Problem 27 - Applied Calculus

Question

Write the definite integral for the area under the graph of \(f(x)=6 x^{2}+1\) between \(x=0\) and \(x=2\). Use the Fundamental Theorem of Calculus to evaluate it.

Step-by-Step Solution

Answer

The area under the graph is 18.

Step 1: Write the Definite Integral

To find the area under the graph of the function \( f(x) = 6x^2 + 1 \) from \( x = 0 \) to \( x = 2 \), write the definite integral as follows: \[ \int_{0}^{2} (6x^2 + 1) \, dx. \] This integral represents the area under the curve of \( f(x) \) from \( x = 0 \) to \( x = 2 \).

Step 2: Find the Antiderivative

The next step is to find the antiderivative (indefinite integral) of the function \( f(x) = 6x^2 + 1 \). The antiderivative of \( 6x^2 \) is \( 2x^3 \), and the antiderivative of \( 1 \) is \( x \). Therefore, the antiderivative \( F(x) \) is: \[ F(x) = 2x^3 + x + C, \] where \( C \) is the constant of integration.

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, the definite integral of \( f(x) \) from \( a \) to \( b \) can be found using \( F(b) - F(a) \). Here, \( a = 0 \) and \( b = 2 \). Substitute these values into \( F(x) = 2x^3 + x \).

Step 4: Evaluate the Antiderivative at Upper and Lower Limits

First, evaluate the antiderivative at the upper limit \( x = 2 \): \[ F(2) = 2(2)^3 + 2 = 16 + 2 = 18. \] Then, evaluate the antiderivative at the lower limit \( x = 0 \): \[ F(0) = 2(0)^3 + 0 = 0. \]

Step 5: Calculate the Definite Integral Result

Subtract the value of the antiderivative at the lower limit from the value at the upper limit: \[ \int_{0}^{2} (6x^2 + 1) \, dx = F(2) - F(0) = 18 - 0 = 18. \]

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between differentiation and integration, two core concepts in calculus. This theorem shows how these two operations are essentially inverse processes. In its simplest form, the theorem says that if you have a continuous function, its definite integral over a specific interval can be found using its antiderivative. The theorem has two main parts:

• Part 1: If a function is continuous on a closed interval, then it has an antiderivative.
• Part 2: To find the definite integral of a function over an interval [a, b], evaluate its antiderivative at b and a, then subtract these values: \( F(b) - F(a) \).

This is incredibly powerful because it simplifies the calculation of definite integrals significantly.
In the exercise provided, we use this theorem to compute the area under the curve of the function \( f(x) = 6x^2 + 1 \) from \( x = 0 \) to \( x = 2 \). By finding and evaluating the antiderivative at these points, we effortlessly determine the area to be 18.

Antiderivative

An antiderivative of a function is another function whose derivative yields the original function. In simple terms, it's the reverse process of differentiation.When given the function \( f(x) = 6x^2 + 1 \), finding the antiderivative involves integrating each term separately:

• The integral of \( 6x^2 \) leads to \( 2x^3 \), because differentiating \( 2x^3 \) gives back \( 6x^2 \).
• Similarly, the integral of \( 1 \) is simply \( x \), given that differentiating \( x \) returns 1.

Therefore, the antiderivative \( F(x) \) is \( 2x^3 + x + C \), where \( C \) is the constant of integration. This constant doesn't affect the calculation of definite integrals since it cancels out in the subtraction step of the Fundamental Theorem of Calculus.
Here, recognizing the antiderivative allows us to apply the theorem to find the exact area under a curve for a specific interval.

Area under a curve

Calculating the area under a curve on a graph is a fundamental application of integrals in calculus. The area provides significant insights in various fields, from physics to economics, indicating quantities such as total distance traveled or accumulated cost.The expression \( \int_{0}^{2} (6x^2 + 1) \, dx \) represents this concept directly, asking for the area under the curve of \( f(x) = 6x^2 + 1 \) from \( x = 0 \) to \( x = 2 \).
The process involves:

• Writing the definite integral: This sets the bounds for the area calculation.
• Finding the antiderivative: We use it to apply the Fundamental Theorem of Calculus.
• Evaluating at boundaries: Plugging in the upper and lower limits (2 and 0, in this case) and subtracting gives us the total area.

This approach ensures precise calculation of the area, in this scenario, resulting in an area of 18, reflecting the overall magnitude of space beneath the curve between the specified bounds.