Problem 31 - 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. $$ f(x)=8-4 \sqrt[3]{x} \text { from } x=0 \text { to } x=8 $$

Step-by-Step Solution

Answer

The area under the curve from \( x = 0 \) to \( x = 8 \) is 16 square units.

Step 1: Understand the Problem

We need to find the area between the curve given by the function \( f(x) = 8 - 4 \sqrt[3]{x} \) and the x-axis, from \( x = 0 \) to \( x = 8 \). This requires us to compute a definite integral over this interval.

Step 2: Set Up the Definite Integral

The area under the curve from \( x = 0 \) to \( x = 8 \) is given by the definite integral \( \int_{0}^{8} (8 - 4 \sqrt[3]{x}) \ dx \).

Step 3: Integrate the Function

To solve the integral \( \int (8 - 4 \sqrt[3]{x}) \ dx \), we split it into two separate integrals: \( \int 8 \ dx - 4 \int \sqrt[3]{x} \ dx \). The first integral is straightforward, and the second one requires using the power rule for integrals.

Step 4: Apply the Power Rule for Integration

The integral of a constant, \( \int 8 \ dx \), is \( 8x \). For the term \( 4 \int \sqrt[3]{x} \ dx \), rewrite it as \( 4 \int x^{1/3} \ dx \) and use the power rule: \( \int x^{n} \ dx = \frac{x^{n+1}}{n+1} + C \). Therefore, \( \int x^{1/3} \ dx = \frac{x^{4/3}}{4/3} = \frac{3}{4}x^{4/3} \). Multiplying by 4 gives \( 3x^{4/3} \).

Step 5: Evaluate the Integral at the Limits

Now, combine the integrated results: \[ 8x - 3x^{4/3} \]. Evaluate this expression from \( x = 0 \) to \( x = 8 \), giving: \[ \left[ 8x - 3x^{4/3} \right]_{0}^{8} = \left( 8(8) - 3(8)^{4/3} \right) - \left( 8(0) - 3(0)^{4/3} \right) \].

Step 6: Simplify the Calculations

Calculate: \( 8(8) = 64 \) and \( (8)^{4/3} = 16 \). So, the expression simplifies to \( 64 - 3(16) = 64 - 48 = 16 \).

Step 7: Conclusion and Sketch of Region

The area under the curve from \( x = 0 \) to \( x = 8 \) is 16 square units. Sketch the graph of \( f(x) \) showing the curve and shading the region from \( x = 0 \) to \( x = 8 \) underneath it.

Key Concepts

Area under a Curve

Finding the area under a curve involves calculating the region between the graph of a function and the x-axis, over a specified interval. In this case, the function is given by:\[ f(x) = 8 - 4 \sqrt[3]{x} \]We want to find the area from \(x = 0\) to \(x = 8\). Using a definite integral, we can determine this area precisely. The integral helps us add up an infinite number of vertical slices of infinitesimal width under the curve. This approach lets us compute complex areas that would be difficult to measure using simple geometrical shapes. In this exercise, the definite integral is:\[ \int_{0}^{8} (8 - 4 \sqrt[3]{x}) \, dx \]Thus, the area under the curve from \(x = 0\) to \(x = 8\) has been calculated to be 16 square units.

Integration by Parts

Integration by parts is an essential technique in calculus for calculating integrals. It is particularly useful when dealing with products of functions. The technique is based on the product rule for differentiation and is given by the formula:\[ \int u \, dv = uv - \int v \, du \]However, in this particular exercise, integration by parts was not necessary because the function \(f(x) = 8 - 4 \sqrt[3]{x}\) could be handled using simpler integration techniques. The original function was split into two components that could be more easily integrated using basic rules, such as the power rule for integrals. Integration by parts would become necessary for more complex expressions involving products of functions like polynomials and exponential or trigonometric functions.

Evaluating Integrals

Evaluating an integral involves finding the antiderivative of a function. For the given curve, we need to evaluate:\[ \int (8 - 4 \sqrt[3]{x}) \, dx \]First, we separate this into two individual integrals:

• \(\int 8 \, dx\)
• \(- 4 \int \sqrt[3]{x} \, dx\)

The first integral is straightforward because the integral of a constant \(8\) is simply \(8x\). For the second, we use the power rule which states:\[ \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \]Applying this rule to \(x^{1/3}\), we get \(\frac{x^{4/3}}{4/3}\), simplifying it to \(\frac{3}{4}x^{4/3}\). Multiplying by \(-4\), we find:\[-3 x^{4/3}\]Hence, the antiderivative is \(8x - 3x^{4/3}\), which, when evaluated from \(x = 0\) to \(x = 8\), gives the total area of 16 square units.

Sketching Functions

Sketching functions is a valuable skill in understanding the behavior of a curve over a given interval. By visualizing, you can see where the curve lies relative to the x-axis and identify regions of interest, such as those needing area calculation. To sketch the function:\[ f(x) = 8 - 4 \sqrt[3]{x} \]consider several key points:

• When \(x = 0\), \(f(x) = 8\).
• When \(x = 8\), \(f(x) = 8 - 4 \times 2 = 0\).

The function starts high on the y-axis and decreases down to zero at \(x = 8\). This negative cubic root relationship creates a downward-sloping curve. As you sketch, mark these points and connect them with a smooth curve. Shade the region between the curve and the x-axis from \(x = 0\) to \(x = 8\). This gives a visual representation of the area you calculated, helping to reinforce your understanding of integral calculus.