Problem 41 - Thomas Calculus

Question

In Exercises \(37-42,\) find the total area between the region and the \(x\) -axis. $$ y=x^{1 / 3}, \quad-1 \leq x \leq 8 $$

Step-by-Step Solution

Answer

The total area is 11.25.

Step 1: Identify the function and limits

The function provided is \(y = x^{1/3}\). We need to find the total area between this curve and the x-axis over the interval \(-1 \leq x \leq 8\).

Step 2: Set up the integral

To find the area between the curve \(y = x^{1/3}\) and the x-axis, we set up the definite integral from \(x = -1\) to \(x = 8\): \[\text{Area} = \int_{-1}^{8} x^{1/3} \, dx\]

Step 3: Integrate the function

Find the antiderivative of \(x^{1/3}\). The antiderivative of \(x^{n}\) is \((x^{n+1})/(n+1)\), so the antiderivative of \(x^{1/3}\) is \[\frac{x^{4/3}}{4/3} = \frac{3}{4}x^{4/3}\]

Step 4: Evaluate the integral

Evaluate the definite integral by substituting the upper and lower limits:\[\int_{-1}^{8} x^{1/3} \, dx = \left[ \frac{3}{4}x^{4/3} \right]_{-1}^{8} = \frac{3}{4}(8^{4/3}) - \frac{3}{4}((-1)^{4/3})\]

Step 5: Simplify the expression

Calculate the numeric values:- \(8^{1/3} = 2\), so \(8^{4/3} = (8^{1/3})^4 = 2^4 = 16\)- \((-1)^{1/3} = -1\), so \((-1)^{4/3} = ((-1)^{1/3})^4 = (-1)^4 = 1\)Substitute these back:\[\frac{3}{4}(16) - \frac{3}{4}(1) = 12 - \frac{3}{4} = \frac{48}{4} - \frac{3}{4} = \frac{45}{4} = 11.25\]

Step 6: Conclusion

Thus, the total area between the curve \(y = x^{1/3}\) and the x-axis from \(x = -1\) to \(x = 8\) is \(11.25\).

Key Concepts

Area Under a Curve

When we talk about finding the area under a curve in calculus, we're essentially looking for the space that a curve encloses above the x-axis between two points. In this case, the curve is defined by the function \(y = x^{1/3}\), and the bounds of this region are from \(x = -1\) to \(x = 8\). To calculate this area, we use definite integrals, which help us sum up infinitely small slices of area beneath the curve.

The concept relies on integrating the curve over the specified interval:

• Start by identifying the function and the interval: Here, \(y = x^{1/3}\) and \(-1 \leq x \leq 8\).
• Set up the integral to calculate the area: \(\int_{-1}^{8} x^{1/3} \, dx\).
• Integrate the function and evaluate the limits to find the total area.

This approach ensures we account for any curvature in the graph, explaining why integration is the go-to technique for finding areas under curves.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the gap between differentiation and integration and confirms that these processes are essentially inverses of each other. In practical terms, it allows us to evaluate definite integrals using antiderivatives.

To understand its application in finding the area under a curve:

• Find the antiderivative of the function: For \(y = x^{1/3}\), the antiderivative is \(\frac{3}{4}x^{4/3}\).
• Use it to evaluate the definite integral over the interval: Substitute the upper and lower limits into the antiderivative function, then subtract to find the definite integral's value.
• The result gives the exact area under the curve from one point to another.

This theorem provides the tools to turn a potentially complex calculation into a straightforward problem-solving approach.

Integration Techniques

When solving for the area under the curve \(y = x^{1/3}\), choosing the right integration technique is crucial. For polynomial functions, like \(x^{1/3}\), finding antiderivatives is manageable using simple power rules.

Here's how it's done:

• Recognize \(x^{1/3}\) as a power of \(x\). The rule for finding antiderivatives of \(x^n\) is \(\frac{x^{n+1}}{n+1}\).
• Apply this rule: The antiderivative of \(x^{1/3}\) is calculated as \(\frac{x^{4/3}}{4/3} = \frac{3}{4}x^{4/3}\).
• Evaluate using the antiderivative over the given interval, incorporating any simplification needed to get the exact area.

Mastering these techniques allows efficient solving of similar problems across various functions, making integration a powerful tool in calculus.