Problem 3 - Thomas Calculus

Question

The integrals in Exercises \(1-34\) converge. Evaluate the integrals without using tables. $$\int_{0}^{1} \frac{d x}{\sqrt{x}}$$

Step-by-Step Solution

Answer

The integral evaluates to 2.

Step 1: Rewrite the Integral

The integral \( \int_{0}^{1} \frac{d x}{\sqrt{x}} \) can be rewritten using the exponent form of the square root. Note that \( \frac{1}{\sqrt{x}} = x^{-1/2} \). Thus, the integral becomes:\[\int_{0}^{1} x^{-1/2} \, dx\]

Step 2: Use the Power Rule for Integration

To integrate \( x^{-1/2} \), apply the power rule for integration, which states \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \). Here, \( n = -\frac{1}{2} \), so:\[\int x^{-1/2} \, dx = \frac{x^{1/2}}{1/2} = 2x^{1/2} + C\]

Step 3: Evaluate the Definite Integral

We now evaluate the definite integral from 0 to 1. Using the antiderivative found in Step 2, compute:\[2x^{1/2}\bigg|_{0}^{1} = \left[ 2(1)^{1/2} - 2(0)^{1/2} \right] = 2(1) - 0 = 2\]

Step 4: Conclude the Evaluation

After evaluating the expression, we can conclude that the value of the integral \( \int_{0}^{1} \frac{d x}{\sqrt{x}} \) is 2. This concludes our solution.

Key Concepts

Power Rule for Integration

The power rule for integration is a fundamental technique in calculus used to integrate functions of the form \( x^n \). It is especially useful when dealing with polynomials or expressions that can be expressed using exponents. This rule allows us to find the antiderivative of a function by increasing the power by one and then dividing by the new exponent. Here's the formula for the power rule:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

where \( n eq -1 \) and \( C \) is the constant of integration. This rule is applicable only when \( n \) is not equal to -1 because this case would result in division by zero. For our specific problem, we had \( x^{-1/2} \) as the integrand, which fits perfectly within the realm of the power rule. By applying this rule, we found that the integral of \( x^{-1/2} \) is \( 2x^{1/2} + C \).

Antiderivative

An antiderivative is essentially the reverse of taking a derivative. It is a function whose derivative is the original function. For instance, if the derivative of function \( F(x) \) is \( f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \). In integration, finding an antiderivative is crucial because it enables us to compute integrals.

When given a specific interval, as in a definite integral, the antiderivative allows us to evaluate the total area under the curve from the lower to the upper limit. In our problem, the antiderivative of \( x^{-1/2} \) calculates the area under the curve from 0 to 1, giving us a value of 2. The antiderivative is thus a key part of solving definite integrals.

Exponent Notation

Exponent notation is a way of expressing numbers where a base number is raised to a power or exponent. This method simplifies the representation of very large or very small numbers and is critical in calculus for simplifying expressions before integration.

• The exponent tells us how many times to multiply the base by itself.
• For example, \( x^{1/2} \) is the square root of \( x \), and \( x^{-1/2} \) is the reciprocal of the square root of \( x \).

In our problem, we initially had \( \frac{1}{\sqrt{x}} \). By rewriting it in exponent notation as \( x^{-1/2} \), we made it compatible with the power rule for integration. This conversion is a common approach to simplify functions and make integration processes more straightforward. Mastering exponent notation can greatly ease dealing with complicated integral calculations.