Problem 179 - Calculus Volume 1

Question

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{1}^{4} \frac{1}{2 \sqrt{x}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to 1.

Step 1: Identify the Integrand

The given integrand is \( \frac{1}{2 \sqrt{x}} \). Before applying the integration, it helps to rewrite this in a more workable form. The square root \( \sqrt{x} \) can be rewritten as \( x^{1/2} \), making the integrand \( \frac{1}{2} x^{-1/2} \) for easier computation.

Step 2: Find the Antiderivative

According to the rules of integration, the antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \). Therefore, the antiderivative of \( \frac{1}{2}x^{-1/2} \) is \( \frac{1}{2} \cdot \frac{x^{1/2}}{1/2} = x^{1/2} = \sqrt{x} \).

Step 3: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus, Part 2, states that if \( F(x) \) is an antiderivative of \( f(x) \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). In this case, our antiderivative \( F(x) = \sqrt{x} \).

Step 4: Evaluate at the Upper Limit

Substitute \( x = 4 \) into the antiderivative: \( F(4) = \sqrt{4} = 2 \).

Step 5: Evaluate at the Lower Limit

Substitute \( x = 1 \) into the antiderivative: \( F(1) = \sqrt{1} = 1 \).

Step 6: Compute the Definite Integral

Using the results from the upper and lower limits, calculate \( F(4) - F(1) = 2 - 1 = 1 \).

Key Concepts

Definite Integral

A definite integral is a fundamental tool in calculus that calculates the accumulation of quantities, expressed as the area under a curve between two points. In this exercise, the definite integral \( \int_{1}^{4} \frac{1}{2 \sqrt{x}} \, dx \) measures the total area beneath the function from \( x = 1 \) to \( x = 4 \). It provides a way to sum infinitely small changes in the function's value over this range.
The process of finding a definite integral involves three key steps:

• Rewriting the function into a more manageable form if necessary. For instance, converting \( \frac{1}{2 \sqrt{x}} \) to \( \frac{1}{2} x^{-1/2} \) for easier integration.
• Determining the antiderivative or the original function that gives the derivative of the integrand.
• Applying the Fundamental Theorem of Calculus to evaluate the integral.

This theorem asserts that we can calculate a definite integral by evaluating the antiderivative at the upper and lower bounds, and finding their difference.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the given function. In the context of our exercise, it involves reversing the process of differentiation to find the original function from a given derivative.
For example, we found the integrand \( \frac{1}{2}x^{-1/2} \) and determined its antiderivative using

• The rule that states the antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \), where \( n eq -1 \).
• Applying this rule provided \( x^{1/2} \), which simplifies back to \( \sqrt{x} \). This function, \( \sqrt{x} \), returns \( \frac{1}{2 \sqrt{x}} \) when differentiated.

Finding an antiderivative is crucial because it allows us to use the Fundamental Theorem of Calculus to compute the exact value of definite integrals.

Integration Rules

Integration rules are a set of guidelines that simplify the process of finding integrals, much like differentiation rules help finding derivatives. These rules are essential for efficiently computing antiderivatives, especially for functions that are more complex.
Here are a few key rules:

• Power Rule: The antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} \), provided \( n eq -1 \).
• Constant Multiple Rule: The antiderivative of \( c \cdot f(x) \) is \( c \cdot F(x) \), where \( c \) is a constant.
• Sum Rule: The antiderivative of \( f(x) + g(x) \) is \( F(x) + G(x) \).

This exercise primarily utilized the Power Rule to transform and find the antiderivative of the function. Mastery of these rules enables one to tackle more complex integrals with confidence, reinforcing the foundational concepts of calculus.