Problem 17 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=x-4 \sqrt{x}, y=0$$

Step-by-Step Solution

Answer

The area bounded by the curves is 42.67 square units.

Step 1: Understand the Problem

The task is to find the area between the curve given by the equation \( y = x - 4 \sqrt{x} \) and the x-axis (\( y = 0 \)). This requires us to determine where these two curves intersect and integrate over the interval.

Step 2: Find Points of Intersection

To find points of intersection, set the two equations equal: \( x - 4\sqrt{x} = 0 \). Solve for \( x \) to find the points of intersection:\[ x - 4\sqrt{x} = 0 \implies \sqrt{x}(\sqrt{x} - 4) = 0 \]This gives solutions \( \sqrt{x} = 0 \) (\( x = 0 \)) and \( \sqrt{x} = 4 \) (\( x = 16 \)). Thus, the curves intersect at \( x = 0 \) and \( x = 16 \).

Step 3: Set Up the Integral

Since the area is bounded by \( y = x - 4\sqrt{x} \) and \( y = 0 \), we need to integrate \( x - 4\sqrt{x} \) from \( x = 0 \) to \( x = 16 \).\[ \text{Area} = \int_{0}^{16} (x - 4\sqrt{x}) \, dx\]

Step 4: Integrate the Function

First, simplify the integrand:\( x - 4\sqrt{x} = x - 4x^{1/2} \).Integrate term by term:\[ \int (x - 4x^{1/2}) \, dx = \int x \, dx - 4 \int x^{1/2} \, dx\]This gives:\[ \frac{x^2}{2} - 4\left(\frac{x^{3/2}}{3/2}\right) \]\[ = \frac{x^2}{2} - \frac{8}{3}x^{3/2} + C\]

Step 5: Evaluate the Definite Integral

Now, evaluate the definite integral from \( 0 \) to \( 16 \):\[ \left[ \frac{x^2}{2} - \frac{8}{3}x^{3/2} \right]_{0}^{16}\]Calculate:\[ \left[ \frac{16^2}{2} - \frac{8}{3}(16)^{3/2} \right] - \left[ \frac{0^2}{2} - \frac{8}{3}(0)^{3/2} \right] \]Simplify:\[ = \frac{256}{2} - \frac{8}{3}(64) = 128 - \frac{512}{3} = 128 - 170.67 = -42.67\]Since area cannot be negative, use the absolute value: 42.67 square units.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that are used to calculate the area under a curve between two points on the x-axis. When we talk about finding the area bounded by curves, we use definite integrals to determine this specific area.

Imagine a curve on the coordinate plane. The definite integral considers the space below this curve and above the x-axis, within the limits defined by the points of intersection or other specified boundaries. The integration sums up tiny rectangles under the curve from one point to the other to find the total area.

Here's what happens step-by-step when using a definite integral:

• First, identify the function that defines the curve.
• Next, find the limits of integration, often determined by the points where the curve intersects another curve or the axes.
• Set up the integral of the function over these limits. This involves expressing the area as the integral of the function minus any other function if needed.
• Finally, evaluate the integral from the lower limit to the upper limit to obtain the bounded area.

In the solution above, the definite integral was set from 0 to 16 to find the area between the curve and the x-axis.

Points of Intersection

When two or more curves intersect, they meet at specific points on the Cartesian plane. These are the points where their equations equate to the same y-value for a given x. To find these points of intersection, it is necessary to set the equations of the curves equal to each other and solve for x.

In the given exercise, the task was to find the intersection of the curve described by the equation \(y = x - 4\sqrt{x}\) and the line \(y = 0\). Solving \(x - 4\sqrt{x} = 0\) gives us the points \(x = 0\) and \(x = 16\). These points are crucial because they determine the limits of integration for calculating the area under the curve.

Here's a recap of the process for finding points of intersection:

• Set the two equations equal to each other.
• Solve the equation for x. This may involve factoring or using algebraic techniques to simplify the problem.
• The solutions are your intersection points along the x-axis.

Finding these points is essential for correctly setting up areas and understanding how the curves relate to each other.

Integration Techniques

Integrating a function involves finding the antiderivative, which is a function whose derivative is the given function. However, different functions require different integration techniques, which can range from simple power rules to more complex methods like integration by parts or substitution.

For the exercise, the expression \(x - 4\sqrt{x}\) was integrated. Simplifying this expression is a key step:

• Recognize that \( \sqrt{x} \) can be expressed as \(x^{1/2}\).
• Apply the power rule for integration: Add one to the exponent and divide by the new exponent.

The integral then becomes a straightforward calculation of \[\int x \, dx - 4 \int x^{1/2} \, dx\].

When using the power rule:

• The integral of \(x^n\) is \(\frac{x^{n+1}}{n+1}\), for \(n eq -1\).
• This concept simplifies integrating powers of x, like \(x\) and \(x^{1/2}\), into expressions that can be easily evaluated over an interval.

Practicing various integration techniques helps in mastering calculus and understanding how to find areas and solve real-world problems.