Problem 32 - Applied Calculus

Question

Use a definite integral to find the area under each curve between the given \(x\) -values. For Exercises \(19-24\) also make a sketch of the curve showing the region. $$ f(x)=9-3 \sqrt{x} \text { from } x=0 \text { to } x=9 $$

Step-by-Step Solution

Answer

The area under the curve is 27 square units.

Step 1: Set up the Integral

To find the area under the curve between the given values, we need to set up a definite integral for the function \(f(x) = 9 - 3\sqrt{x}\). The integral will be from \(x = 0\) to \(x = 9\): \[\int_{0}^{9} (9 - 3\sqrt{x}) \, dx\].

Step 2: Integrate the Function

Now, solve the integral: \(\int (9 - 3\sqrt{x}) \, dx\). Split the integral into two parts: \(\int 9 \, dx - \int 3\sqrt{x} \, dx\).

Step 3: Evaluate Each Part Separately

The first part: \(\int 9 \, dx = 9x\). The second part: \(\int 3\sqrt{x} \, dx = \int 3x^{1/2} \, dx\). Use the power rule to integrate: \[3 \cdot \frac{x^{3/2}}{3/2} = 2x^{3/2}\].

Step 4: Combine Integrals

Combine the results of the integration: \[9x - 2x^{3/2}\].

Step 5: Evaluate the Definite Integral

Now substitute the boundaries 0 and 9 into the combined function: \[\left[ 9x - 2x^{3/2} \right]_{0}^{9}\].

Step 6: Evaluate at Upper and Lower Bound

First, evaluate at \(x = 9\): \(9(9) - 2(9^{3/2}) = 81 - 2 \times 27 = 27\). Then, evaluate at \(x = 0\): \(9(0) - 2(0^{3/2}) = 0\).

Step 7: Calculate the Area

Subtract the value at the lower bound from the value at the upper bound to find the area: \(27 - 0 = 27\). This is the area under the curve from \(x = 0\) to \(x = 9\).

Key Concepts

Area Under Curve

The area under a curve in calculus is a way to measure the space between the x-axis and the function graph over a specific interval. This measurement is crucial especially when dealing with functions that are not straight lines. The mathematical tool for finding this area is called the definite integral.

Calculating the area under a curve involves interpreting the integral as a sum of infinitesimally small rectangles. Each of these rectangles is as tall as the function value and as wide as the smallest possible dx, effectively covering the area under the curve. In our specific exercise, we calculate the area under the curve represented by the function \( f(x) = 9 - 3\sqrt{x} \) from \( x = 0 \) to \( x = 9 \).

The definite integral of this function over the specified x-values gives us this area, which we determined to be 27 square units. This area signifies the total space beneath the curve and above the x-axis within the given interval.

Integration Techniques

Integration is a fundamental concept in calculus, essential for solving many problems associated with finding areas under curves, volumes, and other things. In solving the function \( f(x) = 9 - 3\sqrt{x} \), we employ the technique of splitting the integral and using basic integration rules.

Here's a quick breakdown of what these techniques involve:

• First, we split the integral \( \int_{0}^{9} (9 - 3\sqrt{x}) \, dx \) into two separate integrals: \( \int_{0}^{9} 9 \, dx \) and \( \int_{0}^{9} -3\sqrt{x} \, dx \).
• The integration of a constant, \( \int_{0}^{9} 9 \, dx \), results in \(9x\).
• For the integral of \( \sqrt{x} \), we rewrite it as \( x^{1/2} \) and apply the power rule of integration: \( \int x^{n} \, dx = \frac{x^{n+1}}{n+1} \).
• Applying this rule, we get \( 3 \cdot \frac{x^{3/2}}{3/2} = 2x^{3/2} \).

Through these techniques, we simplify the process of integration into manageable steps, which leads us to the solution of our definite integral.

Calculus Problem Solving

Calculus problem-solving often starts with understanding what is being asked. Typically, students face the challenge of breaking down complex expressions and dealing with unknowns represented by variables. In this exercise, our goal was to find the area under a curve using definite integration.

To tackle such problems:

• Begin by clearly identifying the function and the interval over which you need to integrate.
• Set up your integral correctly, ensuring that you've accounted for all parts of the function.
• Use integration techniques to find the antiderivative, simplifying where possible.
• Evaluate the antiderivative at the upper and lower bounds of the interval.
• Subtract the value at the lower bound from the value at the upper bound.

These steps guide you through calculus problems systematically, ensuring no part of the process is overlooked. By following this logical flow, problem-solving in calculus transforms from a daunting task to a structured approach that yields precise results.