Problem 31 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the area of the region under the graph of \(f\) on \([a, b]\). $$f(x)=\frac{1}{x^{2}} ;[1,2]$$

Step-by-Step Solution

Answer

The area of the region under the graph of the function \(f(x)=\frac{1}{x^2}\) on the interval \([1, 2]\) is \(\frac{1}{2}\).

Step 1: Identify the function and the interval

The given function is \(f(x) = \frac{1}{x^2}\) and the interval we're interested in is \([a, b] = [1, 2]\).

Step 2: Calculate the indefinite integral

First, let's find the indefinite integral of the function. The function is \(f(x) = \frac{1}{x^2} = x^{-2}\), so we need to find its integral with respect to x: \[\int f(x)~dx = \int x^{-2}~dx\] We apply the power rule (add 1 to the exponent and divide by the new exponent): \[F(x) = \frac{x^{-1}}{-1} + C\] Simplifying the expression, \[F(x) = -\frac{1}{x} + C\] Now, we have found the antiderivative (indefinite integral) of \(f(x)\).

Step 3: Calculate the definite integral on the interval [1, 2]

Now, we need to evaluate the definite integral of \(f(x)\) on the given interval \([1, 2]\). We do this by finding the difference of the antiderivative at the endpoints of the interval: \[\int_{1}^{2} f(x)~dx = F(b) - F(a) = F(2) - F(1)\] Plugging the values of \(a=1\) and \(b=2\) into the antiderivative: \[F(2) - F(1) = -\frac{1}{2} + C - (-\frac{1}{1} + C)\] Since the constant \(C\) appears in both terms, it cancels out: \[-\frac{1}{2} + \frac{1}{1} = \frac{1}{2}\]

Step 4: Find the area under the curve

Finally, we have found the definite integral of \(f(x)\) on the interval \([1, 2]\), which represents the area under the curve. The area is: \[Area = \int_{1}^{2} f(x)~dx = \frac{1}{2}\] Therefore, the area of the region under the graph of \(f(x) = \frac{1}{x^2}\) on the interval \([1, 2]\) is \(\frac{1}{2}\).

Key Concepts

Indefinite Integral

An indefinite integral is essentially the reverse process of differentiation and is represented by the integral sign without bounds. When we find an indefinite integral, we're looking for a function that, when differentiated, gives us back the original function we started with. This is also known as finding the antiderivative.
The process involves taking a function like \( f(x) = \frac{1}{x^2} \) and determining its integral. In this case, it's beneficial to rewrite \( f(x) \) as \( x^{-2} \) to make the integration process smoother using the power rule:

• Add 1 to the exponent \( n \) of \( x^n \).
• Divide by the new exponent.

For \( f(x) = x^{-2} \), we get:
\[ \int x^{-2} \ dx = \frac{x^{-1}}{-1} + C \]
This simplifies to:
\[ -\frac{1}{x} + C \]
where \( C \) is the constant of integration, representing the family of all possible antiderivatives of the function.

Antiderivative

An antiderivative of a function is a function whose derivative is the original function. It's important to realize that finding an antiderivative is essentially performing the process of indefinite integration.
For our function \( f(x) = \frac{1}{x^2} \), the antiderivative is calculated as:
\[ F(x) = -\frac{1}{x} + C \]
Here, \( F(x) \) is the antiderivative of \( f(x) \). The constant \( C \) is included because differentiating a constant gives zero, meaning any constant \( C \) can be a part of the general solution. Knowing the antiderivative is essential for finding definite integrals because it allows us to apply the Fundamental Theorem of Calculus, which links antiderivatives and definite integrals, helping us find specific values like the area under a curve.

Area Under a Curve

The area under a curve in the context of calculus is often computed using definite integrals. This area can represent many things in different applications, such as the total accumulated quantity.
To find this area between a function and the x-axis, you use a definite integral over a specific interval \([a, b]\). For the function \( f(x) = \frac{1}{x^2} \), the area under the curve from \( x = 1 \) to \( x = 2 \) is calculated by evaluating the integral:
\[ \int_{1}^{2} \frac{1}{x^2} \, dx \]
By finding the antiderivative, which we determined as \( F(x) = -\frac{1}{x} + C \), we then apply the limits of integration:

• Calculate \( F(2) = -\frac{1}{2} \)
• Calculate \( F(1) = -1 \)

The difference gives us the area:
\[ F(2) - F(1) = -\frac{1}{2} - (-1) = \frac{1}{2} \]
This \\( \frac{1}{2} \) represents the exact area between the curve and the x-axis from 1 to 2.