Problem 41 - 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 \(\int \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\) ). $$ f(x)=12-3 x^{2} \text { from } x=1 \text { to } x=2 $$

Step-by-Step Solution

Answer

The area under the curve is 5.

Step 1: Set up the integral

To find the area under the curve of the function \( f(x) = 12 - 3x^2 \) from \( x = 1 \) to \( x = 2 \), you need to set up the definite integral: \[ \int_{1}^{2} (12 - 3x^2) \, dx \]

Step 2: Integrate the function

Find the antiderivative of the function. The antiderivative of \( 12 \) is \( 12x \) and for \( -3x^2 \) it is \( -x^3 \). So, the antiderivative is: \[ F(x) = 12x - x^3 \]

Step 3: Evaluate the definite integral

Use the Fundamental Theorem of Calculus to evaluate the definite integral from 1 to 2. First, plug in the upper limit \( x = 2 \): \[ F(2) = 12(2) - (2)^3 = 24 - 8 = 16 \] Then, plug in the lower limit \( x = 1 \): \[ F(1) = 12(1) - (1)^3 = 12 - 1 = 11 \] Find the difference: \[ F(2) - F(1) = 16 - 11 = 5 \] Thus, the area under the curve is 5.

Step 4: Verify with a calculator

Use a calculator to graph \( f(x) = 12 - 3x^2 \) and employ a numerical integration function to find the area under the curve from \( x = 1 \) to \( x = 2 \). The calculator should confirm the area is approximately 5.

Key Concepts

Antiderivative

The antiderivative, also known as the indefinite integral, is a function whose derivative is the given function. Think of it like going backwards from a derivative to find the original function. It's an essential tool in calculus when you're finding the area under curves, like in the problem we just reviewed. For the function \( f(x) = 12 - 3x^2 \), finding the antiderivative means figuring out a function \( F(x) \) such that \( F'(x) = f(x) \).
The antiderivative of \( 12 \) is \( 12x \) because the derivative of \( 12x \) with respect to \( x \) is \( 12 \). Likewise, the antiderivative of \(-3x^2\) is \(-x^3\) because the derivative of \(-x^3\) with respect to \( x \) is \(-3x^2\). Thus, the antiderivative for our function is
\[ F(x) = 12x - x^3 \]
Finding the antiderivative is important in solving definite integrals as it helps us move to evaluating the bounds of integration.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a key concept that connects differentiation and integration. It states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b] \), then the definite integral of \( f \) from \( a \) to \( b \) is \( F(b) - F(a) \). This theorem allows us to evaluate the area under a curve by simply finding the values of the antiderivative at the upper and lower limits.
In our case, after finding the antiderivative \( F(x) = 12x - x^3 \), we apply the Fundamental Theorem of Calculus:

• Evaluate \( F(x) \) at the upper limit \( x = 2 \): \( F(2) = 12 \times 2 - 2^3 = 24 - 8 = 16 \).
• Evaluate \( F(x) \) at the lower limit \( x = 1 \): \( F(1) = 12 \times 1 - 1^3 = 12 - 1 = 11 \).
• The definite integral is \( F(2) - F(1) = 16 - 11 = 5 \).

This calculation shows that the area under the curve from \( x = 1 \) to \( x = 2 \) is 5, all thanks to this powerful theorem.

Numerical Integration

Numerical integration comes into play when we can't easily find an antiderivative analytically. It's essentially a method to approximate the area under a curve using various techniques.
In this exercise, we used a calculator to perform numerical integration as a way to verify the area we found by hand.

• Graph the function \( f(x) = 12 - 3x^2 \).
• Use functions like FnInt or \( \int f(x) \, dx \) on the calculator to compute the area approximately.
• Make sure the calculator's result (approximately 5 in our case) matches the analytical result.

This provides a good check on the accuracy of our analytical answer. Numerical integration is useful because it offers a practical way to handle functions that are too complex for simple analytical methods, providing a reliable verification tool.