Problem 20 - Applied Calculus

Question

Using the Fundamental Theorem, evaluate the definite integrals in Problems \(1-20\) exactly. $$\int_{4}^{9} \sqrt{x} d x$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{38}{3} \).

Step 1: Identify the Problem

We need to evaluate the definite integral \( \int_{4}^{9} \sqrt{x} \, dx \). This involves finding the antiderivative (or indefinite integral) of \( \sqrt{x} \) and then applying the limits of integration.

Step 2: Express the Integrand in a Simpler Form

The function \( \sqrt{x} \) can be rewritten as \( x^{1/2} \). This change will make it easier to apply the power rule for integration.

Step 3: Find the Antiderivative

Use the power rule for integration, which states that the antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. Thus, the antiderivative of \( x^{1/2} \) is \( \frac{x^{3/2}}{3/2} \). Simplifying, this becomes \( \frac{2}{3}x^{3/2} \).

Step 4: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, the definite integral \( \int_{a}^{b} f(x) \, dx \) is equal to \( F(b) - F(a) \), where \( F \) is an antiderivative of \( f \). We apply this to \( \frac{2}{3}x^{3/2} \).

Step 5: Evaluate at the Upper Limit

Substitute the upper limit of integration into the antiderivative: \( F(9) = \frac{2}{3}(9)^{3/2} \). Calculating this gives \( 18 \).

Step 6: Evaluate at the Lower Limit

Substitute the lower limit of integration into the antiderivative: \( F(4) = \frac{2}{3}(4)^{3/2} \). Calculating this gives \( \frac{16}{3} \).

Step 7: Compute the Definite Integral

Subtract the value at the lower limit from the value at the upper limit: \( 18 - \frac{16}{3} = \frac{54}{3} - \frac{16}{3} = \frac{38}{3} \).

Step 8: Provide the Final Answer

The exact value of the definite integral \( \int_{4}^{9} \sqrt{x} \, dx \) is \( \frac{38}{3} \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a crucial link in calculus that connects differentiation and integration, two core concepts. It asserts that the definite integral of a function over a particular interval can be found using any of its antiderivatives. This makes solving definite integrals significantly more straightforward. Particularly, the theorem consists of two parts:

• First, it states that if a function is continuous over a segment, then its indefinite integral is differentiable, and its derivative is the original function.
• Second, it establishes that for any antiderivative \( F \) of a continuous function \( f \), the integral from \( a \) to \( b \) of \( f \) is given by \( F(b) - F(a) \).

In our particular problem, we utilized this theorem by first finding an antiderivative for the function \( \sqrt{x} \), which, when evaluated at the specified bounds, yielded the result of the definite integral. This efficient approach transforms the often difficult task of integrating directly into a simpler problem of evaluation.

Antiderivative

An antiderivative is essentially a reverse operation to differentiation. It is a function whose derivative returns the original function. Understanding antiderivatives is vital for solving integral problems, particularly indefinite integrals, which do not have limits of integration.

• Finding an antiderivative can sometimes be straightforward, especially with functions like polynomials or those with well-known integration rules.
• In our problem, the antiderivative of \( \sqrt{x} \) was determined using known formulas, specifically the power rule for integration.

Once the antiderivative is found, it can be used within the Fundamental Theorem of Calculus to evaluate definite integrals. This process involves using the function obtained to evaluate the area under the curve between two points, which is the essence of the definite integral.

Power Rule for Integration

The Power Rule for Integration is a fundamental technique in calculus used to find antiderivatives of polynomial functions. The rule simplifies the integration of power functions, and it states that for any real number \( n eq -1 \), \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \\)Here, \( C \) is the constant of integration, which accounts for any vertical shifts in the original function.

• The power rule is particularly useful because it provides a direct method to handle polynomials and functions expressed as powers.
• In the given exercise, rewriting \( \sqrt{x} \) as \( x^{1/2} \) allowed us to directly apply the power rule. This resulted in \( \frac{x^{3/2}}{3/2} \) or equivalently \( \frac{2}{3}x^{3/2} \).

This method not only provides a systematic way to find antiderivatives but also demonstrates one of the many ways calculus can simplify seemingly complex algebraic expressions to solve integral problems effectively.