The result of a definite integral is a specific value, indicating the total accumulated area.

Understanding definite integrals allows us to solve problems involving real-world measurements, such as distances, areas, and even cumulative quantities.

Antiderivative

An antiderivative of a function is essentially the reverse of differentiation. It is a function whose derivative yields the original function. When we talk about finding an antiderivative, we are performing the integral operation in which we seek a function \( F(x) \) such that \( F'(x) = f(x) \).

In the context of the exercise problem, the antiderivative of \( \sqrt{t} \) plays a crucial role in computing the definite integral. By rewriting \( \sqrt{t} \) as \( t^{1/2} \), we use the power rule for integration:\[ \int t^{n} \, dt = \frac{t^{n+1}}{n+1} + C \]Here, \( C \) denotes the integration constant, which is typically omitted for definite integrals since they result in numerical values.

• The process of finding an antiderivative is integral to solving definite integrals.
• An antiderivative is not unique; the set of all antiderivatives of a function is called an indefinite integral.
• For the definite integral, we specifically consider the change in the antiderivative over the given interval.

Mastering antiderivatives opens up a range of problem-solving abilities within calculus, especially in contexts involving continuous functions.

Limits of Integration

Limits of integration are the values at which the integration process starts and ends in a definite integral. These limits define the interval over which the area under the curve is accumulated. In the expression \( \int_{0}^{4} \sqrt{t} \, dt \), the numbers \( 0 \) and \( 4 \) are the limits of integration.

Understanding and properly applying limits of integration is crucial for evaluating definite integrals. They determine the exact bounds of integration and directly influence the resulting value.

• The lower limit indicates where the integration begins.
• The upper limit shows where the integration concludes.
• By applying these limits to the antiderivative, it's possible to find the definite integral's precise value.

Additionally, the evaluation involves substituting these limits into the antiderivative and then taking the difference, a fundamental step emphasized by the Second Fundamental Theorem of Calculus.

Calculus Problem-Solving

Solving calculus problems, like evaluating definite integrals, requires a blend of understanding multiple concepts and procedures. The process typically includes:

• Identifying the function and type of the integral.
• Determining the antiderivative of the given function.
• Applying the Second Fundamental Theorem of Calculus.
• Calculating the values using the limits of integration.
• Computing the final result.

This problem-solving method is essential not just for academic exercises but also for practical applications in physics, engineering, and economics.

By building a strong foundation in these methods, students can tackle complex real-world problems that involve continuous functions and their integrals. The blend of technique and practice solidifies understanding and prepares learners for advanced topics in calculus. Patience and practice are key in mastering these problems, as each step intertwines with fundamental calculus principles.