Problem 6 - Applied Calculus

Question

Using the Fundamental Theorem, evaluate the definite integrals in problem exactly. $$ \int_{1}^{4} \frac{1}{\sqrt{x}} d x $$

Step-by-Step Solution

Answer

The value of the integral is 2.

Step 1: Identify the Integral

The integral we need to evaluate is \( \int_{1}^{4} \frac{1}{\sqrt{x}} \, dx \). This is a definite integral with limits of integration from 1 to 4.

Step 2: Find the Antiderivative

The function \( \frac{1}{\sqrt{x}} \) can be rewritten as \( x^{-1/2} \). The antiderivative of \( x^{-1/2} \) is \( 2x^{1/2} \). Thus, the antiderivative of \( \frac{1}{\sqrt{x}} \) is \( 2\sqrt{x} \).

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, to evaluate \( \int_{a}^{b} f(x) \, dx \), we find the antiderivative \( F(x) \) and compute \( F(b) - F(a) \). Here, \( F(x) = 2\sqrt{x} \), so we calculate \( 2\sqrt{4} - 2\sqrt{1} \).

Step 4: Calculate and Simplify

Substitute the limits of integration: \( 2\sqrt{4} = 2 \times 2 = 4 \) and \( 2\sqrt{1} = 2 \times 1 = 2 \). Therefore, \( 4 - 2 = 2 \).

Key Concepts

Definite Integral

The concept of a definite integral is a fundamental idea in calculus. It represents the accumulation of quantities, such as area under a curve, between two specific points, known as the limits of integration. In essence, the definite integral over an interval \[ [a, b] \] provides the net area between the graph of a function and the x-axis.

• For the integral \( \int_{a}^{b} f(x)\, dx \), \( a \) and \( b \) are the limits of integration.
• The result is a number, indicating the total accumulated quantity between \( x = a \) and \( x = b \).
• Unlike indefinite integrals, a definite integral has boundaries and yields a specific numerical result.

When solving definite integrals, the Fundamental Theorem of Calculus plays a crucial role, connecting derivatives with antiderivatives to simplify calculations and find exact values.

Antiderivative

Antiderivatives are the opposite of derivatives. While a derivative gives the rate at which a function changes, an antiderivative gives us the original function, or a family of functions, that caused that rate of change.

• The antiderivative, sometimes represented by \( F(x) \), is a function such that \( F'(x) = f(x) \).
• To find an antiderivative of \( \frac{1}{\sqrt{x}} \), we re-write it as \( x^{-1/2} \).
• The antiderivative of \( x^{-1/2} \) is \( 2x^{1/2} \) because when we differentiate \( 2x^{1/2} \), we get back to \( x^{-1/2} \).

Understanding how to find antiderivatives is essential for evaluating definite integrals since it allows us to use the Fundamental Theorem of Calculus effectively.

Limits of Integration

The limits of integration are crucial when dealing with definite integrals.They define the interval over which integration occurs, essentially framing the area or quantity of interest.

• The lower limit is \( a \) and the upper limit is \( b \), specifying the interval \([a, b]\).
• In our example, \( a = 1 \) and \( b = 4 \), which tells us that we're evaluating the integral from \( x = 1 \) to \( x = 4 \).
• When applying the antiderivative over the limits of integration, we compute the difference: \( F(b) - F(a) \).

This difference gives us the net accumulation between these two points, which is a principle directly derived from the Fundamental Theorem of Calculus. It's essential to substitute the limits correctly to ensure accurate results.

Evaluate Integrals

Evaluating integrals involves a few steps, especially when dealing with definite integrals. We apply the antiderivative and compute the difference over the specified limits of integration.

Steps to Evaluate a Definite Integral:

• Find the antiderivative \( F(x) \) of the integrand \( f(x) \).
• Use the Fundamental Theorem of Calculus: calculate \( F(b) - F(a) \).
• Substitute the upper and lower limits into the antiderivative and perform the arithmetic.

In our example, this evaluation process led us to calculate \( 2\sqrt{4} - 2\sqrt{1} \), resulting in a total accumulation of 2. This method illustrates how powerful calculus can be in solving real-world problems efficiently.