Problem 19 - 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. $$ f(x)=x^{2} \text { from } x=0 \text { to } x=3 $$

Step-by-Step Solution

Answer

The area under the curve is 9 square units.

Step 1: Understand the Problem

We need to find the area under the curve of the function \( f(x) = x^2 \) from \( x = 0 \) to \( x = 3 \). This area can be calculated using a definite integral.

Step 2: Set Up the Integral

To find the area under the curve \( f(x) = x^2 \) from \( x=0 \) to \( x=3 \), we need to evaluate the definite integral: \[ \int_{0}^{3} x^2 \, dx \].

Step 3: Integrate the Function

Next, we integrate the function \( x^2 \). The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \). Therefore, \[ \int x^2 \, dx = \frac{x^3}{3} + C \], where \( C \) is the constant of integration.

Step 4: Evaluate the Definite Integral

Now, evaluate the definite integral from \( x=0 \) to \( x=3 \):\[ \left[ \frac{x^3}{3} \right]_{0}^{3} = \frac{3^3}{3} - \frac{0^3}{3} = \frac{27}{3} - 0 = 9 \].

Step 5: Interpret the Result

The result from evaluating the integral is 9. This means the area under the curve \( f(x) = x^2 \) from \( x = 0 \) to \( x = 3 \) is 9 square units.

Step 6: Sketch the Graph

Sketch the graph of \( f(x) = x^2 \) from \( x = 0 \) to \( x = 3 \). The graph is a parabola opening upwards, and the area under the curve between these points is the region between the parabola and the x-axis over the interval [0, 3].

Key Concepts

Area Under a Curve

The concept of finding the area under a curve is crucial in calculus. This technique determines the total space enclosed between the curve and the x-axis over a specific interval. It's especially useful for calculating physical quantities such as distance, where the curve represents speed.

• Definite Integrals: To find this area, we use definite integrals. A definite integral calculates the accumulation of quantities, such as areas, over a given interval. It is denoted by the integral sign with limits of integration, like this: \( \int_{a}^{b} f(x) \, dx \).
• Bounding Intervals: The limits of integration, 'a' and 'b', define the exact interval on the x-axis over which you're calculating the area. For this exercise, the bounds are from \(x = 0\) to \(x = 3\).

Understanding this empowers us to evaluate how much 'space' the curve covers in the specified interval, an essential aspect in various fields like physics and economics.

Integration Techniques

Integration techniques are methods used to calculate the integral of a function, playing a key role in finding areas under curves.

• Power Rule: One fundamental technique is the power rule, which simplifies finding integrals of polynomials. For a function \(x^n\), its integral is \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the integration constant. This was used in our exercise to integrate \(x^2\), yielding \(\frac{x^3}{3} + C\).
• Definite vs. Indefinite Integrals: While indefinite integrals contain a constant \(C\), definite integrals eliminate it by considering specific boundaries. This results in a particular numerical result rather than a family of curves. In our task, we moved from an indefinite to a definite integral by applying limits from \(x=0\) to \(x=3\).

By mastering these techniques, you can swiftly tackle various integral problems, helping you compute complex areas and solve differential equations.

Parabolic Functions

Parabolic functions, like \(f(x) = x^2\), are a standard type of quadratic polynomial known for their characteristic U-shaped curve. These functions are fundamental in algebra and calculus due to their simplicity and wide applicability.

• Basic Shape: The graph of \(f(x) = x^2\) is a parabola opening upwards. Every parabola has a vertex, which is the bottom point of the curve for upward opening parabolas. Here, the vertex is at \((0,0)\).
• Symmetry: Parabolas have a line of symmetry, which in this case is the y-axis. This means the curve is mirrored equally on either side of this axis.
• Application in Integration: When determining the area under a parabolic curve between two points, the integration gives the exact measure of the space below the arc of the parabola, down to the x-axis. In our example, the area between \(x=0\) and \(x=3\) was found to be 9 square units.

Parabolas help simplify many mathematical models and are key in highlighting the practical utility of calculus techniques.