Problem 10 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the area of the region under the graph of the function \(f\) on the interval \([a, b]\). $$f(x)=\frac{1}{x^{2}},[2,4]$$

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 \([2, 4]\) is \(\frac{1}{4}\).

Step 1: Find the antiderivative of the function

To integrate the function \(f(x)=\frac{1}{x^2}\), we first rewrite it as \(f(x)=x^{-2}\). Now we can find the antiderivative, \(F(x)\), by using the power rule of integration: \[\int x^n dx = \frac{x^{n+1}}{n+1} + C\] Applying the power rule to the function: \[F(x) = \int x^{-2} dx = \frac{x^{-2 + 1}}{-2 + 1} + C = -x^{-1} + C = -\frac{1}{x} + C\]

Step 2: Evaluate the antiderivative at the upper and lower limits

Now we need to evaluate the antiderivative, \(F(x)\), at the upper and lower limits of the interval, and then find the difference. First, evaluate at the upper limit, 4: \[F(4) = -\frac{1}{4} + C\] Next, evaluate at the lower limit, 2: \[F(2) = -\frac{1}{2} + C\]

Step 3: Calculate the definite integral by finding the difference between the evaluations

Now, subtract the evaluation of the antiderivative at the lower limit, \(F(2)\), from the evaluation at the upper limit, \(F(4)\), to find the area under the curve: \[\int_2^4 \frac{1}{x^2} dx = F(4) - F(2) = \left(-\frac{1}{4} + C\right) - \left(-\frac{1}{2} + C\right) = -\frac{1}{4} + \frac{1}{2} = \frac{1}{4}\] The area of the region under the graph of the function \(f(x)=\frac{1}{x^2}\) on the interval \([2, 4]\) is \(\frac{1}{4}\).

Key Concepts

Antiderivative

An antiderivative is essentially a backward derivative. It helps us find the original function that was differentiated to get what we have now. When you integrate a function, you are finding its antiderivative. Think of it as reversing the process of differentiation.
For the function \(f(x) = \frac{1}{x^2}\), we first rewrite it into a form that makes integration easier: \(f(x) = x^{-2}\). This reveals the power the variable is raised to, which is important for the next step using the power rule.
The antiderivative of a function \(f(x)\) often includes a constant \(C\). This constant represents any vertical shifts in the family of antiderivative functions and is found when you have additional information, like boundary conditions. However, for definite integrals, the constant cancels out during evaluation at limits.

Power Rule of Integration

The power rule of integration is a basic tool for finding antiderivatives. It's easy to remember and very helpful when you deal with polynomial expressions. This rule states: if you have a function \(x^n\), the integral is \(\frac{x^{n+1}}{n+1} + C\).
When using the power rule, first check if the exponent \(n\) is a number that makes the formula valid. The rule doesn't apply to \(n = -1\) directly as it leads to an undefined expression, but it does apply nicely when \(f(x)\) is \(x^{-2}\).
Applying the power rule of integration to \(x^{-2}\), we add 1 to the exponent (\(-2 + 1 = -1\)) and divide by the new exponent \(-1\), resulting in the antiderivative \(-x^{-1} + C\) or \(-\frac{1}{x} + C\).

Area Under a Curve

Finding the area under a curve is a common application of integration. This concept connects geometry and calculus because it gives us the exact amount of space between the function's graph and the x-axis over a certain interval.
For a given function, the definite integral over an interval \([a, b]\) finds this area. Think of it as adding up an infinite number of infinitely thin rectangles under the curve. In our specific example, we are finding the definite integral of \(f(x) = \frac{1}{x^2}\) from 2 to 4.
To determine this area, we performed these steps:

• First, find the antiderivative.
• Evaluate the antiderivative at the upper limit \(b = 4\) and the lower limit \(a = 2\).
• Subtract the value at the lower limit from the value at the upper limit to get the area.

The result from this process tells us the area under the curve from 2 to 4, which is \(\frac{1}{4}\). It's like measuring how much space the curve covers above the x-axis within that interval.