Problem 44 - Applied Calculus

Question

For each function: a. Integrate ("by hand") to find the area under the curve between the given \(x\) -values. b. Verify your answer to part (a) by having your calculator graph the function and find the area (using a command like FnInt or \(\int \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\) ). $$ f(x)=\frac{1}{\sqrt[3]{x}} \text { from } x=8 \text { to } x=27 $$

Step-by-Step Solution

Answer

The area under the curve is \( 7.5 \).

Step 1: Understand the Problem

We need to find the area under the curve for the function \( f(x) = \frac{1}{\sqrt[3]{x}} \) from \( x = 8 \) to \( x = 27 \). This involves integrating the function over the specified interval.

Step 2: Rewrite the Function for Integration

Rewrite the function in an easier form for integration: \( f(x) = x^{-1/3} \). This will help us use integration rules for power functions.

Step 3: Integrate the Function

To find the indefinite integral of \( f(x) = x^{-1/3} \), use the power rule: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \text{ for } n eq -1 \]Applying it here:\[ \int x^{-1/3} \, dx = \frac{x^{2/3}}{2/3} + C = \frac{3}{2}x^{2/3} + C \]

Step 4: Evaluate the Definite Integral

Apply the limits of integration from \( x = 8 \) to \( x = 27 \):\[ \left[ \frac{3}{2}x^{2/3} \right]_8^{27} = \frac{3}{2} (27^{2/3}) - \frac{3}{2} (8^{2/3}) \]Calculate the cube roots and powers:- \( 27^{1/3} = 3 \), thus \( 27^{2/3} = 9 \)- \( 8^{1/3} = 2 \), thus \( 8^{2/3} = 4 \)Now compute the values:\[ \frac{3}{2}(9) - \frac{3}{2}(4) = \frac{27}{2} - \frac{12}{2} = \frac{15}{2} \]

Step 5: Verify Using a Calculator

Plot the function \( f(x) = x^{-1/3} \) on a calculator and use the calculator's function integration command (like FnInt or similar) over the interval from \( x = 8 \) to \( x = 27 \). The numerical integration should confirm that the area under the curve is approximately \( 7.5 \).

Key Concepts

Integration by Hand

To integrate a function like \( f(x) = \frac{1}{\sqrt[3]{x}} \) by hand, it's often helpful to first rewrite the function in a form that's easier to work with. In this case, we can express it as \( f(x) = x^{-1/3} \). This transformation sets us up to use integration techniques for power functions more straightforwardly.

The goal is to determine the area under this curve from \( x = 8 \) to \( x = 27 \). By hand, this means applying the rules for integration to find an antiderivative: a function \( F(x) \) such that \( F'(x) = f(x) \).

Once the antiderivative is determined, we use the concept of definite integrals. This involves evaluating \( F(x) \) at the upper limit \( x = 27 \) and subtracting the value of \( F(x) \) at the lower limit \( x = 8 \). This method provides the exact area under the curve within the chosen interval, known as the net area when function values are positive.

Power Rule for Integration

The power rule for integration is incredibly useful for integrating polynomial expressions and functions that can be rewritten as powers of \( x \). The rule states:

\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \text{, provided } n eq -1\]

In the exercise, this rule is applied to \( f(x) = x^{-1/3} \), converting to the integral:

\[\int x^{-1/3} \, dx = \frac{x^{2/3}}{2/3} + C = \frac{3}{2}x^{2/3} + C\]

Remember, \( C \) represents the constant of integration, which is not needed in definite integrals since it cancels out in the evaluation process.

To apply the definite integral from \( 8 \) to \( 27 \), we substitute these bounds into \( \frac{3}{2}x^{2/3} \) to find the net area. Calculating:

• \( 27^{1/3} = 3 \) leads to \( 27^{2/3} = 9 \)
• \( 8^{1/3} = 2 \) leads to \( 8^{2/3} = 4 \)

Resulting in:
\[\frac{3}{2}(9) - \frac{3}{2}(4) = \frac{27}{2} - \frac{12}{2} = \frac{15}{2}\]

Verification with Graphing Calculator

Verification using a graphing calculator provides a practical check for the results obtained by hand calculations. By plotting the function \( f(x) = x^{-1/3} \) on a calculator, students can visualize the curve and the region that represents the area under it between the specified limits.

Most graphing calculators have a function integration command like FnInt or \( \int \text{f}(x) \text{d}x \) that allows users to select an interval for numeric integration. In this scenario, using the interval from \( x = 8 \) to \( x = 27 \), this command should yield a numerical approximation.

The calculator should confirm the area as approximately \( 7.5 \), verifying the accuracy of the manual integration method. This dual approach—both analytical and graphical—ensures that any miscalculations can be caught and corrected, bolstering understanding and confidence in both integration techniques and technology's role in mathematics.